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|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Submitted Solution:
```
for _ in range(int(input())):
n,k = map(int,input().split())
a = list(input())
flag = False
dic = [-1 for i in range(k)]
for i in range(n):
if a[i] == '1' and dic[i%k] != 0:
dic[i%k] = 1
elif a[i] == '0' and dic[i%k] != 1:
dic[i%k] = 0
elif a[i] == '?':
if dic[i%k] != -1:
continue
else:
dic[i%k] = '?'
else:
flag = True
break
if flag:
print('NO')
else:
s = count = 0
for i in dic:
if i == '?':
count += 1
elif i == 1:
s += 1
if s > k//2 or (s<k//2 and count == 0):
print('NO')
elif s == k//2 and count ==0:
print('YES')
elif count != 0:
if s == k//2:
print('YES')
elif (k//2)-s <= count:
print('YES')
else:
print('NO')
```
|
instruction
| 0
| 12,643
| 0
| 25,286
|
Yes
|
output
| 1
| 12,643
| 0
| 25,287
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Submitted Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
N, K = map(int, input().split())
a = list(input())[: -1]
cs = 0
q = 0
for i in range(K):
t = 0
if a[i] == "1": t = 1
elif a[i] == "0": t = -1
else: q += 1
cs += t
q -= abs(cs)
if q < 0:
print("NO")
continue
for i in range(N - K):
if a[i] != "?" and (a[i + K] != "?") and (a[i] != a[i + K]):
print("NO")
break
elif a[i] != a[i + K]:
if a[i] == "0": a[i + K] = "0"
elif a[i] == "1": a[i + K] = "1"
elif a[i + K] == "0": a[i] = "0"
elif a[i + K] == "1": a[i] = "1"
else:
cs = 0
q = 0
for i in range(K):
t = 0
if a[i] == "1": t = 1
elif a[i] == "0": t = -1
else: q += 1
cs += t
q -= abs(cs)
if q < 0:
print("NO")
continue
for i in range(N - K):
if a[i] != "?" and (a[i + K] != "?") and (a[i] != a[i + K]):
print("NO")
break
elif a[i] != a[i + K]:
if a[i] == "0": a[i + K] = "0"
elif a[i] == "1": a[i + K] = "1"
elif a[i + K] == "0": a[i] = "0"
elif a[i + K] == "1": a[i] = "1"
else:
cs = [0] * (N + 1)
qs = [0] * (N + 1)
for i in range(N):
t = 0
if a[i] == "1": t = 1
elif a[i] == "0": t = -1
else: qs[i + 1] += 1
cs[i + 1] = cs[i] + t
qs[i + 1] += qs[i]
for i in range(N - K + 1):
if (abs(cs[i + K] - cs[i]) - (qs[i + K] - qs[i])) % 2:
print("NO")
break
elif (abs(cs[i + K] - cs[i]) - (qs[i + K] - qs[i])) > 0:
print("NO")
break
else: print("YES")
```
|
instruction
| 0
| 12,644
| 0
| 25,288
|
Yes
|
output
| 1
| 12,644
| 0
| 25,289
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Submitted Solution:
```
from sys import stdin
N = int(stdin.readline())
for case in range(N):
length, k = map(int, stdin.readline().split())
string = str(stdin.readline())
count = {
"1":0,
"0":0,
"?":0
}
flag = True
for i in range(k):
res = string[i]
for j in range(i+k, length, k):
if res != "?" and string[j] != res and string[j] != "?":
flag = False
break
if res == "?" and string[j] != "?":
res = string[j]
if flag:
count[res] += 1
if count["1"] <= k//2 and count["0"] <= k//2 and flag:
print("YES")
else:
print("NO")
```
|
instruction
| 0
| 12,645
| 0
| 25,290
|
Yes
|
output
| 1
| 12,645
| 0
| 25,291
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sun Sep 6 21:22:04 2020
@author: Dark Soul
"""
t=int(input(''))
arr=[]
s=[]
for i in range(t):
arr.append(list(map(int,input().split())))
s.append(input(''))
for j in range(t):
[n,k]=arr[j]
tst=list(s[j])
flag=0
f1=0
f2=0
q=0
for i in range(k):
if tst[i]=='1':
f1+=1
elif tst[i]=='0':
f2+=1
else:
q+=1
if q==0:
if f1!=k//2 or f2!=k//2:
flag=1
if f1>k//2 or f2>k//2:
flag=1
if flag:
print('NO')
continue
if q:
if f1==k//2:
for i in range(k):
if tst[i]=='?':
tst[i]='0'
if i+k<n:
if tst[i+k]!=tst[i]:
flag=1
break
elif f2==k//2:
for i in range(k):
if tst[i]=='?':
tst[i]='1'
if i+k<n:
if tst[i+k]!=tst[i]:
flag=1
break
else:
f1=abs(f1-k//2)
f2=abs(f2-k//2)
for i in range(k):
if tst[i]=='?':
if i+k<n:
if tst[i+k]!='?':
tst[i]=tst[i+k]
else:
if f1:
tst[i]='1'
f1-=1
elif f2:
tst[i]='0'
f2-=1
else:
flag=1
break
else:
if f1:
tst[i]='1'
f1-=1
elif f2:
tst[i]='0'
f2-=1
else:
flag=1
break
if flag:
print('NO')
continue
for i in range(n-k):
if tst[i]!=tst[i+k]:
if tst[i+k]=='?':
tst[i+k]=tst[i]
else:
flag=1
break
if flag:
print('NO')
else:
print('YES')
```
|
instruction
| 0
| 12,646
| 0
| 25,292
|
No
|
output
| 1
| 12,646
| 0
| 25,293
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=998244353
EPS=1e-6
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def check():
return True
for _ in range(Int()):
n,k=value()
s=[i for i in input()]
ok="YES"
ind=[i for i in range(n) if s[i]=='?']
id=0
need=0
have=0
# INITIALS ------------------------------------/
for i in range(k):
if(s[i]=='1'):need+=1
elif(s[i]=='0'):need-=1
else:have+=1
if(abs(need)>have or (have-abs(need))%2):
ok="NO"
elif(have==abs(need)):
for i in range(have):
s[ind[id]]=str(int(need<0))
id+=1
have=0
need=0
# SLIDE ---------------------------------------/
# print(ok)
for i in range(1,n-k+1):
if(id<len(ind) and ind[id]<i):
id=bisect_left(ind,i)
if(s[i-1]=='0'): need+=1
elif(s[i-1]=='1'): need-=1
else: have-=1
if(s[i+k-1]=='0'):need-=1
elif(s[i+k-1]=='1'):need+=1
else: have+=1
# print(id,have,need,ind)
if(abs(need)>have or (have-abs(need))%2):
ok="NO"
break
elif(have==abs(need)):
for i in range(have):
s[ind[id]]=str(int(need<0))
id+=1
have=0
need=0
# print(s)
print(ok)
```
|
instruction
| 0
| 12,647
| 0
| 25,294
|
No
|
output
| 1
| 12,647
| 0
| 25,295
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Submitted Solution:
```
t=int(input())
for _ in range(t):
n,k=map(int,input().split())
a=list(input())
flag=1
for i in range(n):
if(a[i]=="?"):
if(i+k<n):
a[i]=a[i+k]
else:
if(a[i]=="0"):
if(i+k<n):
if(a[i]!=a[i+k] and a[i+k]=="1"):
flag=0
break
a[i+k]=a[i]
if(i-k>=0 ):
if(a[i]!=a[i-k]and a[i-k]=="1"):
flag=0
break
a[i-k]=a[i]
else:
if(i+k<n):
if(a[i]!=a[i+k]and a[i+k]=="0"):
flag=0
break
a[i+k]=a[i]
#print(i+k)
if(i-k>=0):
if(a[i]!=a[i-k]and a[i-k]=="1"):
flag=0
break
a[i-k]=a[i]
#print(a)
one=0
zero=0
doubt=0
#print(a)
for i in range(k):
if(a[i]=="1"):
one+=1
elif(a[i]=="0"):
zero+=1
else:
doubt=0
if(one>k/2 or zero>k/2):
print("NO")
continue
oneneed=k/2-one
zeroneed=k/2-zero
for i in range(k,n):
if(a[i-k]=='1'):
one-=1
if(a[i-k]=='0'):
zero-=1
if(a[i]=='1'):
one+=1
else:
zero+=1
if(one>k/2 or zero>k/2):
flag=0
break
if(flag==0):
print("NO")
else:
print("YES")
```
|
instruction
| 0
| 12,648
| 0
| 25,296
|
No
|
output
| 1
| 12,648
| 0
| 25,297
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bitstring is a string consisting only of the characters 0 and 1. A bitstring is called k-balanced if every substring of size k of this bitstring has an equal amount of 0 and 1 characters (k/2 of each).
You are given an integer k and a string s which is composed only of characters 0, 1, and ?. You need to determine whether you can make a k-balanced bitstring by replacing every ? characters in s with either 0 or 1.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
The first line of each test case contains two integers n and k (2 ≤ k ≤ n ≤ 3 ⋅ 10^5, k is even) — the length of the string and the parameter for a balanced bitstring.
The next line contains the string s (|s| = n). It is given that s consists of only 0, 1, and ?.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, print YES if we can replace every ? in s with 0 or 1 such that the resulting bitstring is k-balanced, or NO if it is not possible.
Example
Input
9
6 4
100110
3 2
1?1
3 2
1?0
4 4
????
7 4
1?0??1?
10 10
11??11??11
4 2
1??1
4 4
?0?0
6 2
????00
Output
YES
YES
NO
YES
YES
NO
NO
YES
NO
Note
For the first test case, the string is already a 4-balanced bitstring.
For the second test case, the string can be transformed into 101.
For the fourth test case, the string can be transformed into 0110.
For the fifth test case, the string can be transformed into 1100110.
