message stringlengths 2 23.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has recently learned at school what a number's divisor is and decided to determine a string's divisor. Here is what he came up with.
String a is the divisor of string b if and only if there exists a positive integer x such that if we write out string a consecutively x times, we get string b. For example, string "abab" has two divisors β "ab" and "abab".
Now Vasya wants to write a program that calculates the number of common divisors of two strings. Please help him.
Input
The first input line contains a non-empty string s1.
The second input line contains a non-empty string s2.
Lengths of strings s1 and s2 are positive and do not exceed 105. The strings only consist of lowercase Latin letters.
Output
Print the number of common divisors of strings s1 and s2.
Examples
Input
abcdabcd
abcdabcdabcdabcd
Output
2
Input
aaa
aa
Output
1
Note
In first sample the common divisors are strings "abcd" and "abcdabcd".
In the second sample the common divisor is a single string "a". String "aa" isn't included in the answer as it isn't a divisor of string "aaa".
Submitted Solution:
```
def is_div(s1, s2):
s = s2
while len(s)<len(s1):
s += (s2)
if s==s1:
return True
else:
return False
a = input()
b = input()
for i in range(1,len(a)+1):
if len(a)%i != 0:
continue
if is_div(a,a[:i]):
root_a = a[:i]
break
for i in range(1,len(b)+1):
if len(b)%i != 0:
continue
if is_div(b,b[:i]):
root_b = b[:i]
break
if root_a != root_b:
print(0)
else:
p = len(root_a)
q = len(root_b)
la = len(a)
lb = len(b)
sieve = [0 for i in range(max(la,lb)+1)]
res = 0
for i in range(p, la+1, p):
sieve[i] += 1
for i in range(q, lb+1, q):
sieve[i] += 1
for i in range(1, min(la,lb)+1):
if sieve[i]>=2 and la%i==0 and lb%i==0:
res += 1
print(res)
``` | instruction | 0 | 46,546 | 0 | 93,092 |
Yes | output | 1 | 46,546 | 0 | 93,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has recently learned at school what a number's divisor is and decided to determine a string's divisor. Here is what he came up with.
String a is the divisor of string b if and only if there exists a positive integer x such that if we write out string a consecutively x times, we get string b. For example, string "abab" has two divisors β "ab" and "abab".
Now Vasya wants to write a program that calculates the number of common divisors of two strings. Please help him.
Input
The first input line contains a non-empty string s1.
The second input line contains a non-empty string s2.
Lengths of strings s1 and s2 are positive and do not exceed 105. The strings only consist of lowercase Latin letters.
Output
Print the number of common divisors of strings s1 and s2.
Examples
Input
abcdabcd
abcdabcdabcdabcd
Output
2
Input
aaa
aa
Output
1
Note
In first sample the common divisors are strings "abcd" and "abcdabcd".
In the second sample the common divisor is a single string "a". String "aa" isn't included in the answer as it isn't a divisor of string "aaa".
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**9
def factorize(n):
if n < 2:
return []
arr = []
def _add(n):
if len(arr) > 0 and arr[-1][0] == n:
arr[-1][1] += 1
else:
arr.append([n,1])
for i in [2, 3]:
while (n % i == 0):
_add(i)
n = n / i
i, add = 5, 2
while (n >= i * i):
while (n % i == 0):
_add(i)
n = n / i
i = i + add
add = 6 - add
if (n > 1):
_add(i)
return arr
def f():
a = input()
b = input()
if len(a) > len(b):
a,b = b,a
al = len(a)
bl = len(b)
if al > 4 and a[4] == 'u':
return [al,bl]
if a != b[:al]:
return 0
g = fractions.gcd(al, bl)
t = factorize(g)
tk = 1
for n,i in t:
tk *= i+1
r = 0
if len(set(b)) == 1:
r += 1
for n,i in t:
ri = 0
s = b
ni = n**i
while ni > 1:
si = s[:ni]
sf = False
for j in range(1,n):
if si != s[ni*j:ni*j+ni]:
sf = True
break
if sf:
break
ri += 1
ni //= n
s = si
if ri > 0:
tk //= i+1
r += ri * tk
return r
print(f())
``` | instruction | 0 | 46,547 | 0 | 93,094 |
No | output | 1 | 46,547 | 0 | 93,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has recently learned at school what a number's divisor is and decided to determine a string's divisor. Here is what he came up with.
String a is the divisor of string b if and only if there exists a positive integer x such that if we write out string a consecutively x times, we get string b. For example, string "abab" has two divisors β "ab" and "abab".
Now Vasya wants to write a program that calculates the number of common divisors of two strings. Please help him.
Input
The first input line contains a non-empty string s1.
The second input line contains a non-empty string s2.
Lengths of strings s1 and s2 are positive and do not exceed 105. The strings only consist of lowercase Latin letters.
Output
Print the number of common divisors of strings s1 and s2.
Examples
Input
abcdabcd
abcdabcdabcdabcd
Output
2
Input
aaa
aa
Output
1
Note
In first sample the common divisors are strings "abcd" and "abcdabcd".
In the second sample the common divisor is a single string "a". String "aa" isn't included in the answer as it isn't a divisor of string "aaa".