Submitted Solution:
```
import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools,itertools,operator,bisect,fractions
from collections import deque,defaultdict,OrderedDict,Counter
from fractions import Fraction
from decimal import Decimal
def main():
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
for _ in range(ri()):
n,k=ria()
a=rs()
a=list(a)
q=deque(a[:k])
n1=0
nq=0
n0=0
for i in q:
if i=='?':
nq+=1
if i=='1':
n1+=1
if i=='0':
n0+=1
if max(n1,n0)>k//2:
print("NO")
else:
t=0
if n1==k//2:
for e in range(k):
if q[e]=='?':
q[e]='0'
n0=nq
nq=0
t=1
if n0==k//2 and t==0:
for e in range(k):
if q[e]=='?':
q[e]='1'
n1=nq
nq=0
c=True
for i in range(k,n):
if q[0]=='?':
nq-=1
if q[0]=='1':
n1-=1
if q[0]=='0':
n0-=1
q.popleft()
q.append(a[i])
if q[-1]=='?':
nq+=1
if q[-1]=='1':
n1+=1
if q[-1]=='0':
n0+=1
if max(n1,n0)>k//2:
c=False
break
t=0
if n1==k//2:
for e in range(k):
if q[e]=='?':
q[e]='0'
n0=nq
nq=0
t=1
if n0==k//2 and t==0:
for e in range(k):
if q[e]=='?':
q[e]='1'
n1=nq
nq=0
if c:
print("YES")
else:
print("NO")
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
```
|
instruction
| 0
| 12,649
| 0
| 25,298
|
No
|
output
| 1
| 12,649
| 0
| 25,299
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
|
instruction
| 0
| 12,654
| 0
| 25,308
|
Tags: dp, strings
Correct Solution:
```
def make_array(n, m):
new_arr = []
for _ in range(n):
new_arr += [[0] * m]
return new_arr
def catching_cheaters():
n, m = [int(x) for x in input().split(' ')]
str_a = input()
str_b = input()
dp = make_array(n + 1, m + 1)
ans = 0
for i in range(1, n + 1):
for j in range(1, m + 1):
if str_a[i - 1] == str_b[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 2
else:
dp[i][j] = max(0, max(dp[i - 1][j], dp[i][j - 1]) - 1)
ans = max(ans, dp[i][j])
print(ans)
if __name__ == "__main__":
catching_cheaters()
```
|
output
| 1
| 12,654
| 0
| 25,309
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
|
instruction
| 0
| 12,655
| 0
| 25,310
|
Tags: dp, strings
Correct Solution:
```
import sys
import os
from io import BytesIO, IOBase
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
n, m = map(int, input().split())
a, b = input(), input()
dp = [[0] * (m+2) for _ in range(n+2)]
ans = 0
for i in range(n-1, -1, -1):
for j in range(m-1, -1, -1):
if a[i] == b[j]:
dp[i][j] = dp[i+1][j+1] + 2
else:
dp[i][j] = max(0, max(dp[i][j+1], dp[i+1][j]) -1)
ans = max(ans, dp[i][j])
print(ans)
```
|
output
| 1
| 12,655
| 0
| 25,311
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
|
instruction
| 0
| 12,656
| 0
| 25,312
|
Tags: dp, strings
Correct Solution:
```
from collections import deque
n,m = list(map(int,input().split()))
a = input()
b = input()
l1 = len(a)
l2 = len(b)
dp = [[0 for j in range(l2+1)] for i in range(l1+1)]
mv = 0
for i in range(1,l1+1):
for j in range(1,l2+1):
if a[i-1]==b[j-1]:
dp[i][j] = dp[i-1][j-1]+2
else:
dp[i][j] = max(max(dp[i-1][j],dp[i][j-1])-1,0)
mv = max(mv,dp[i][j])
print(mv)
```
|
output
| 1
| 12,656
| 0
| 25,313
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
|
instruction
| 0
| 12,657
| 0
| 25,314
|
Tags: dp, strings
Correct Solution:
```
# coding: utf-8
n, m = map(int, input().split())
a = input()
b = input()
dp = [[0 for j in range(m + 1)] for i in range(n + 1)]
maximal = 0
for i in range(1, n + 1):
for j in range(1, m + 1):
dp[i][j] = max(dp[i - 1][j] - 1, dp[i][j - 1] - 1, dp[i - 1][j - 1] + 2 if a[i - 1] == b[j - 1] else 0)
maximal = max(maximal, dp[i][j])
print(maximal)
```
|
output
| 1
| 12,657
| 0
| 25,315
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
|
instruction
| 0
| 12,658
| 0
| 25,316
|
Tags: dp, strings
Correct Solution:
```
n,m=map(int,input().split())
a=input()
b=input()
dp=[[0 for i in range(m+1)] for j in range(n+1)]
ans=0
for i in range(1,n+1):
for j in range(1,m+1):
if a[i-1]==b[j-1]:
dp[i][j]=max(dp[i][j],dp[i-1][j-1]+2)
ans=max(ans,dp[i][j])
else:
dp[i][j]=max(dp[i-1][j], dp[i][j-1],1)-1
print(ans)
```
|
output
| 1
| 12,658
| 0
| 25,317
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
|
instruction
| 0
| 12,659
| 0
| 25,318
|
Tags: dp, strings
Correct Solution:
```
''' ===============================
-- @uthor : Kaleab Asfaw
-- Handle : kaleabasfaw2010
-- Bio : High-School Student
==============================='''
# Fast IO
import sys
import os
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file): self._fd = file.fileno(); self.buffer = BytesIO(); self.writable = "x" in file.mode or "r" not in file.mode; self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b: break
ptr = self.buffer.tell(); self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0; return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)); self.newlines = b.count(b"\n") + (not b); ptr = self.buffer.tell(); self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1; return self.buffer.readline()
def flush(self):
if self.writable: os.write(self._fd, self.buffer.getvalue()); self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file): self.buffer = FastIO(file); self.flush = self.buffer.flush; self.writable = self.buffer.writable; self.write = lambda s: self.buffer.write(s.encode("ascii")); self.read = lambda: self.buffer.read().decode("ascii"); self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout); input = lambda: sys.stdin.readline().rstrip("\r\n")
def solve(n, m, a, b):
dp = dp = [[0] * (m + 1) for _ in range(n + 1)]
maxx = 0
for i in range(1, n+1):
for j in range(1, m+1):
if a[i-1] == b[j-1]:
dp[i][j] = max(dp[i][j], dp[i-1][j-1] + 2)
else:
dp[i][j] = max(dp[i][j], dp[i-1][j] - 1, dp[i][j-1] - 1)
maxx = max(maxx, dp[i][j])
return maxx
n, m = list(map(int, input().split()))
a = input()
b = input()
print(solve(n, m, a, b))
```
|
output
| 1
| 12,659
| 0
| 25,319
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
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instruction
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Tags: dp, strings
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
from bisect import bisect_right
from math import gcd,log
from collections import Counter,defaultdict
from pprint import pprint
def lcs(a,b):
ans=0
dp=[[0]*(len(b)+1) for i in range(len(a)+1)]
for i in range(1,1+len(a)):
for j in range(1,1+len(b)):
if a[i-1]==b[j-1]:
dp[i][j]=dp[i-1][j-1]+2
dp[i][j]=max(0,dp[i][j],dp[i][j-1]-1,dp[i-1][j]-1)
ans=max(ans,dp[i][j])
# pprint(dp)
return ans
def main(case_no):
n,m=map(int,input().split())
a=input()
b=input()
ans=lcs(a,b)
print(ans)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
for _ in range(1):
main(_+1)
```
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| 25,321
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
|
instruction
| 0
| 12,661
| 0
| 25,322
|
Tags: dp, strings
Correct Solution:
```
#include <CodeforcesSolutions.h>
#include <ONLINE_JUDGE <solution.cf(contestID = "1447",questionID = "A",method = "GET")>.h>
"""
Author : thekushalghosh
Team : CodeDiggers
I prefer Python language over the C++ language :p :D
Visit my website : thekushalghosh.github.io
"""
import sys,math,cmath,time
start_time = time.time()
##########################################################################
################# ---- THE ACTUAL CODE STARTS BELOW ---- #################
def solve():
n,m = invr()
a = insr()
b = insr()
dp = [[0] * (m + 1) for i in range(n + 1)]
q = 0
for i in range(n):
for j in range(m):
if a[i] == b[j]:
dp[i + 1][j + 1] = max(dp[i + 1][j] - 1,dp[i][j] + 2,dp[i][j + 1] - 1)
q = max(q,dp[i + 1][j + 1])
else:
dp[i + 1][j + 1] = max(dp[i][j + 1] - 1,dp[i + 1][j] - 1,0)
print(q)
################## ---- THE ACTUAL CODE ENDS ABOVE ---- ##################
##########################################################################
def main():
global tt
if not ONLINE_JUDGE:
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
t = 1
for tt in range(1,t + 1):
solve()
if not ONLINE_JUDGE:
print("Time Elapsed :",time.time() - start_time,"seconds")
sys.stdout.close()