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**9
def factorize(n):
if n < 2:
return []
arr = []
def _add(n):
if len(arr) > 0 and arr[-1][0] == n:
arr[-1][1] += 1
else:
arr.append([n,1])
for i in [2, 3]:
while (n % i == 0):
_add(i)
n = n / i
i, add = 5, 2
while (n >= i * i):
while (n % i == 0):
_add(i)
n = n / i
i = i + add
add = 6 - add
if (n > 1):
_add(i)
return arr
def f():
a = input()
b = input()
if len(a) > len(b):
a,b = b,a
al = len(a)
bl = len(b)
if a != b[:al]:
return 0
g = fractions.gcd(al, bl)
if g == 1:
if len(set(b)) > 1:
return 0
else:
return 1
t = factorize(g)
tk = 1
for n,i in t:
tk *= i+1
r = 0
for n,i in t:
ri = 0
tk //= i+1
s = b
ni = n**i
while ni > 1:
si = s[:ni]
sf = False
for j in range(1,n):
if si != s[ni*j:ni*j+ni]:
sf = True
break
if sf:
break
ri += 1
ni //= n
s = si
r += ri * tk
return r
print(f())
``` | instruction | 0 | 46,548 | 0 | 93,096 |
No | output | 1 | 46,548 | 0 | 93,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has recently learned at school what a number's divisor is and decided to determine a string's divisor. Here is what he came up with.
String a is the divisor of string b if and only if there exists a positive integer x such that if we write out string a consecutively x times, we get string b. For example, string "abab" has two divisors β "ab" and "abab".
Now Vasya wants to write a program that calculates the number of common divisors of two strings. Please help him.
Input
The first input line contains a non-empty string s1.
The second input line contains a non-empty string s2.
Lengths of strings s1 and s2 are positive and do not exceed 105. The strings only consist of lowercase Latin letters.
Output
Print the number of common divisors of strings s1 and s2.
Examples
Input
abcdabcd
abcdabcdabcdabcd
Output
2
Input
aaa
aa
Output
1
Note
In first sample the common divisors are strings "abcd" and "abcdabcd".
In the second sample the common divisor is a single string "a". String "aa" isn't included in the answer as it isn't a divisor of string "aaa".
Submitted Solution:
```
import os,sys,math
from io import BytesIO, IOBase
from collections import defaultdict,deque,OrderedDict
import bisect as bi
def yes():print('YES')
def no():print('NO')
def I():return (int(input()))
def In():return(map(int,input().split()))
def ln():return list(map(int,input().split()))
def Sn():return input().strip()
BUFSIZE = 8192
#complete the main function with number of test cases to complete greater than x
def find_gt(a, x):
i = bi.bisect_left(a, x)
if i != len(a):
return i
else:
return len(a)
def printDivisors(n) :
# Note that this loop runs till square root
i = 1
ans=[]
while i <= math.sqrt(n):
if (n % i == 0) :
# If divisors are equal, print only one
if (n / i == i) :
ans.append(i)
else :
# Otherwise print both
ans.append(i)
ans.append(n//i)
i = i + 1
return(ans)
def solve():
a=Sn()
b=Sn()
n1,n2=len(a),len(b)
d=defaultdict(list)
for i in range(n1):
d[a[i]].append(i)
mx1=0
for x in d:
if len(d[x])==1:
mx1=n1
break
for i in range(len(d[x])-1):
mx1=max(mx1,d[x][i+1]-d[x][i])
d1=defaultdict(list)
for i in range(n2):
d1[b[i]].append(i)
mx2=0
for x in d1:
if len(d1[x])==1:
mx2=n2
break
for i in range(len(d1[x])-1):
mx2=max(mx2,d1[x][i+1]-d1[x][i])
if mx1!=mx2:
print(0)
return
for i in range(mx1):
if a[i]!=b[i]:
print(0)
return
cnt=0
# for i in range(1,min(n1,n2)+1):
# if n1%(i*mx1)==0 and len(b)%(i*mx1)==0:
# cnt+=1
# print(cnt)
if n1>n2:
n1,n2=n2,n1
p1,p2=printDivisors(n1),printDivisors(n2)
cn1,cn2=0,0
for x in p1:
if n2%x==0 and x%mx1==0:
cn1+=1
print(cn1)
pass
def main():
T=1
for i in range(T):
solve()
M = 998244353
P = 1000000007
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__":
main()
``` | instruction | 0 | 46,549 | 0 | 93,098 |
No | output | 1 | 46,549 | 0 | 93,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has recently learned at school what a number's divisor is and decided to determine a string's divisor. Here is what he came up with.
String a is the divisor of string b if and only if there exists a positive integer x such that if we write out string a consecutively x times, we get string b. For example, string "abab" has two divisors β "ab" and "abab".
Now Vasya wants to write a program that calculates the number of common divisors of two strings. Please help him.
Input
The first input line contains a non-empty string s1.
The second input line contains a non-empty string s2.
Lengths of strings s1 and s2 are positive and do not exceed 105. The strings only consist of lowercase Latin letters.
Output
Print the number of common divisors of strings s1 and s2.
Examples
Input
abcdabcd
abcdabcdabcdabcd
Output
2
Input
aaa
aa
Output
1
Note
In first sample the common divisors are strings "abcd" and "abcdabcd".
In the second sample the common divisor is a single string "a". String "aa" isn't included in the answer as it isn't a divisor of string "aaa".
Submitted Solution:
```
import os,sys,math
from io import BytesIO, IOBase
from collections import defaultdict,deque,OrderedDict
import bisect as bi
def yes():print('YES')
def no():print('NO')
def I():return (int(input()))
def In():return(map(int,input().split()))
def ln():return list(map(int,input().split()))
def Sn():return input().strip()
BUFSIZE = 8192
#complete the main function with number of test cases to complete greater than x
def find_gt(a, x):
i = bi.bisect_left(a, x)
if i != len(a):
return i
else:
return len(a)
def printDivisors(n) :
# Note that this loop runs till square root
i = 1
ans=[]
while i <= math.sqrt(n):
if (n % i == 0) :
# If divisors are equal, print only one
if (n / i == i) :
ans.append(i)
else :
# Otherwise print both
ans.append(i)
ans.append(n//i)
i = i + 1
return(ans)
def solve():
a=Sn()
b=Sn()
tp='wutmmwxxnjymqpcbarigedelmjnlcjgkdncjvpmzhaimacqm'
n1,n2=len(a),len(b)
ok=False
if n1>=len(tp) and a[:len(tp)]==tp:
ok=True
d=defaultdict(list)
for i in range(n1):
d[a[i]].append(i)
mx1=0
for x in d:
if len(d[x])==1:
mx1=n1
break
for i in range(len(d[x])-1):
mx1=max(mx1,d[x][i+1]-d[x][i])
d1=defaultdict(list)
for i in range(n2):
d1[b[i]].append(i)
mx2=0
for x in d1:
if len(d1[x])==1:
mx2=n2
break
for i in range(len(d1[x])-1):
mx2=max(mx2,d1[x][i+1]-d1[x][i])
if mx1!=mx2:
if ok:
print('break1')
print(0)
return
for i in range(mx1):
if a[i]!=b[i]:
if ok:
print('break1')
print(0)
return
cnt=0
# for i in range(1,min(n1,n2)+1):
# if n1%(i*mx1)==0 and len(b)%(i*mx1)==0:
# cnt+=1
# print(cnt)
if n1>n2:
n1,n2=n2,n1
p1,p2=printDivisors(n1),printDivisors(n2)
cn1,cn2=0,0
for x in p1:
if n2%x==0 and x%mx1==0:
cn1+=1
print(cn1)
if ok:
print('break3',n1,n2,mx1,mx2)
pass
def main():
T=1
for i in range(T):
solve()
M = 998244353
P = 1000000007
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__":
main()
``` | instruction | 0 | 46,550 | 0 | 93,100 |
No | output | 1 | 46,550 | 0 | 93,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Santa Claus likes palindromes very much. There was his birthday recently. k of his friends came to him to congratulate him, and each of them presented to him a string si having the same length n. We denote the beauty of the i-th string by ai. It can happen that ai is negative β that means that Santa doesn't find this string beautiful at all.
Santa Claus is crazy about palindromes. He is thinking about the following question: what is the maximum possible total beauty of a palindrome which can be obtained by concatenating some (possibly all) of the strings he has? Each present can be used at most once. Note that all strings have the same length n.
Recall that a palindrome is a string that doesn't change after one reverses it.
Since the empty string is a palindrome too, the answer can't be negative. Even if all ai's are negative, Santa can obtain the empty string.
Input
The first line contains two positive integers k and n divided by space and denoting the number of Santa friends and the length of every string they've presented, respectively (1 β€ k, n β€ 100 000; nΒ·k β€ 100 000).
k lines follow. The i-th of them contains the string si and its beauty ai ( - 10 000 β€ ai β€ 10 000). The string consists of n lowercase English letters, and its beauty is integer. Some of strings may coincide. Also, equal strings can have different beauties.
Output
In the only line print the required maximum possible beauty.
Examples
Input
7 3
abb 2
aaa -3
bba -1
zyz -4
abb 5
aaa 7
xyx 4
Output
12
Input
3 1
a 1
a 2
a 3
Output
6
Input
2 5
abcde 10000
abcde 10000
Output
0
Note
In the first example Santa can obtain abbaaaxyxaaabba by concatenating strings 5, 2, 7, 6 and 3 (in this order).
Submitted Solution:
```
from sys import stdin
n, k = map(int, stdin.readline().split())
strings = dict()
for i in range(n):
string, beauty = stdin.readline().split()
beauty = int(beauty)
if string not in strings:
strings[string] = []
strings[string].append(beauty)
for string in strings:
strings[string].sort(reverse=True)
used = set()
best = 0
mid = 0
for string in strings:
if string not in used:
pal = string[::-1]
used.add(pal)
if string == pal:
i = 0
while i < len(strings[pal]) and strings[pal][i] > 0:
i += 1
if i % 2 == 1:
best += sum(strings[pal][:i - 1])
if mid < strings[pal][i - 1]:
mid = strings[pal][i - 1]
else:
best += sum(strings[pal][:i])
elif pal in strings:
i = 0
while i < min(len(strings[pal]), len(strings[string])) and strings[pal][i] + strings[string][i] > 0:
best += strings[pal][i] + strings[string][i]
i += 1
print(best + mid)
``` | instruction | 0 | 46,744 | 0 | 93,488 |
No | output | 1 | 46,744 | 0 | 93,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Santa Claus likes palindromes very much. There was his birthday recently. k of his friends came to him to congratulate him, and each of them presented to him a string si having the same length n. We denote the beauty of the i-th string by ai. It can happen that ai is negative β that means that Santa doesn't find this string beautiful at all.
Santa Claus is crazy about palindromes. He is thinking about the following question: what is the maximum possible total beauty of a palindrome which can be obtained by concatenating some (possibly all) of the strings he has? Each present can be used at most once. Note that all strings have the same length n.
Recall that a palindrome is a string that doesn't change after one reverses it.
Since the empty string is a palindrome too, the answer can't be negative. Even if all ai's are negative, Santa can obtain the empty string.
Input
The first line contains two positive integers k and n divided by space and denoting the number of Santa friends and the length of every string they've presented, respectively (1 β€ k, n β€ 100 000; nΒ·k β€ 100 000).
k lines follow. The i-th of them contains the string si and its beauty ai ( - 10 000 β€ ai β€ 10 000). The string consists of n lowercase English letters, and its beauty is integer. Some of strings may coincide. Also, equal strings can have different beauties.
Output
In the only line print the required maximum possible beauty.
Examples
Input
7 3
abb 2
aaa -3
bba -1
zyz -4
abb 5
aaa 7
xyx 4
Output
12
Input
3 1
a 1
a 2
a 3
Output
6
Input
2 5
abcde 10000
abcde 10000
Output
0
Note
In the first example Santa can obtain abbaaaxyxaaabba by concatenating strings 5, 2, 7, 6 and 3 (in this order).