#----------------- USER DEFINED INPUT - OUTPUT FUNCTIONS -----------------#
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
return(input().strip())
def invr():
return(map(int,input().split()))
def prints(a,qw = " "):
sys.stdout.write(qw.join(map(str,a)) + "\n")
#------------------ USER DEFINED PROGRAMMING FUNCTIONS ------------------#
def counter(a):
q = [0] * max(a)
for i in range(len(a)):
q[a[i] - 1] = q[a[i] - 1] + 1
return(q)
def counter_elements(a):
q = dict()
for i in range(len(a)):
if a[i] not in q:
q[a[i]] = 0
q[a[i]] = q[a[i]] + 1
return(q)
def string_counter(a):
q = [0] * 26
for i in range(len(a)):
q[ord(a[i]) - 97] = q[ord(a[i]) - 97] + 1
return(q)
def factorial(n,m = 1000000007):
q = 1
for i in range(n):
q = (q * (i + 1)) % m
return(q)
def factors(n):
q = []
for i in range(1,int(n ** 0.5) + 1):
if n % i == 0: q.append(i); q.append(n // i)
return(list(sorted(list(set(q)))))
def prime_factors(n):
q = []
while n % 2 == 0: q.append(2); n = n // 2
for i in range(3,int(n ** 0.5) + 1,2):
while n % i == 0: q.append(i); n = n // i
if n > 2: q.append(n)
return(list(sorted(q)))
def transpose(a):
n,m = len(a),len(a[0])
b = [[0] * n for i in range(m)]
for i in range(m):
for j in range(n):
b[i][j] = a[j][i]
return(b)
def power_two(x):
return (x and (not(x & (x - 1))))
def ceil(a, b):
return -(-a // b)
def seive(n):
a = [1]
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p ** 2,n + 1,p):
prime[i] = False
p = p + 1
for p in range(2,n + 1):
if prime[p]:
a.append(p)
return(a)
def ncr(n,r):
return(math.factorial(n) // (math.factorial(n - r) * math.factorial(r)))
def npr(n,r):
return(math.factorial(n) // math.factorial(n - r))
#-----------------------------------------------------------------------#
ONLINE_JUDGE = __debug__
if ONLINE_JUDGE:
#import io,os
#input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
input = sys.stdin.readline
main()
```
|
output
| 1
| 12,661
| 0
| 25,323
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
Submitted Solution:
```
n, m = map(int, input().split());a = input();b = input();LR = [0 for _ in range(n+1)];best = 0
for c in b:
NR = [0 for _ in range(n+1)]
for i in range(1,n+1):
back = NR[i-1] - 1;up = LR[i] - 1;diag = LR[i-1] - 2
if c == a[i-1]: diag += 4
NR[i] = max(back, up, diag, 0);best = max(best, NR[i])
LR = NR
print(best)
```
|
instruction
| 0
| 12,662
| 0
| 25,324
|
Yes
|
output
| 1
| 12,662
| 0
| 25,325
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
import math
import itertools as iter
from collections import defaultdict as ddic
from collections import Counter as Co
from collections import deque
import threading
def increase_stack():
sys.setrecursionlimit(2**32//2-1)
threading.stack_size(1 << 27)
#sys.setrecursionlimit(10**6)
#threading.stack_size(10**8)
t = threading.Thread(target=main)
t.start()
t.join()
#? Region Funtions
def binary(n):
return (bin(n).replace("0b", ""))
def decimal(s):
return (int(s, 2))
def pow2(n):
p = 0
while (n > 1):
n //= 2
p += 1
return (p)
def primeFactors(n):
l=[]
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3, int(math.sqrt(n)) + 1, 2):
while n % i == 0:
l.append(i)
n = n / i
if n > 2:
l.append(int(n))
return (l)
def isPrime(n):
if (n == 1):
return (False)
else:
root = int(n ** 0.5)
root += 1
for i in range(2, root):
if (n % i == 0):
return (False)
return (True)
def maxPrimeFactors(n):
maxPrime = -1
while n % 2 == 0:
maxPrime = 2
n >>= 1
for i in range(3, int(math.sqrt(n)) + 1, 2):
while n % i == 0:
maxPrime = i
n = n / i
if n > 2:
maxPrime = n
return int(maxPrime)
def countchar(s,i):
c=0
ch=s[i]
for i in range(i,len(s)):
if(s[i]==ch):
c+=1
else:
break
return(c)
def lis(arr):
n = len(arr)
lis = [1] * n
maximum=0
for i in range(1, n):
for j in range(0, i):
if arr[i] > arr[j] and lis[i] < lis[j] + 1:
lis[i] = lis[j] + 1
maximum=max(maximum,lis[i])
return maximum
def lcm(arr):
a=arr[0]; val=arr[0]
for i in range(1,len(arr)):
gcd=gcd(a,arr[i])
a=arr[i]; val*=arr[i]
return val//gcd
#? Region Taking Input
def inint():
return int(input())
def inarr():
return list(map(int,input().split()))
def invar():
return map(int,input().split())
def instr():
s=input()
return list(s)
#? Region Fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#!==================================== Write The Useful Code Here ============================
def solver():
pass
#! This is the Main Function
def main():
n,m=invar()
s1=input()
s2=input()
ans=0
dp=[[0]*(m+1) for i in range(0,n+1)]
#print(dp)
for i in range(0,n):
for j in range(0,m):
if(s1[i]==s2[j]):
dp[i+1][j+1]=max(0,dp[i][j]+2)
else:
dp[i+1][j+1]=max(0,dp[i+1][j]-1,dp[i][j+1]-1)
ans=max(ans,dp[i+1][j+1])
print(ans)
#? End Region
if __name__ == "__main__":
#? Incresing Stack Limit
#increase_stack()
#! Calling Main Function
main()
```
|
instruction
| 0
| 12,663
| 0
| 25,326
|
Yes
|
output
| 1
| 12,663
| 0
| 25,327
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
Submitted Solution:
```
n, m = map(int, input().split())
A = input()
B = input()
dp = []
for i, a in enumerate(A):
dp.append([])
for j, b in enumerate(B):
if i == 0 and j == 0:
if a == b:
dp[i].append(2)
else:
dp[i].append(0)
elif i == 0:
if a == b:
dp[i].append(2)
else:
dp[i].append(max(0, dp[i][j-1] - 1))
elif j == 0:
if a == b:
dp[i].append(2)
else:
dp[i].append(max(0, dp[i-1][j] - 1))
else:
if a != b:
dp[i].append(max(0, dp[i-1][j]-1, dp[i][j-1]-1))
else:
dp[i].append(max(0, dp[i-1][j-1]+2, dp[i-1][j]-1, dp[i][j-1]-1))
maxi = max([max(val) for val in dp])
print(maxi)
```
|
instruction
| 0
| 12,664
| 0
| 25,328
|
Yes
|
output
| 1
| 12,664
| 0
| 25,329
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
Submitted Solution:
```
import pprint
n, m = map(int, input().split())
s = input()
t = input()
ans = 0
X = s
Y = t
m = len(X)
n = len(Y)
L = [[0]*(n + 1) for i in range(m + 1)]
i = 1
j = 1
while i < m + 1:
j = 1
while j < n + 1:
v = 0
if X[i-1] == Y[j-1]:
v = max(L[i - 1][j - 1] + 2, 2)
if L[i][j - 1] > v:
v = L[i][j - 1] - 1
if L[i - 1][j] > v:
v = L[i - 1][j] - 1
L[i][j] = v
if v > ans:
ans = v
j += 1
i += 1
print(ans)
```
|
instruction
| 0
| 12,665
| 0
| 25,330
|
Yes
|
output
| 1
| 12,665
| 0
| 25,331
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
Submitted Solution:
```
import sys
from math import gcd,sqrt,ceil,log2
from collections import defaultdict,Counter,deque
from bisect import bisect_left,bisect_right
import math
sys.setrecursionlimit(2*10**5+10)
import heapq
from itertools import permutations
# input=sys.stdin.readline
# def print(x):
# sys.stdout.write(str(x)+"\n")
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
aa='abcdefghijklmnopqrstuvwxyz'
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# import sys
# import io, os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def get_sum(bit,i):
s = 0
i+=1
while i>0:
s+=bit[i]
i-=i&(-i)
return s
def update(bit,n,i,v):
i+=1
while i<=n:
bit[i]+=v
i+=i&(-i)
def modInverse(b,m):
g = math.gcd(b, m)
if (g != 1):
return -1
else:
return pow(b, m - 2, m)
def primeFactors(n):
sa = []
# sa.add(n)
while n % 2 == 0:
sa.append(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
sa.append(i)
n = n // i
# sa.add(n)
if n > 2:
sa.append(n)
return sa
def seive(n):
pri = [True]*(n+1)
p = 2
while p*p<=n:
if pri[p] == True:
for i in range(p*p,n+1,p):
pri[i] = False
p+=1
return pri
def check_prim(n):
if n<0:
return False
for i in range(2,int(sqrt(n))+1):
if n%i == 0:
return False
return True
def getZarr(string, z):
n = len(string)
# [L,R] make a window which matches
# with prefix of s
l, r, k = 0, 0, 0
for i in range(1, n):
# if i>R nothing matches so we will calculate.
# Z[i] using naive way.
if i > r:
l, r = i, i
# R-L = 0 in starting, so it will start
# checking from 0'th index. For example,
# for "ababab" and i = 1, the value of R
# remains 0 and Z[i] becomes 0. For string
# "aaaaaa" and i = 1, Z[i] and R become 5
while r < n and string[r - l] == string[r]:
r += 1
z[i] = r - l
r -= 1
else:
# k = i-L so k corresponds to number which
# matches in [L,R] interval.
k = i - l
# if Z[k] is less than remaining interval
# then Z[i] will be equal to Z[k].
# For example, str = "ababab", i = 3, R = 5
# and L = 2
if z[k] < r - i + 1:
z[i] = z[k]
# For example str = "aaaaaa" and i = 2,
# R is 5, L is 0
else:
# else start from R and check manually
l = i
while r < n and string[r - l] == string[r]:
r += 1
z[i] = r - l
r -= 1
def search(text, pattern):
# Create concatenated string "P$T"
concat = pattern + "$" + text
l = len(concat)
z = [0] * l
getZarr(concat, z)
ha = []
for i in range(l):
if z[i] == len(pattern):
ha.append(i - len(pattern) - 1)
return ha
# n,k = map(int,input().split())
# l = list(map(int,input().split()))
#
# n = int(input())
# l = list(map(int,input().split()))
#
# hash = defaultdict(list)
# la = []
#
# for i in range(n):
# la.append([l[i],i+1])
#
# la.sort(key = lambda x: (x[0],-x[1]))
# ans = []
# r = n
# flag = 0
# lo = []
# ha = [i for i in range(n,0,-1)]
# yo = []
# for a,b in la:
#
# if a == 1:
# ans.append([r,b])
# # hash[(1,1)].append([b,r])
# lo.append((r,b))
# ha.pop(0)
# yo.append([r,b])
# r-=1
#
# elif a == 2:
# # print(yo,lo)
# # print(hash[1,1])
# if lo == []:
# flag = 1
# break
# c,d = lo.pop(0)
# yo.pop(0)
# if b>=d:
# flag = 1
# break
# ans.append([c,b])
# yo.append([c,b])
#
#
#
# elif a == 3:
#
# if yo == []:
# flag = 1
# break
# c,d = yo.pop(0)
# if b>=d:
# flag = 1
# break
# if ha == []:
# flag = 1
# break
#
# ka = ha.pop(0)
#
# ans.append([ka,b])
# ans.append([ka,d])
# yo.append([ka,b])
#
# if flag:
# print(-1)
# else:
# print(len(ans))
# for a,b in ans:
# print(a,b)
def mergeIntervals(arr):
# Sorting based on the increasing order
# of the start intervals
arr.sort(key = lambda x: x[0])
# array to hold the merged intervals
m = []
s = -10000
max = -100000
for i in range(len(arr)):
a = arr[i]
if a[0] > max:
if i != 0:
m.append([s,max])
max = a[1]
s = a[0]
else:
if a[1] >= max:
max = a[1]
#'max' value gives the last point of
# that particular interval
# 's' gives the starting point of that interval
# 'm' array contains the list of all merged intervals
if max != -100000 and [s, max] not in m:
m.append([s, max])
return m
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def sol(n):
seti = set()
for i in range(1,int(sqrt(n))+1):
if n%i == 0:
seti.add(n//i)
seti.add(i)
return seti
def lcm(a,b):
return (a*b)//gcd(a,b)
#
# n,p = map(int,input().split())
#
# s = input()
#
# if n <=2:
# if n == 1:
# pass
# if n == 2:
# pass
# i = n-1
# idx = -1
# while i>=0:
# z = ord(s[i])-96
# k = chr(z+1+96)
# flag = 1
# if i-1>=0:
# if s[i-1]!=k:
# flag+=1
# else:
# flag+=1
# if i-2>=0:
# if s[i-2]!=k:
# flag+=1
# else:
# flag+=1
# if flag == 2:
# idx = i
# s[i] = k
# break
# if idx == -1:
# print('NO')
# exit()
# for i in range(idx+1,n):
# if
#
def moore_voting(l):
count1 = 0
count2 = 0
first = 10**18
second = 10**18
n = len(l)
for i in range(n):
if l[i] == first:
count1+=1
elif l[i] == second:
count2+=1
elif count1 == 0:
count1+=1
first = l[i]
elif count2 == 0:
count2+=1
second = l[i]
else:
count1-=1
count2-=1
for i in range(n):
if l[i] == first:
count1+=1
elif l[i] == second:
count2+=1
if count1>n//3:
return first
if count2>n//3:
return second
return -1
def find_parent(u,parent):
if u!=parent[u]:
parent[u] = find_parent(parent[u],parent)
return parent[u]
def dis_union(n):
par = [i for i in range(n+1)]
rank = [1]*(n+1)
k = int(input())
for i in range(k):
a,b = map(int,input().split())
z1,z2 = find_parent(a,par),find_parent(b,par)
if z1!=z2:
par[z1] = z2
rank[z2]+=rank[z1]
def dijkstra(n,tot,hash):
hea = [[0,n]]
dis = [10**18]*(tot+1)
dis[n] = 0
boo = defaultdict(bool)
check = defaultdict(int)
while hea:
a,b = heapq.heappop(hea)
if boo[b]:
continue
boo[b] = True
for i,w in hash[b]:
if b == 1:
c = 0
if (1,i,w) in nodes:
c = nodes[(1,i,w)]
del nodes[(1,i,w)]
if dis[b]+w<dis[i]:
dis[i] = dis[b]+w
check[i] = c
elif dis[b]+w == dis[i] and c == 0:
dis[i] = dis[b]+w
check[i] = c
else:
if dis[b]+w<=dis[i]:
dis[i] = dis[b]+w
check[i] = check[b]
heapq.heappush(hea,[dis[i],i])
return check
def power(x,y,p):
res = 1
x = x%p
if x == 0:
return 0
while y>0:
if (y&1) == 1:
res*=x
x = x*x
y = y>>1
return res
n,m = map(int,input().split())
s1 = input()
s2 = input()
dp = defaultdict(int)
maxi = 0
for i in range(n+1):
for j in range(m+1):
if i == 0 or j == 0:
continue
if s1[i-1] == s2[j-1]:
dp[(i,j)] = max(dp[(i,j)],dp[(i-1,j-1)]+2)
else:
dp[(i,j)] = max(dp[(i-1,j)],dp[(i,j-1)]) - 1
maxi = max(dp[(i,j)],maxi)
print(maxi)
```
|
instruction
| 0
| 12,666
| 0
| 25,332
|
No
|
output
| 1
| 12,666
| 0
| 25,333
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
Submitted Solution:
```
n,m = list(map(int,input().split()))
a = input()
b = input()
l1 = len(a)
l2 = len(b)
lcs = [[0 for j in range(l2+1)] for i in range(l1+1)]
for i in range(1,l1+1):
for j in range(1,l2+1):
if a[i-1]==b[j-1]:
lcs[i][j] = lcs[i-1][j-1]+1
else:
lcs[i][j] = max(lcs[i-1][j],lcs[i][j-1])
i,j = l1,l2
ilist = []
jlist = []
while i>0 and j>0:
if a[i-1]==b[j-1]:
ilist.append(i)
jlist.append(j)
i-=1
j-=1
else:
if lcs[i][j]==lcs[i][j-1]:
j-=1
else:
i-=1
out = lcs[-1][-1]
la,lb = 0,0
if ilist and jlist:
la = ilist[0]-ilist[-1]+1
lb = jlist[0]-jlist[-1]+1
print(4*out-la-lb)
```
|
instruction
| 0
| 12,667
| 0
| 25,334
|
No
|
output
| 1
| 12,667
| 0
| 25,335
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
Submitted Solution:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
import time
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
#start_time = time.time()
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
def getMat(n):
return [getInts() for _ in range(n)]
MOD = 10**9+7
"""
4*L(C,D) - |C| - |D|
5 7 8
5
3 4 6
2
4 6 7
3
abba
babab
Substrings can have length 0 to len(S) in each
dp = [[0 for j in range(M+1)] for i in range(N+1)]
dp[1][0] = dp[0][1] = 0
dp[1][1] = 0 if S[0] != T[0] else 2
"""
def solve():
N, M = getInts()
S = [0]+listStr()
T = [0]+listStr()
ans = 0
dp = [[0 for m in range(M+1)] for n in range(N+1)]
for i in range(1,N+1):
for j in range(1,M+1):
dp[i][j] = max(dp[i-1][j]-1,dp[i][j-1]-1)
if S[i] == T[j]:
dp[i][j] = max(dp[i][j],dp[i-1][j-1]+2)
ans = max(ans,dp[i][j])
return ans
#for _ in range(getInt()):
print(solve())
#print(time.time()-start_time)
```
|
instruction
| 0
| 12,668
| 0
| 25,336
|
No
|
output
| 1
| 12,668
| 0
| 25,337
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings A and B representing essays of two students who are suspected cheaters. For any two strings C, D we define their similarity score S(C,D) as 4⋅ LCS(C,D) - |C| - |D|, where LCS(C,D) denotes the length of the Longest Common Subsequence of strings C and D.
You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.
Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
If X is a string, |X| denotes its length.
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.
You may wish to read the [Wikipedia page about the Longest Common Subsequence problem](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem).
Input
The first line contains two positive integers n and m (1 ≤ n, m ≤ 5000) — lengths of the two strings A and B.
The second line contains a string consisting of n lowercase Latin letters — string A.
The third line contains a string consisting of m lowercase Latin letters — string B.
Output
Output maximal S(C, D) over all pairs (C, D), where C is some substring of A, and D is some substring of B.
Examples
Input
4 5
abba
babab
Output
5
Input
8 10
bbbbabab
bbbabaaaaa
Output
12
Input
7 7
uiibwws
qhtkxcn
Output
0
Note
For the first case:
abb from the first string and abab from the second string have LCS equal to abb.
The result is S(abb, abab) = (4 ⋅ |abb|) - |abb| - |abab| = 4 ⋅ 3 - 3 - 4 = 5.