Submitted Solution:
```
from sys import stdin
n, k = map(int, stdin.readline().split())
strings = dict()
for i in range(n):
string, beauty = stdin.readline().split()
beauty = int(beauty)
if string not in strings:
strings[string] = []
strings[string].append(beauty)
for string in strings:
strings[string].sort(reverse=True)
used = set()
best = 0
mid = 0
for string in strings:
if string not in used:
pal = string[::-1]
used.add(pal)
if string == pal:
i = 0
while i + 1 < len(strings[pal]) and strings[pal][i] + strings[pal][i + 1] > 0:
best += strings[pal][i] + strings[pal][i + 1]
i += 2
if i < len(strings[pal]) and strings[pal][i] > mid:
mid = strings[pal][i]
elif pal in strings:
i = 0
while i < min(len(strings[pal]), len(strings[string])) and strings[pal][i] + strings[string][i] > 0:
best += strings[pal][i] + strings[string][i]
i += 1
print(best + mid)
``` | instruction | 0 | 46,745 | 0 | 93,490 |
No | output | 1 | 46,745 | 0 | 93,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Santa Claus likes palindromes very much. There was his birthday recently. k of his friends came to him to congratulate him, and each of them presented to him a string si having the same length n. We denote the beauty of the i-th string by ai. It can happen that ai is negative β that means that Santa doesn't find this string beautiful at all.
Santa Claus is crazy about palindromes. He is thinking about the following question: what is the maximum possible total beauty of a palindrome which can be obtained by concatenating some (possibly all) of the strings he has? Each present can be used at most once. Note that all strings have the same length n.
Recall that a palindrome is a string that doesn't change after one reverses it.
Since the empty string is a palindrome too, the answer can't be negative. Even if all ai's are negative, Santa can obtain the empty string.
Input
The first line contains two positive integers k and n divided by space and denoting the number of Santa friends and the length of every string they've presented, respectively (1 β€ k, n β€ 100 000; nΒ·k β€ 100 000).
k lines follow. The i-th of them contains the string si and its beauty ai ( - 10 000 β€ ai β€ 10 000). The string consists of n lowercase English letters, and its beauty is integer. Some of strings may coincide. Also, equal strings can have different beauties.
Output
In the only line print the required maximum possible beauty.
Examples
Input
7 3
abb 2
aaa -3
bba -1
zyz -4
abb 5
aaa 7
xyx 4
Output
12
Input
3 1
a 1
a 2
a 3
Output
6
Input
2 5
abcde 10000
abcde 10000
Output
0
Note
In the first example Santa can obtain abbaaaxyxaaabba by concatenating strings 5, 2, 7, 6 and 3 (in this order).
Submitted Solution:
```
from sys import stdin
n, k = map(int, stdin.readline().split())
strings = dict()
for i in range(n):
string, beauty = stdin.readline().split()
beauty = int(beauty)
if string not in strings:
strings[string] = []
strings[string].append(beauty)
for string in strings:
strings[string].sort(reverse=True)
used = set()
best = 0
mid = 0
for string in strings:
if string not in used:
pal = string[::-1]
used.add(pal)
if string == pal:
i = 0
while i + 1 < len(strings[pal]) and strings[pal][i] + strings[pal][i + 1] > 0:
best += strings[pal][i] + strings[pal][i + 1]
i += 2
if i < len(strings[pal]) and strings[pal][i] > mid:
mid = strings[pal][i]
continue
if pal in strings:
i = 0
while i < min(len(strings[pal]), len(strings[string])) and strings[pal][i] + strings[string][i] > 0:
best += strings[pal][i] + strings[string][i]
i += 1
print(best + mid)
``` | instruction | 0 | 46,747 | 0 | 93,494 |
No | output | 1 | 46,747 | 0 | 93,495 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3 | instruction | 0 | 46,776 | 0 | 93,552 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
s=input()
s1=set()
for i in s:
s1.add(i)
l1=list(s1)
an=len(s)
for i in l1:
st=-1
mx=0
for j in range(len(s)):
if s[j]==i:
mx=max(mx,j-st)
st=j
mx=max(mx,len(s)-st)
an=min(an,mx)
print(an)
``` | output | 1 | 46,776 | 0 | 93,553 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3 | instruction | 0 | 46,777 | 0 | 93,554 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
import sys,math
s=input()
n=len(s)
a=[[0] for i in range(26)]
for i in range(n):
a[ord(s[i])-97].append(i+1)
for i in range(26):
a[i].append(n+1)
ans=n+1
for i in range(26):
cur=0
for j in range(1,len(a[i])):
cur=max(cur,a[i][j]-a[i][j-1])
ans=min(ans,cur)
print(ans)
``` | output | 1 | 46,777 | 0 | 93,555 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3 | instruction | 0 | 46,778 | 0 | 93,556 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
from math import ceil
s=input()
def isgood(k):
d={}
q=set(s[:k])
for x in range(26):
d[chr(x+ord('a'))]=0
for x in range(k):
d[s[x]]+=1
for x in range(k,len(s)):
d[s[x-k]]-=1
d[s[x]]+=1
tmp=q.copy()
for y in q:
if d[y]==0:
tmp.remove(y)
q=tmp
if len(q)==0:
break
return len(q)>0
l=0
r=len(s)
while r-l>1:
mid=ceil((r+l)/2)
if isgood(mid):
r=mid
else:
l=mid
print(r)
``` | output | 1 | 46,778 | 0 | 93,557 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3 | instruction | 0 | 46,779 | 0 | 93,558 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
"""
Author - Satwik Tiwari .
19th Jan , 2021 - Tuesday
"""
#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
# from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt,log2
from copy import deepcopy
from collections import deque
from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect
#==============================================================================================
#fast I/O 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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def testcase(t):
for pp in range(t):
solve(pp)
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
y%=(p-1) #not so sure about this. used when y>p-1. if p is prime.
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
inf = pow(10,20)
mod = 10**9+7
#===============================================================================================
# code here ;))
def solve(case):
a = list(inp());n = len(a)
ans = n
for i in range(97,97+26):
curr = chr(i)
prev = -1
k = -1
for j in range(n):
if(j == n-1):
k = max(k,j-prev if a[j] == curr else j-prev+1)
else:
if(a[j] == curr):
k = max(k,j-prev)
prev = j
ans = min(ans,k)
print(ans)
testcase(1)
# testcase(int(inp()))
``` | output | 1 | 46,779 | 0 | 93,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3 | instruction | 0 | 46,780 | 0 | 93,560 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
alphavite = 'abcdefghijklmnopqrstyvwxyz'
s=input()
minimal= len(s)+1
for i in alphavite:
l= -1
maxl=0
for j in range (len(s)):
if s[j]== i:
maxl=max(maxl,j-l)
l=j
maxl=max(maxl,len(s)-l)
minimal = min(minimal,maxl)
print(minimal)
``` | output | 1 | 46,780 | 0 | 93,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3 | instruction | 0 | 46,781 | 0 | 93,562 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
s=input()
c=float('Inf')
for chr in set(s):
l=0
for j in s.split(chr):
l=max(l,len(j))
c=min(l,c)
print(c+1)
``` | output | 1 | 46,781 | 0 | 93,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3 | instruction | 0 | 46,782 | 0 | 93,564 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
s = input().strip()
memomin = dict()
memolast = dict()
for i,e in enumerate(s):
if e not in memomin:
memomin[e] = i+1
memolast[e] = i
else:
memomin[e] = max(memomin[e], i - memolast[e])
memolast[e] = i
for e in memomin:
memomin[e] = max(memomin[e], len(s) - memolast[e])
print(min(memomin.values()))
``` | output | 1 | 46,782 | 0 | 93,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3 | instruction | 0 | 46,783 | 0 | 93,566 |
Tags: binary search, implementation, two pointers
Correct Solution:
```
import sys
import math
def check(L):
vis = [True for i in range(26)]
cnt = [0 for i in range(26)]
for i in range(L):
val = s[i]
cnt[val] += 1
for i in range(len(cnt)):
if not cnt[i]:
vis[i] = False
for i in range(L, len(s)):
val = s[i]
cnt[val] += 1
val = s[i-L]
cnt[val] -= 1
if not cnt[val]:
vis[val] = False
for i in vis:
if i:
return True
return False
s = list(map(lambda x: ord(x)-ord('a'), input()))
l = 0
r = len(s)
while r > l:
mid = (l + r) >> 1
if check(mid):
r = mid
else:
l = mid+1
print(l)
``` | output | 1 | 46,783 | 0 | 93,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3
Submitted Solution:
```
s = input()
chars = list(set(s))
print(min(max(map(len, s.split(c))) + 1 for c in chars))
``` | instruction | 0 | 46,784 | 0 | 93,568 |
Yes | output | 1 | 46,784 | 0 | 93,569 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3
Submitted Solution:
```
MOD = 1000000007
ii = lambda : int(input())
si = lambda : input()
dgl = lambda : list(map(int, input()))
f = lambda : map(int, input().split())
il = lambda : list(map(int, input().split()))
ls = lambda : list(input())
s=si()
n=len(s)
mn=n+10
for i in 'abcdefghijklmnopqrstuvwxyz':
pi=0
mx=0
for j in range(n):
if s[j]==i:
mx=max(mx,j-pi+1)
pi=j+1
mx=max(mx,n-pi+1)
mn=min(mn,mx)
print(mn)
``` | instruction | 0 | 46,785 | 0 | 93,570 |
Yes | output | 1 | 46,785 | 0 | 93,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3
Submitted Solution:
```
word = input()
ans = []
for c in set(list(word)):
maior_dist = 0
atual_dist = 0
last_pos = 0
for i, let in enumerate(word):
if let == c:
atual_dist += 1
if atual_dist > maior_dist: maior_dist = atual_dist
atual_dist = 0
last_pos = i
else:
atual_dist += 1
if atual_dist > maior_dist: maior_dist = atual_dist
ans.append(max(maior_dist, len(word) - last_pos))
print(min(ans))
``` | instruction | 0 | 46,786 | 0 | 93,572 |
Yes | output | 1 | 46,786 | 0 | 93,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3
Submitted Solution:
```
s = str(input())
n = len(s)
ans = 10**18
for i in range(26):
c = chr(i+ord('a'))
temp = -1
p = []
for j in range(n):
if s[j] == c:
p.append(j-temp)
temp = j
else:
p.append(n-temp)
#print(p)
if p:
ans = min(ans, max(p))
print(ans)
``` | instruction | 0 | 46,787 | 0 | 93,574 |
Yes | output | 1 | 46,787 | 0 | 93,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3
Submitted Solution:
```
s=input()
se=set(s)
if len(se)==len(s):
print((len(s)//2)+1)
elif len(se)==1:
print(1)
else:
dif={}
d={}
for index,i in enumerate(s):
if d.get(i,[])!=[]:
if len(d[i])>=2:
dif[i]=max(index-d[i][-1],dif[i])
else:
dif[i] =index - d[i][-1]
d[i].append(index)
else:
d[i]=[]
d[i].append(index)
dif[i]=index+1
need=sorted(d.keys(),key=lambda k:dif[k],reverse=False)
req=True
num = (len(s) // 2) + 1
for i in need:
if len(d[i])>=(len(s)//dif[i]) and dif[i]<=num:
print(dif[i])
req=False
break
if req:
print(num)
``` | instruction | 0 | 46,788 | 0 | 93,576 |
No | output | 1 | 46,788 | 0 | 93,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3
Submitted Solution:
```
from collections import defaultdict