Submitted Solution:
```
import sys
import math
def II():
return int(sys.stdin.readline())
def LI():
return list(map(int, sys.stdin.readline().split()))
def MI():
return map(int, sys.stdin.readline().split())
def SI():
return sys.stdin.readline().strip()
def FACT(n, mod):
s = 1
facts = [1]
for i in range(1,n+1):
s*=i
s%=mod
facts.append(s)
return facts[n]
def C(n, k, mod):
return (FACT(n,mod) * pow((FACT(k,mod)*FACT(n-k,mod))%mod,mod-2, mod))%mod
n,m = MI()
a = list(SI())
b = list(SI())
n1 = len(a)
n2 = len(b)
ans = 0
dp = [[0 for i in range(n2 + 1)] for i in range(n1 + 1)]
dp1 = [[[1,1] for i in range(n2 + 1)] for i in range(n1 + 1)]
for i in range(1,n+1):
for j in range(1,m+1):
if a[i-1] == b[j-1]:
dp[i][j] = dp[i-1][j-1]+1
dp1[i][j] = dp1[i-1][j-1]
if dp[i][j] == 1:
dp1[i][j] = [i,j]
x,y = dp1[i][j]
ans = max(ans,4*dp[i][j]-(i-x+1)-(j-y+1))
else:
if dp[i-1][j]>dp[i][j-1]:
dp[i][j] = dp[i-1][j]
dp1[i][j] = dp1[i-1][j]
elif dp[i][j-1]>dp[i-1][j]:
dp[i][j] = dp[i][j-1]
dp1[i][j] = dp1[i][j-1]
else:
x,y = dp1[i][j-1]
x2,y2 = dp1[i-1][j]
if x2-x+y2-y > 0:
dp1[i][j] = dp1[i-1][j]
else:
dp1[i][j] = dp1[i][j-1]
print(ans)
```
|
instruction
| 0
| 12,669
| 0
| 25,338
|
No
|
output
| 1
| 12,669
| 0
| 25,339
|
Provide a correct Python 3 solution for this coding contest problem.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
|
instruction
| 0
| 13,160
| 0
| 26,320
|
"Correct Solution:
```
N = int(input())
*A, = map(int, input().split())
def check(K):
st = []
for i, a in enumerate(A):
#print(K, i, st, a)
while st and a < st[-1][0]:
st.pop()
u = i > 0 and A[i-1] >= A[i]
if not st:
st.append([a, u])
continue
l = st[-1]
if l[0] == a:
l[1] += 1
while st and l[1] == K:
st.pop(); a -= 1
if a == 0:
return 0
if not st:
st.append([a, 1])
break
l = st[-1]
if l[0] == a:
l[1] += 1
else:
st.append([a, 1])
break
#print(l, st)
else:
st.append([a, u])
#print(st)
return 1
ok = 1
for i in range(N-1):
if not A[i] < A[i+1]:
ok = 0
if ok:
print(1)
exit(0)
left = 1; right = N+1
while left+1 < right:
mid = (left + right) // 2
if check(mid):
right = mid
else:
left = mid
print(right)
```
|
output
| 1
| 13,160
| 0
| 26,321
|
Provide a correct Python 3 solution for this coding contest problem.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
|
instruction
| 0
| 13,161
| 0
| 26,322
|
"Correct Solution:
```
import sys
stdin = sys.stdin
sys.setrecursionlimit(10**5)
def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x)-1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())
def shave(stack:list, nex:int, prv:int):
res = prv-nex
while res > 0:
if res >= stack[-1][1]:
_,y = stack.pop()
res -= y
else:
stack[-1][1] -= res
res = 0
def normalize(stack: list):
if stack[-1][0] == stack[-2][0]:
stack[-2][1] += stack[-1][1]
stack.pop()
def add(stack:list):
if stack[-1][1] == 1:
stack[-1][0] += 1
else:
stack[-1][1] -= 1
stack.append([stack[-1][0]+1, 1])
def kuriagari(stack:list):
_, y = stack.pop()
add(stack)
normalize(stack)
stack.append([1,y])
def check(a:list, upto:int):
stack = [[0,0]]
for i in range(n):
if a[i+1] > a[i]:
stack.append([1, a[i+1]-a[i]])
normalize(stack)
else:
shave(stack, a[i+1], a[i])
if stack[-1][0] == upto and stack[-1][1] == a[i+1]:
return False
elif stack[-1][0] == upto:
kuriagari(stack)
else:
add(stack)
normalize(stack)
return True
def binsearch(a:list):
low = 0
high = 10**9+1
while high-low > 1:
mid = (high+low) // 2
if check(a, mid):
high = mid
else:
low = mid
return high
n = ni()
a = [0] + list(li())
ans = binsearch(a)
print(ans)
```
|
output
| 1
| 13,161
| 0
| 26,323
|
Provide a correct Python 3 solution for this coding contest problem.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
|
instruction
| 0
| 13,162
| 0
| 26,324
|
"Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
ok, ng = n, 0
while ok-ng > 1:
x = (ok+ng)//2
#d = defaultdict(int)
d = dict()
last = 0
valid = True
if x == 1:
for i in range(n-1):
if a[i] >= a[i+1]:
valid = False
break
if valid:
ok = x
else:
ng = x
continue
for i in range(n-1):
dels = []
if a[i] < a[i+1]:
last = a[i+1]
continue
j = a[i+1]
for k in d.keys():
if j < k:
dels.append(k)
while j > 0:
if j in d and d[j] == x-1:
#d[j] = 0
dels.append(j)
j -= 1
else:
break
if j in d:
d[j] += 1
else:
d[j] = 1
for k in dels:
del d[k]
if 0 in d:
valid = False
break
last = j
if valid:
ok = x
else:
ng = x
print(ok)
```
|
output
| 1
| 13,162
| 0
| 26,325
|
Provide a correct Python 3 solution for this coding contest problem.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
|
instruction
| 0
| 13,163
| 0
| 26,326
|
"Correct Solution:
```
import sys
input = sys.stdin.readline
N = int(input())
A = [int(x) for x in input().split()]
A.append(1)
B = [A[0]]
B_cnt = [1]
for i,(x,y) in enumerate(zip(A[:-1],A[1:])):
if x < y and y < A[i+2]:
continue
if x == y:
B_cnt[-1] += 1
else:
B.append(y)
B_cnt.append(1)
def test(x):
digit = [0]
cnt = [0]
for a,a_cnt in zip(B,B_cnt):
if digit[-1] < a:
digit.append(a)
cnt.append(a_cnt)
continue
if digit[-1] == a:
cnt[-1] += a_cnt
continue
while True:
# 繰り上がり処理をしながら左に戻る
if digit[-1] <= a:
break
n = digit.pop()
k = cnt.pop()
if k <= x:
continue
if digit[-1] == n-1:
cnt[-1] += (k-1)//x
else:
digit.append(n-1)
cnt.append((k-1)//x + 1)
if digit[-1] == a:
cnt[-1] += a_cnt
else:
digit.append(a)
cnt.append(1+a_cnt)
return cnt[1] <= x+1
if all(x<y for x,y in zip(A[:-2],A[1:-1])):
print(1)
exit()
left = 1 # 無理
right = 10 # 可能?
while not test(right):
left *= 10
right *= 10
right = min(right,N)
while right > left + 1:
mid = (left+right)//2
if test(mid):
right = mid
else:
left = mid
answer = right
print(answer)
```
|
output
| 1
| 13,163
| 0
| 26,327
|
Provide a correct Python 3 solution for this coding contest problem.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
|
instruction
| 0
| 13,164
| 0
| 26,328
|
"Correct Solution:
```
N = int(input())
A = [int(a) for a in input().split()]
def chk(k):
if k == 1:
for i in range(1, N):
if A[i] <= A[i-1]: return 0
return 1
X = [(0, 0)]
def add(x, y):
if x <= 0: return 0
if x > X[-1][0]:
X.append((x, 0 if x == y else 1))
elif x == X[-1][0]:
if X[-1][1] + 1 < k:
X[-1] = (X[-1][0], X[-1][1] + 1)
else:
if add(x-1, y) == 0: return 0
else:
while X[-1][0] > x:
X.pop()
if x > X[-1][0]:
X.append((x, 1))
elif x == X[-1][0]:
if X[-1][1] + 1 < k:
X[-1] = (X[-1][0], X[-1][1] + 1)
else:
if add(x-1, y) == 0: return 0
if X[-1][0] < y: X.append((y, 0))
return 1
for a in A:
if add(a, a) == 0: return 0
return 1
l, r = 0, 1 << 18
while r - l > 1:
m = (l + r) // 2
if chk(m):
r = m
else:
l = m
print(r)
```
|
output
| 1
| 13,164
| 0
| 26,329
|
Provide a correct Python 3 solution for this coding contest problem.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
|
instruction
| 0
| 13,165
| 0
| 26,330
|
"Correct Solution:
```
def is_possible(x, A):
current = {}
length = 0
for a in A:
if a <= length:
if x == 1:
return False
current = {k: current[k] for k in current if k < a}
i = a - 1
if i not in current:
current[i] = 0
current[i] += 1
while i in current and current[i] >= x:
if i == 0:
return False
if (i-1) not in current:
current[i-1] = 0
current[i-1] += 1
del current[i]
i -= 1
length = a
return True
def main():
N = int(input())
A = list(map(int, input().split()))
left = 1
right = N
while left < right:
center = (left + right) // 2
if is_possible(center, A):
right = center
else:
left = center + 1
print(left)
if __name__ == "__main__":
main()
```
|
output
| 1
| 13,165
| 0
| 26,331
|
Provide a correct Python 3 solution for this coding contest problem.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
|
instruction
| 0
| 13,166
| 0
| 26,332
|
"Correct Solution:
```
N = int(input())
A_raw = list(map(int, input().split()))
flg = 1
prev = 0
A = []
for i in range(N):
if A_raw[i] <= prev:
flg = 0
prev = A_raw[i]
if A_raw[i] <= 50:
A.append(A_raw[i])
if flg:
print(1)
exit()
N = len(A)
if N <= 1:
print(2)
exit()
ok = N
ng = 1
mid = (ok + ng) // 2
while ok - ng > 1:
ng_flg = 0
word = [1] * A[0]
for i in range(1, N):
if A[i] > A[i-1]:
word.extend([1] * (A[i] - A[i-1]))
else:
word_prev = word[:A[i]]
k = A[i] - 1
while True:
if k == -1:
ng_flg = 1
break
if word_prev[k] != mid:
word_prev[k] += 1
word = word_prev
break