s=input()
d={}
d=defaultdict(lambda:0,d)
d1={}
d1=defaultdict(lambda:-1,d1)
k=1
for i in s:
if d[i]==0:
d[i]=k
else:
d1[i]=max(k-d[i],d1[i])
d[i]=k
k+=1
#print(d)
#print(d1)
ans=[]
for i in d1.keys():
ans.append(d1[i])
if len(ans)==0:
print(len(s)//2+1)
else:
print(min(ans))
``` | instruction | 0 | 46,789 | 0 | 93,578 |
No | output | 1 | 46,789 | 0 | 93,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3
Submitted Solution:
```
a = input()
k = 999999
o = k
m = 1
err = 0
for i in range(len(a)):
for j in range(i+1,len(a)):
if a[i] == a[j]:
err = 1
break
if err == 1:
for i in range(len(a)):
t = a[i]
j = i + 1
while (j < len(a)) and (a[j] != t):
m += 1
j += 1
if (m < k):
if (j <= len(a)):
if (a[j] == a[i]):
k = m
#Π±Π΅ΡΠ΅ΠΌ a[i] ΡΠΈΠΌΠ²ΠΎΠ» ΠΈ ΠΈΡΠ΅ΠΌ Π΄Π»Ρ Π½Π΅Π³ΠΎ ΠΌΠ°ΠΊΡ Π΄Π»ΠΈΠ½Ρ ΠΏΠΎΠ΄ΡΡΡΠΎΠΊΠΈ
if err == 0:
print(len(a)//2 + 1)
else:
print(k)
``` | instruction | 0 | 46,790 | 0 | 93,580 |
No | output | 1 | 46,790 | 0 | 93,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.
You have to find minimum k such that there exists at least one k-dominant character.
Input
The first line contains string s consisting of lowercase Latin letters (1 β€ |s| β€ 100000).
Output
Print one number β the minimum value of k such that there exists at least one k-dominant character.
Examples
Input
abacaba
Output
2
Input
zzzzz
Output
1
Input
abcde
Output
3
Submitted Solution:
```
s=input()
t=set(s)
t=list(t)
def verif(c,s,k):
compt=0
for i in range(len(s)):
compt+=1
if (c==s[i]):
compt=0
if(compt==k):
return False
return True
v=[]
right=len(s)
left=0
while(right-left>1):
mid=(right+left)//2
if(verif(t[0],s,mid)):
right=mid
else:
left=mid
v.append(right)
for i in range(1,len(t)):
if(verif(s[i],s,min(v)-1)):
left=0
right=len(s)
while(right-left>1):
mid=(right+left)//2
if(verif(t[i],s,mid)):
right=mid
else:
left=mid
v.append(right)
print(min(v))
``` | instruction | 0 | 46,791 | 0 | 93,582 |
No | output | 1 | 46,791 | 0 | 93,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring. | instruction | 0 | 47,173 | 0 | 94,346 |
Tags: brute force, implementation, strings
Correct Solution:
```
#!/usr/bin/env python3
target = "abacaba"
def count(s):
c = 0
for i in range(len(s) - len(target) + 1):
c += (s[i:(i + len(target))] == target)
return c
def main():
t = int(input())
for _ in range(t):
n = int(input())
s = input()
c = count(s)
if c == 0:
for i in range(len(s) - len(target) + 1):
if all(i == "?" or i == j for (i, j) in zip(s[i:(i + len(target))], target)):
temp = s[:i] + target + s[i + len(target):]
if count(temp) == 1:
print("YES")
print(temp.replace("?", "z"))
break
else:
print("NO")
elif c == 1:
print("YES")
print(s.replace("?", "z"))
else:
print("NO")
if __name__ == "__main__":
main()
``` | output | 1 | 47,173 | 0 | 94,347 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring. | instruction | 0 | 47,174 | 0 | 94,348 |
Tags: brute force, implementation, strings
Correct Solution:
```
import math
import collections
def check(n,s):
cnt = 0
for i in range(len(s)):
if "".join(s[i:i+7]) == 'abacaba':
cnt+=1
return cnt
t = int(input())
for i in range(t):
n = int(input())
s = input()
T = 'abacaba'
f = False
for i in range(n-7+1):
ss = [k for k in s]
ok = True
for j in range(7):
if ss[i+j]!="?" and ss[i+j]!=T[j]:
ok = False
break
ss[i+j] = T[j]
if ok and check(n,ss) == 1:
for j in range(n):
if ss[j] == "?":
ss[j] = "d"
f = True
print("Yes")
print("".join(ss))
break
if not f:
print("NO")
``` | output | 1 | 47,174 | 0 | 94,349 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring. | instruction | 0 | 47,175 | 0 | 94,350 |
Tags: brute force, implementation, strings
Correct Solution:
```
import re
t='abacaba'
for s in[*open(0)][2::2]:
i=0;f=1
while 1:
q,f=re.subn(''.join(rf'({x}|\?)'for x in t),t,s[i:],1);q=s[:i]+q;i+=1
if f<1or q.find(t,q.find(t)+1)<0:break
print(*(['NO'],['YES',q.replace('?','d')])[f])
``` | output | 1 | 47,175 | 0 | 94,351 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring. | instruction | 0 | 47,176 | 0 | 94,352 |
Tags: brute force, implementation, strings
Correct Solution:
```
t = "abacaba"
def count(s):
cnt = 0
s = "".join(map(str,s))
for i in range(len(s)):
if s[i:i+len(t)] == t:
cnt += 1
return cnt
def solve(n,s):
ss = []
for i in range(n):
ss.append(s[i])
temp = ss
flag = False
okay = True
for i in range(n-len(t)+1):
ss = temp.copy()
okay = True
for j in range(0,len(t)):
if ss[i+j] != "?" and ss[i+j] != t[j]:
okay = False
break
ss[i+j] = t[j]
if okay and count(ss) == 1:
for j in range(0,n):
if ss[j] == "?":
ss[j] = "z"
flag = True
print("YES")
print("".join(map(str,ss)))
break
if flag == False:
print("NO")
return
if __name__ == '__main__':
q = int(input())
for _ in range(q):
n = int(input())
s = input()
solve(n,s)
``` | output | 1 | 47,176 | 0 | 94,353 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring. | instruction | 0 | 47,177 | 0 | 94,354 |
Tags: brute force, implementation, strings
Correct Solution:
```
import sys
import math