else:
word_prev[k] = 1
k -= 1
if ng_flg == 1:
break
if not ng_flg:
ok = mid
else:
ng = mid
mid = (ok + ng) // 2
print(ok)
```
|
output
| 1
| 13,166
| 0
| 26,333
|
Provide a correct Python 3 solution for this coding contest problem.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
|
instruction
| 0
| 13,167
| 0
| 26,334
|
"Correct Solution:
```
def solve(aaa):
l = 1
r = 200000
while l < r:
m = (l + r) // 2
if check(m, aaa):
r = m
else:
l = m + 1
return r
def check(d, aaa):
if d == 1:
return all(a1 < a2 for a1, a2 in zip(aaa, aaa[1:]))
l = 0
word = [[0, 0]]
for a in aaa:
if l >= a:
if not carry(d, word, a):
return False
l = a
return True
def carry(d, word, i):
while word[-1][0] > i:
word.pop()
while word[-1] == [i, d]:
word.pop()
i -= 1
if i == 0:
return False
if word[-1][0] == i:
word[-1][1] += 1
else:
word.append([i, 2])
return True
n = int(input())
aaa = list(map(int, input().split()))
print(solve(aaa))
```
|
output
| 1
| 13,167
| 0
| 26,335
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
Submitted Solution:
```
def main():
import heapq
n = int(input())
a = list(map(int, input().split()))
for i in range(n-1):
if a[i] >= a[i+1]:
break
else:
print(1)
return
h = []
heapq.heapify(h)
kind = 1
length = 0
d = dict()
for i in a:
if length >= i:
while h:
hh = -heapq.heappop(h)
if hh > i:
d.pop(hh)
else:
heapq.heappush(h, -hh)
break
for j in range(i, 0, -1):
if j not in d:
d[j] = 2
heapq.heappush(h, -j)
kind = max(kind, 2)
while h:
hh = -heapq.heappop(h)
if hh > j:
d.pop(hh)
else:
heapq.heappush(h, -hh)
break
break
else:
if d[j] < kind:
d[j] += 1
while h:
hh = -heapq.heappop(h)
if hh > j:
d.pop(hh)
else:
heapq.heappush(h, -hh)
break
break
else:
d[i] += 1
kind += 1
length = i
def value(kind):
length = 0
d = dict()
for i in a:
if length >= i:
del_list = []
for j in d:
if j > i:
del_list.append(j)
for j in del_list:
d.pop(j)
for j in range(i, 0, -1):
if j not in d:
d[j] = 2
break
else:
if d[j] < kind:
d[j] += 1
break
else:
d.pop(j)
else:
return False
length = i
return True
def b_search(ok, ng, value):
while abs(ok-ng) > 1:
mid = (ok+ng)//2
if value(mid):
ok = mid
else:
ng = mid
return ok
print(b_search(kind, max(1, kind-30), value))
main()
```
|
instruction
| 0
| 13,168
| 0
| 26,336
|
Yes
|
output
| 1
| 13,168
| 0
| 26,337
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
Submitted Solution:
```
import random
input()
A = [int(_) for _ in input().split()]
A = [A[0]] + [j for i, j in zip(A, A[1:]) if i >= j]
N = len(A)
def cut(array, index):
if index < 1:
return []
if index <= array[0][0]:
return [(index, array[0][1])]
for _ in range(len(array)-1, 0, -1):
if array[_ - 1][0] < index:
return array[:_] + [(index, array[_][1])]
def is_possible(K):
dp = [(A[0], 0)]
for a in A[1:]:
if a <= dp[-1][0]:
dp = cut(dp, a)
else:
dp += [(a, 0)]
is_added = False
for j in range(len(dp) - 1, -1, -1):
if dp[j][1] < K - 1:
dp = cut(dp, dp[j][0] - 1) + [(dp[j][0], dp[j][1] + 1)]
if dp[-1][0] < a:
dp += [(a, 0)]
is_added = True
break
if not is_added:
return False
return True
def bis(x, y):
if y == x + 1:
return y
elif is_possible((x + y) // 2):
return bis(x, (x + y) // 2)
else:
return bis((x + y) // 2, y)
print(bis(0, N))
```
|
instruction
| 0
| 13,169
| 0
| 26,338
|
Yes
|
output
| 1
| 13,169
| 0
| 26,339
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
Submitted Solution:
```
import sys
input = sys.stdin.readline
N = int(input())
A = list(map(int, input().split()))
# import random
# N = 10**5
# A = [random.randrange(1,10**9+1) for _ in range(N)]
# import time
def main():
l = 0
r = N
while r-l > 1:
m = (r+l)//2
ok = True
if m == 1:
pre = -1
for a in A:
if pre >= a:
ok = False
break
pre = a
else:
dp = {}
needreset = 10**15
exception = set()
pre = -1
for a in A:
ind = a
for _ in range(40):
# 修正
if needreset >= ind or ind in exception:
if not ind in dp:
dp[ind] = 0 if pre < a else 1
else:
dp[ind] = 0 if pre < a else 1
exception.add(ind)
# たす
if dp[ind] < m:
dp[ind] += 1
break
elif ind == 1:
ok = False
break
else:
# 繰り上がり
dp[ind] = 1
ind -= 1
if not ok: break
# 更新
if pre > a:
needreset = a
exception = set()
pre = a
if ok:
r = m
else:
l = m
print(r)
if __name__ == "__main__":
#dt = time.time()
main()
#print(time.time()-dt)
```
|
instruction
| 0
| 13,170
| 0
| 26,340
|
Yes
|
output
| 1
| 13,170
| 0
| 26,341
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
Submitted Solution:
```
import bisect
from collections import deque
"""
def pl(d,k):
now = d
ind = bisect.bisect_left(dlis,d)
while ind > 0:
dic[dlis[ind]] += 1
if dic[dlis[ind]] // k == 0:
dic[dlis[ind]] = 0
else:
break
ind -= 1
def able(k):
ndep = A[0]
for i in range(len(dlis)):
dic[dlis[i]] = 0
for i in range(N-1):
i += 1
if ndep < A[i]:
dic[A[i]] = 0
ndep = A[i]
else:
ndep = A[i]
pl(ndep,k)
print (dic,k)
if dic[dlis[0]] < k:
return True
else:
return False
"""
def updiv(a,b):
if a % b == 0:
return a // b
else:
return a // b + 1
def able(k):
nq = deque([0])
num = deque([A[0]])
for i in range(N-1):
i += 1
if A[i] > A[i-1]:
if nq[-1] == 0:
num[-1] += A[i] - A[i-1]
else:
nq.append(0)
num.append(A[i] - A[i-1])
else:
if A[i] < A[i-1]:
ds = A[i-1] - A[i]
while len(num) > 0 and ds >= num[-1]:
nq.pop()
ds -= num[-1]
num.pop()
if len(num) == 0:
return False
num[-1] -= ds
now = 0
while len(nq) > 0 and nq[-1] == k-1:
now += num[-1]
nq.pop()
num.pop()
if len(nq) == 0:
return False
if num[-1] == 1:
nq[-1] += 1
else:
num[-1] -= 1
nq.append(nq[-1] + 1)
num.append(1)
if now > 0:
nq.append(0)
num.append(now)
while len(nq) > 1 and nq[-1] == nq[-2]:
num[-2] += num[-1]
nq.pop()
num.pop()
#print (k,A[i])
#print (nq)
#print (num)
return True
N = int(input())
A = list(map(int,input().split()))
A.append(1)
l = 0
r = N
while r - l != 1 :
m = (l + r) // 2
if able(m):
r = m
else:
l = m
print (r)
```
|
instruction
| 0
| 13,171
| 0
| 26,342
|
Yes
|
output
| 1
| 13,171
| 0
| 26,343
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
Submitted Solution:
```
from collections import defaultdict
from copy import copy
n = int(input())
a = list(map(int, input().split()))
ok, ng = n, 0
while ok-ng > 1:
x = (ok+ng)//2
d = defaultdict(int)
last = 0
valid = True
if x == 1:
for i in range(n-1):
if a[i] >= a[i+1]:
valid = False
break
if valid:
ok = x
else:
ng = x
continue
for i in range(n-1):
if a[i] > a[i+1]:
if a[i+1] > last:
d[a[i+1]] = 1
else:
j = a[i+1]
while d[j] == x-1 and j > 0:
del d[j]
j -= 1
d[j] += 1
last = a[i+1]
elif a[i] == a[i+1]:
j = a[i+1]
while d[j] == x-1 and j > 0:
del d[j]
j -= 1
d[j] += 1
last = a[i+1]
e = copy(d)
for k in d.keys():
if k > last:
del e[k]
d = copy(e)
#if x < 5:
#print(x, last, d)
if d[0] > 0:
valid = False
break
if valid:
ok = x
else:
ng = x
print(ok)
```
|
instruction
| 0
| 13,172
| 0
| 26,344
|
No
|
output
| 1
| 13,172
| 0
| 26,345
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
Submitted Solution:
```
N = int(input())
A = list(map(int, input().split()))
ans = 0
for i in range(N - 1):
if A[i] > A[i + 1]:
ans += 1
print(ans)
```
|
instruction
| 0
| 13,173
| 0
| 26,346
|
No
|
output
| 1
| 13,173
| 0
| 26,347
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
Submitted Solution:
```
def main():
n = int(input())
a = list(map(int, input().split()))
for i in range(n-1):
if a[i] >= a[i+1]:
break
else:
print(1)
return
def value(kind):
length = 0
d = []
for i in a:
if length >= i:
while d:
j, k = d[-1]
if j > i:
d.pop()
else:
break
l = len(d)-1
if d:
jj, k = d[-1]
else:
jj, k = -1, -1
for j in range(i, 0, -1):
if jj != j:
d.append([j, 2])
kind = max(kind, 2)
break
else:
if k < kind:
d[l][1] += 1
for jjj in range(l+1, len(d)):
d.pop()
else:
l -= 1
else:
return False
length = i
return True
def b_search(ok, ng, value):
while abs(ok-ng) > 1:
mid = (ok+ng)//2
if value(mid):
ok = mid
else:
ng = mid
return ok
print(b_search(n, 1, value))
main()
```
|
instruction
| 0
| 13,174
| 0
| 26,348
|
No
|
output
| 1
| 13,174
| 0
| 26,349
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N strings arranged in a row. It is known that, for any two adjacent strings, the string to the left is lexicographically smaller than the string to the right. That is, S_1<S_2<...<S_N holds lexicographically, where S_i is the i-th string from the left.
At least how many different characters are contained in S_1,S_2,...,S_N, if the length of S_i is known to be A_i?
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq A_i \leq 10^9
* A_i is an integer.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 ... A_N
Output
Print the minimum possible number of different characters contained in the strings.
Examples
Input
3
3 2 1
Output
2
Input
5
2 3 2 1 2
Output
2
Submitted Solution:
```
N = int(input())
A = [int(i) for i in input().split()]
def f(al, X):
def g(stack, X):
if not any(stack):
stack.append(-1)
return
l = stack.pop()
if l >= X:
g(stack, X)
stack.append(1)
else:
stack.append(l + 1)
st = []
s = 0
for t in al:
d = t - s
if d > 0:
st.extend([1] * d)
else:
for _ in range(-d):
st.pop()
g(st, X)
if st[0] < 0:
return False
s = t
return True
max_N = N
min_N = len([i for i in A if i == 1])
b_max_N = 0
b_min_N = 0
while max_N != min_N:
nx = (max_N + min_N) // 2
if f(A, nx):
max_N = nx
else:
min_N = nx
if b_max_N == max_N and b_min_N == min_N:
break
b_max_N = max_N
b_min_N = min_N
print(max_N)
```
|
instruction
| 0
| 13,175
| 0
| 26,350
|
No
|
output
| 1
| 13,175
| 0
| 26,351
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya owns three strings s , a and b, each of them consists only of first k Latin letters.
Let a template be such a string of length k that each of the first k Latin letters appears in it exactly once (thus there are k! distinct templates). Application of template p to the string s is the replacement of each character in string s with p_i, i is the index of this letter in the alphabet. For example, applying template "bdca" to a string "aabccd" yields string "bbdcca".
Vasya wants to know if there exists such a template which yields a string lexicographically greater than or equal to string a and lexicographically less than or equal to string b after applying it to s.
If there exist multiple suitable templates, print any of them.
String a is lexicographically less than string b if there is some i (1 ≤ i ≤ n) that a_i < b_i and for any j (1 ≤ j < i) a_j = b_j.
You are required to answer t testcases independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^6) — the number of testcases.
In hacks you can only use t = 1.
Each of the next t lines contains the description of the testcase in the following form:
The first line of the testcase contains a single integer k (1 ≤ k ≤ 26) — the length of the template.
The second line of the testcase contains the string s (1 ≤ |s| ≤ 10^6).
The third line of the testcase contains the string a.
The fourth line of the testcase contains the string b.
Strings s, a and b have the same length (|s| = |a| = |b|) and consist only of the first k Latin letters, all letters are lowercase.
It is guaranteed that string a is lexicographically less than or equal to string b.
It is also guaranteed that the total length of strings over all testcase won't exceed 3 ⋅ 10^6.
Output
Print the answers to all testcases in the following form:
If there exists no suitable template then print "NO" in the first line.
Otherwise print "YES" in the first line and the template itself in the second line (k lowercase letters, each of the first k Latin letters should appear exactly once).
If there exist multiple suitable templates, print any of them.
Example
Input
2
4
bbcb
aada
aada
3
abc
bbb
bbb
Output
YES
badc
NO
Submitted Solution:
```
from itertools import permutations
alpha = [a for a in 'qwertyuiopasdfghjklzxcvbnm']
alpha.sort()
def f(k, s, a, b):
for key in permutations(alpha[:k], k):
d = {alpha[i]: key[i] for i in range(k)}
s = [d[x] for x in s]
if a <= ''.join(s) <= b:
return key, True
return '', False
def main():
t = int(input())
res = []
for i in range(t):
k = int(input())
s = input()
a = input()
b = input()
key, r = f(k, s, a, b)
if r:
res.append('YES\n'+''.join(key))
else:
res.append('NO')
print('\n'.join(res))
if __name__ == '__main__':
main()
```
|
instruction
| 0
| 13,339
| 0
| 26,678
|
No
|
output
| 1
| 13,339
| 0
| 26,679
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya owns three strings s , a and b, each of them consists only of first k Latin letters.
Let a template be such a string of length k that each of the first k Latin letters appears in it exactly once (thus there are k! distinct templates). Application of template p to the string s is the replacement of each character in string s with p_i, i is the index of this letter in the alphabet. For example, applying template "bdca" to a string "aabccd" yields string "bbdcca".
Vasya wants to know if there exists such a template which yields a string lexicographically greater than or equal to string a and lexicographically less than or equal to string b after applying it to s.
If there exist multiple suitable templates, print any of them.
String a is lexicographically less than string b if there is some i (1 ≤ i ≤ n) that a_i < b_i and for any j (1 ≤ j < i) a_j = b_j.
You are required to answer t testcases independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^6) — the number of testcases.
In hacks you can only use t = 1.
Each of the next t lines contains the description of the testcase in the following form:
The first line of the testcase contains a single integer k (1 ≤ k ≤ 26) — the length of the template.
The second line of the testcase contains the string s (1 ≤ |s| ≤ 10^6).
The third line of the testcase contains the string a.
The fourth line of the testcase contains the string b.
Strings s, a and b have the same length (|s| = |a| = |b|) and consist only of the first k Latin letters, all letters are lowercase.
It is guaranteed that string a is lexicographically less than or equal to string b.
It is also guaranteed that the total length of strings over all testcase won't exceed 3 ⋅ 10^6.
Output
Print the answers to all testcases in the following form:
If there exists no suitable template then print "NO" in the first line.
Otherwise print "YES" in the first line and the template itself in the second line (k lowercase letters, each of the first k Latin letters should appear exactly once).
If there exist multiple suitable templates, print any of them.
Example
Input
2
4
bbcb
aada
aada
3
abc
bbb
bbb
Output
YES
badc
NO
Submitted Solution:
```
t = int(input())
list = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
for i in range(t):
l = int(input())
sample = input()
a = input()
b = input()
# print(sample)
# print(a)
# print(b)
panel = dict(zip(list[:l],[0]*l))
print(panel)
for i in range(l):
if a[i]==b[i]:
panel[a[i]] = sample[i]
print(panel)
alphas = []
count = 0
for value in panel.values():
if str(value).isalpha():
alphas.append(value)
else:
count += 1
if count+len(alphas):
print("YES")
panel = sorted(panel.items(), key=lambda x:x[0])
result = []
for item in panel:
result.append(item[1])
a = 0
for i in range(l):
if result[i]==0:
while(a<26):
if list[a] not in result:
result[i] = list[a]
a += 1
break
a += 1
print(result)
else:
print("NO")
```
|
instruction
| 0
| 13,340
| 0
| 26,680
|
No
|
output
| 1
| 13,340
| 0
| 26,681
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya owns three strings s , a and b, each of them consists only of first k Latin letters.
Let a template be such a string of length k that each of the first k Latin letters appears in it exactly once (thus there are k! distinct templates). Application of template p to the string s is the replacement of each character in string s with p_i, i is the index of this letter in the alphabet. For example, applying template "bdca" to a string "aabccd" yields string "bbdcca".
Vasya wants to know if there exists such a template which yields a string lexicographically greater than or equal to string a and lexicographically less than or equal to string b after applying it to s.
If there exist multiple suitable templates, print any of them.
String a is lexicographically less than string b if there is some i (1 ≤ i ≤ n) that a_i < b_i and for any j (1 ≤ j < i) a_j = b_j.
You are required to answer t testcases independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^6) — the number of testcases.
In hacks you can only use t = 1.
Each of the next t lines contains the description of the testcase in the following form:
The first line of the testcase contains a single integer k (1 ≤ k ≤ 26) — the length of the template.
The second line of the testcase contains the string s (1 ≤ |s| ≤ 10^6).
The third line of the testcase contains the string a.
The fourth line of the testcase contains the string b.
Strings s, a and b have the same length (|s| = |a| = |b|) and consist only of the first k Latin letters, all letters are lowercase.
It is guaranteed that string a is lexicographically less than or equal to string b.
It is also guaranteed that the total length of strings over all testcase won't exceed 3 ⋅ 10^6.
Output
Print the answers to all testcases in the following form:
If there exists no suitable template then print "NO" in the first line.
Otherwise print "YES" in the first line and the template itself in the second line (k lowercase letters, each of the first k Latin letters should appear exactly once).
If there exist multiple suitable templates, print any of them.
Example
Input
2
4
bbcb
aada
aada
3
abc
bbb
bbb
Output
YES
badc
NO
Submitted Solution:
```
col = int(input())
sp = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
sp1 = {0: 'a', 1: 'b', 2: 'c', 3: 'd', 4: 'e', 5: 'f', 6: 'g', 7: 'h', 8: 'i', 9: 'j', 10: 'k', 11: 'l', 12: 'm', 13: 'n', 14: 'o', 15: 'p', 16: 'q', 17: 'r', 18: 's', 19: 't', 20: 'u', 21: 'v', 22: 'w', 23: 'x', 24: 'y', 25: 'z'}
def getdiff(a, b):
return abs(sp.get(a) - sp.get(b))
def yorn(a, b, s):
flag = False
for i in range(len(a)):
dif = getdiff(a[i], b[i])
if dif > 1:
return (1, i)
elif dif == 1:
flag = True
return (0, flag)
def pattern(a, b, s, i, leng):
pat = ['a']*leng
for j in range(i):
pat[sp.get(a[j])] = a[j]
pat[sp.get(a[i])] = sp1.get(sp.get(a[i]) + 1)
return ''.join(pat)
def pattern1(a, b, s, i, leng):
pat = list(a)
if sp.get(a[i]) + 1 > 26:
return 0
pat[sp.get(a[i])] = sp1.get(sp.get(a[i]) + 1)
return ''.join(pat)
res = ''
for i in range(col):
leng = int(input())
s = input()
a = input()
b = input()
yes = yorn(a, b, s)
if len(set(a)) < len(set(s)):
res += "NO\n"
#print('NO')
elif yes[0]:
#print('YES')
#print(pattern(a, b, s, yes[1], leng))
res += "YES\n"
res += pattern(a, b, s, yes[1], leng) + "\n"
else:
itog = pattern1(a, b, s, yes[1], leng)
if not itog:
res += "NO\n"
#print('NO')
else:
#print('YES')
#print(itog)
res += "YES\n"
res += itog + "\n"
print(res[:-1])
'''
1
4
abcd
abcd
cdef
'''
'''sch = 0
for i in 'abcdefghijklmnopqrstuvwxyz':
s[i] = sch
sch += 1
print(s)'''
```
|
instruction
| 0
| 13,341
| 0
| 26,682
|
No
|
output
| 1
| 13,341
| 0
| 26,683
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya owns three strings s , a and b, each of them consists only of first k Latin letters.