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log
from collections import defaultdict as dd
from bisect import bisect_left as bl, bisect_right as br
from collections import Counter
#sys.setrecursionlimit(100000000)
inp = lambda: int(input())
strng = lambda: input().strip()
jn = lambda x, l: x.join(map(str, l))
strl = lambda: list(input().strip())
mul = lambda: map(int, input().strip().split())
mulf = lambda: map(float, input().strip().split())
seq = lambda: list(map(int, input().strip().split()))
ceil = lambda x: int(x) if (x == int(x)) else int(x) + 1
ceildiv = lambda x, d: x // d if (x % d == 0) else x // d + 1
flush = lambda: stdout.flush()
stdstr = lambda: stdin.readline()
stdint = lambda: int(stdin.readline())
stdpr = lambda x: stdout.write(str(x))
stdarr = lambda: map(int, stdstr().split())
mod = 1000000007
checkstr = 'abacaba'
def my_count(string, substring):
string_size = len(string)
substring_size = len(substring)
count = 0
for i in range(0,string_size-substring_size+1):
if string[i:i+substring_size] == substring:
count+=1
return count
for _ in range(stdint()):
n = stdint()
s = input()
l = list(s)
# print(checkstr)
if checkstr in s:
c = my_count(s,checkstr)
if c==1:
print('Yes')
if '?' in s:
print(s.replace('?','d'))
else:
print(s)
else:
print('No')
else:
flag = 0
for i in range(n-6):
c = []
for j in range(7):
if l[i+j]=='?':
c.append(checkstr[j])
else:
c.append(l[i+j])
x=''
# print(c)
if x.join(c)==checkstr:
c = l[:i+j-6]+c+l[i+j+1:]
# print(x)
x =x.join(c)
count = my_count(x,checkstr)
if count==1:
flag = 1
print('Yes')
if '?' in x:
print(x.replace('?','d'))
else:
print(x)
break
else:
x.replace('?','d')
if flag == 0:
print('No')
``` | output | 1 | 47,177 | 0 | 94,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring. | instruction | 0 | 47,178 | 0 | 94,356 |
Tags: brute force, implementation, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
want = 'abacaba'
T = int(input())
for _ in range(T):
n = int(input())
s = input().strip()
l = list(s)
for i in range(n - 6):
ll = l[:]
works = True
for c in range(7):
if ll[c + i] == want[c] or ll[c+i] == '?':
ll[c + i] = want[c]
else:
works = False
if works:
for j in range(n - 6):
if i != j and ll[j:j+7] == list(want):
break
else:
print('Yes')
for j in range(n):
if ll[j] == '?':
ll[j] = 'd'
print(''.join(ll))
break
else:
print('No')
``` | output | 1 | 47,178 | 0 | 94,357 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring. | instruction | 0 | 47,179 | 0 | 94,358 |
Tags: brute force, implementation, strings
Correct Solution:
```
import re
cases = int(input())
output = ''
for t in range(cases):
n = int(input())
s = input()
# if t == 178:
# print(n,s)
f1 = s.find('abacaba')
f2 = s.find('abacaba',f1+1)
if f1>-1 and f2>-1:
output += 'No\n'
else:
if f1>-1:
if '?' in s:
s = s.replace('?','m')
output += 'Yes\n'+s+'\n'
else:
rp = "(a|\?)(b|\?)(a|\?)(c|\?)(a|\?)(b|\?)(a|\?)"
f = 0
for i in range(n-6):
s1 = s[i:i+7]
v = re.search(rp,s1)
if v:
vf = v
start = i
end = i+7
s2 = s
s2 = s2.replace('?', 'm')
s2 = s2[:start] + 'abacaba' + s2[end:]
f4 = s2.find('abacaba')
f3 = s2.find('abacaba', f4 + 1)
if f4 > -1 and f3 > -1:
continue
else:
output += 'Yes\n' + s2 + '\n'
f = 1
break
if f == 0:
output += 'No\n'
print(output)
``` | output | 1 | 47,179 | 0 | 94,359 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring. | instruction | 0 | 47,180 | 0 | 94,360 |
Tags: brute force, implementation, strings
Correct Solution:
```
def solve():
target = "abacaba"
target_with_z = "abazaba"
n = int(input())
s = input()
for i in range(n - 6):
possible = True
for j in range(7):
if not (s[i + j] == "?" or s[i + j] == target[j]):
possible = False
if possible and (s[:i] + target_with_z + s[i + 7:]).find(target) == -1:
print("Yes")
print(s[:i].replace("?", "z") + target + s[i + 7:].replace("?", "z"))
return
print("No")
T = int(input())
for i in range(T):
solve()
``` | output | 1 | 47,180 | 0 | 94,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
t = list('abacaba')
s = list(input())
c = 0
for i in range(n):
if i +7 <= n:
check = s[i:i+7]
if check == t:
c += 1
if c > 1:
print('NO')
continue
if c != 1:
flag2 = 0
for i in range(n):
if i+7 <= n:
ss = s[:]
flag = 1
for j in range(i,i+7):
#print(ss)
if ss[j] == t[j-i]:
pass
elif ss[j] == '?':
pass
else:
flag = 0
break
if flag == 1:
for j in range(i,i+7):
ss[j] = t[j-i]
#print(ss)
for j in range(i+7,n):
if ss[j] == '?':
ss[j] = 'z'
c = 0
#print(ss)
for i in range(n):
if i +7 <= n:
check = ss[i:i+7]
if check == t:
c += 1
#print(c)
if c == 1:
s = ss
flag2 = 1
#print(s)
break
if flag2:
break
#break
#print(s,i)
for i in range(n):
if s[i] == '?':
s[i] = 'z'
c = 0
#print(s)
#print(i)
for i in range(n):
if i +7 <= n:
check = s[i:i+7]
#print(check,t)
if check == t:
c += 1
#print(c,i)
if c == 1:
print('YES')
print(''.join(s))
else:
print('NO')
``` | instruction | 0 | 47,181 | 0 | 94,362 |
Yes | output | 1 | 47,181 | 0 | 94,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring.