Let a template be such a string of length k that each of the first k Latin letters appears in it exactly once (thus there are k! distinct templates). Application of template p to the string s is the replacement of each character in string s with p_i, i is the index of this letter in the alphabet. For example, applying template "bdca" to a string "aabccd" yields string "bbdcca".
Vasya wants to know if there exists such a template which yields a string lexicographically greater than or equal to string a and lexicographically less than or equal to string b after applying it to s.
If there exist multiple suitable templates, print any of them.
String a is lexicographically less than string b if there is some i (1 ≤ i ≤ n) that a_i < b_i and for any j (1 ≤ j < i) a_j = b_j.
You are required to answer t testcases independently.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^6) — the number of testcases.
In hacks you can only use t = 1.
Each of the next t lines contains the description of the testcase in the following form:
The first line of the testcase contains a single integer k (1 ≤ k ≤ 26) — the length of the template.
The second line of the testcase contains the string s (1 ≤ |s| ≤ 10^6).
The third line of the testcase contains the string a.
The fourth line of the testcase contains the string b.
Strings s, a and b have the same length (|s| = |a| = |b|) and consist only of the first k Latin letters, all letters are lowercase.
It is guaranteed that string a is lexicographically less than or equal to string b.
It is also guaranteed that the total length of strings over all testcase won't exceed 3 ⋅ 10^6.
Output
Print the answers to all testcases in the following form:
If there exists no suitable template then print "NO" in the first line.
Otherwise print "YES" in the first line and the template itself in the second line (k lowercase letters, each of the first k Latin letters should appear exactly once).
If there exist multiple suitable templates, print any of them.
Example
Input
2
4
bbcb
aada
aada
3
abc
bbb
bbb
Output
YES
badc
NO
Submitted Solution:
```
print("dddddddddddddddddddddddddddd")
print("dddddddddddddddddddddddddddd")
```
|
instruction
| 0
| 13,342
| 0
| 26,684
|
No
|
output
| 1
| 13,342
| 0
| 26,685
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 200) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
|
instruction
| 0
| 13,411
| 0
| 26,822
|
Tags: constructive algorithms, dp, graphs, greedy, sortings
Correct Solution:
```
def main():
n = int(input())
ss = list(zip(input(), range(0,n)))
res = [1]*n
max_colour = 1
nextx = [0] * 26
colours_after = [[True]+[False]*26 for _ in range(26)]
for i in range(n):
ch = ord(ss[i][0]) - ord('a')
opos = ss[i][1]
nx = nextx[ch]
# ss[nx:i + 1] = [ss[i]] + ss[nx:i]
col = colours_after[ch].index(False)
if col > max_colour:
max_colour = col
res[opos] = col
for c in range(ch):
colours_after[c][col] = True
for c in range(ch, 26):
nextx[c] = min(nextx[c], nx+1)
print(max_colour)
print(' '.join(map(str, res)))
if __name__ == "__main__":
main()
```
|
output
| 1
| 13,411
| 0
| 26,823
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 200) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
|
instruction
| 0
| 13,412
| 0
| 26,824
|
Tags: constructive algorithms, dp, graphs, greedy, sortings
Correct Solution:
```
n = int(input())
s = input()
ans = [-1] * n
f = True
for i in range(n):
if ans[i] == -1:
ans[i] = 0
for j in range(i + 1, n):
if (s[j] < s[i]):
if ans[j] == -1:
ans[j] = (ans[i] - 1) ** 2
else:
now = ans[j]
new = (ans[i] - 1) ** 2
if new != now:
f = False
else:
ans[j] = new
if not f:
print("NO")
else:
print("YES")
for i in range(n):
print(ans[i], end='')
```
|
output
| 1
| 13,412
| 0
| 26,825
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 200) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
|
instruction
| 0
| 13,413
| 0
| 26,826
|
Tags: constructive algorithms, dp, graphs, greedy, sortings
Correct Solution:
```
n=int(input())
s=input()
last,last1='a','a'
ans=''
flag=1
for i in s:
if i<last and i<last1:
flag=0
break
elif i>=last:
last=i
ans+='0'
else:
last1=i
ans+='1'
if flag:
print('YES')
print(ans)
else:
print('NO')
```
|
output
| 1
| 13,413
| 0
| 26,827
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 200) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
|
instruction
| 0
| 13,414
| 0
| 26,828
|
Tags: constructive algorithms, dp, graphs, greedy, sortings
Correct Solution:
```
import sys
def main():
n = int(sys.stdin.readline().split()[0])
s = sys.stdin.readline().split()[0]
color = [None]*n
color[0] = 0
for i in range(n):
if color[i] == None:
color[i] = 0
for j in range(i+1, n):
if ord(s[j]) < ord(s[i]):
if color[j] == color[i]:
print("NO")
return
if color[j] == None:
color[j] = color[i]^1
print("YES")
print(*color, sep = "")
main()
```
|
output
| 1
| 13,414
| 0
| 26,829
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 200) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
|
instruction
| 0
| 13,415
| 0
| 26,830
|
Tags: constructive algorithms, dp, graphs, greedy, sortings
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
#threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
#sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**30, func=lambda a, b: min(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b:a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] > k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
#-----------------------------------------trie---------------------------------
class Node:
def __init__(self, data):
self.data = data
self.count=0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count+=1
if not val:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += 1
self.temp.data = pre_xor
def query(self, xor):
self.temp = self.root
for i in range(31, -1, -1):
val = xor & (1 << i)
if not val:
if self.temp.left and self.temp.left.count>0:
self.temp = self.temp.left
elif self.temp.right:
self.temp = self.temp.right
else:
if self.temp.right and self.temp.right.count>0:
self.temp = self.temp.right
elif self.temp.left:
self.temp = self.temp.left
self.temp.count-=1
return xor ^ self.temp.data
#-------------------------bin trie-------------------------------------------
n=int(input())
d=dict()
for i in range(97,124):
d[chr(i)]=0
l=input()
d[l[0]]+=1
ans=[0]*n
ans[0]=d[l[0]]
for i in range(1,n):
for j in sorted(d.keys(),reverse=True):
if j>l[i]:
ans[i]=max(ans[i],d[j]+1)
else:
break
d[l[i]]=ans[i]
print(max(ans))
print(*ans,sep=" ")
```
|
output
| 1
| 13,415
| 0
| 26,831
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 200) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
|
instruction
| 0
| 13,416
| 0
| 26,832
|
Tags: constructive algorithms, dp, graphs, greedy, sortings
Correct Solution:
```
#t = int(input())
t=1
while t!=0:
t-=1
n = int(input())
s = input()
l1, l2 = 'a' , 'a'
ans = ''
for i in range(n):
if s[i]>=l1:
l1 = s[i]
ans+='0'
elif s[i]>=l2:
l2 = s[i]
ans+='1'
else:
ans='NO'
break
if ans=='NO':
print(ans)
else:
print('YES')
print(ans)
```
|
output
| 1
| 13,416
| 0
| 26,833
|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an easy version of the problem. The actual problems are different, but the easy version is almost a subtask of the hard version. Note that the constraints and the output format are different.
You are given a string s consisting of n lowercase Latin letters.
You have to color all its characters one of the two colors (each character to exactly one color, the same letters can be colored the same or different colors, i.e. you can choose exactly one color for each index in s).
After coloring, you can swap any two neighboring characters of the string that are colored different colors. You can perform such an operation arbitrary (possibly, zero) number of times.
The goal is to make the string sorted, i.e. all characters should be in alphabetical order.
Your task is to say if it is possible to color the given string so that after coloring it can become sorted by some sequence of swaps. Note that you have to restore only coloring, not the sequence of swaps.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 200) — the length of s.
The second line of the input contains the string s consisting of exactly n lowercase Latin letters.
Output
If it is impossible to color the given string so that after coloring it can become sorted by some sequence of swaps, print "NO" (without quotes) in the first line.
Otherwise, print "YES" in the first line and any correct coloring in the second line (the coloring is the string consisting of n characters, the i-th character should be '0' if the i-th character is colored the first color and '1' otherwise).
Examples
Input
9
abacbecfd
Output
YES
001010101
Input
8
aaabbcbb
Output
YES
01011011
Input
7
abcdedc
Output
NO
Input
5
abcde
Output
YES
00000
|
instruction
| 0
| 13,417
| 0
| 26,834
|
Tags: constructive algorithms, dp, graphs, greedy, sortings
Correct Solution:
```
def main():
n = int(input())
s = list(input())
colors = [-1]*n
for i in range(len(s)):
if colors[i] == -1:
colors[i] = 0
for j in range(i+1,n):
if s[j] < s[i]:
if colors[j] == colors[i]:
print('NO')
return
if colors[j] == -1:
if colors[i] == 0:
colors[j] = 1
else:
colors[j] = 0
print('YES')
for i in colors:
print(i,end = '')
main()
```
|
output
| 1
| 13,417
| 0
| 26,835
|
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