Submitted Solution:
```
t = int(input())
def cout(t,sub):
c = 0
for i in range(0,len(t)-6):
if t[i:i+7] == sub:
c += 1
if c > 1:
return True
return False
for i in range(t):
n = int(input())
p = input().strip(' ')
sub = 'abacaba'
if sub in p:
if cout(p,sub):
print('No')
continue
print('Yes')
pnew = p.replace('?','z')
print(pnew)
continue
else:
for i in range(0,n-6):
for j in range(i,i+7):
if not((p[j] == sub[j-i]) or (p[j] == '?')):
break
else:
pnew = p[:i] + sub + p[i+7:]
pnew = pnew.replace('?','z')
if cout(pnew,sub):
continue
print('Yes')
print(pnew)
break
else:
print('No')
``` | instruction | 0 | 47,182 | 0 | 94,364 |
Yes | output | 1 | 47,182 | 0 | 94,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring.
Submitted Solution:
```
def get_Z_arr(s):
z = [0 for _ in range(len(s))]
l = r = 0
for i in range(1, len(s)):
if i <= r:
z[i] = min(z[i - l], r - i + 1)
while z[i] + i < len(s) and s[z[i] + i] == s[z[i]]:
z[i] += 1
if r < i + z[i] - 1:
l, r = i, i + z[i] - 1
return z
if __name__ == '__main__':
t = int(input())
for _ in range(t):
n = int(input())
s = input()
t = 'abacaba'
z1 = get_Z_arr("abacaba#" + s)
counter = sum([1 for x in z1 if x >= 7])
if counter >= 1 or n < 7:
if counter == 1:
print("Yes")
print("".join([(s[k] if s[k] != '?' else 'z') for k in range(n)]))
else:
print("No")
else:
for i in range(n - 6):
j = 0
stack = set()
flag = True
while j < 7 and (s[i + j] == '?' or s[i + j] == t[j]):
# if s[i + j] == '?':
if (j == 0 or j == 2) and i + j >= 6:
flag &= (t != s[i + j - 6: i + j] + t[6])
elif (j == 4 or j == 6) and i + j + 6 < len(s):
flag &= (t != t[0] + s[i + j + 1: i + j + 7])
elif j == 1 and i + j >= 5:
flag &= (t != s[i + j - 5: i + j - 1] + t[4:])
elif j == 5 and i + j + 5 < len(s):
flag &= (t != t[:3] + s[i + j + 2: i + j + 6])
if not flag:
break
j += 1
if j == 7 and flag:
print("Yes")
print("".join([t[k - i] if i <= k < i + j else (s[k] if s[k] != '?' else 'z') for k in range(n)]))
break
else:
print('No')
``` | instruction | 0 | 47,183 | 0 | 94,366 |
Yes | output | 1 | 47,183 | 0 | 94,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring.
Submitted Solution:
```
t="abacaba"
def match(s1,s):
for i in range(len(s1)):
if s1[i]!=t[i] and s1[i]!='?':
return False
return True
def solve():
if s.find(t)!=s.rfind(t):
return False
if s.find(t)>=0:
return s.replace('?','z')
else:
for i in range(n-6):
if match(s[i:i+7],s):
k=s[:i]+t+s[i+7:]
if k.find(t)==k.rfind(t):
return k.replace("?","z")
return False
for _ in range(int(input())):
n=int(input())
s=input()
ans=solve()
if ans:
print("Yes")
print(ans)
else:
print("No")
``` | instruction | 0 | 47,184 | 0 | 94,368 |
Yes | output | 1 | 47,184 | 0 | 94,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring.
Submitted Solution:
```
def my_count(string, substring):
string_size = len(string)
substring_size = len(substring)
count = 0
for i in range(0,string_size-substring_size+1):
if string[i:i+substring_size] == substring:
count+=1
return count
x=int(input())
for j in range(x):
y=0
a=int(input())
s=input()
d=my_count(s,'abacaba')
#print(d)
if d==1:
s=s.replace('?','d')
print('Yes')
print(s)
elif d>1 or s.count('?')==0:
print('No')
else:
a=s.count('?')
for i in range(a):
if my_count(s,'abacaba')==1:
s=s.replace('?','d')
print('Yes')
print(s)
y=1
break
n=s.index('?')
if n==0:
s=s.replace('?','a',1)
else:
if s[n-1]=='b' or s[n-1]=='c':
s=s.replace('?','a',1)
elif s[n-2:n]=='ba':
s=s.replace('?','c',1)
else:
s=s.replace('?','b',1)
#print("s= ",s)
if my_count(s,'abacaba')!=1:
print('No')
elif y!=1:
print('Yes')
print(s)
``` | instruction | 0 | 47,185 | 0 | 94,370 |
No | output | 1 | 47,185 | 0 | 94,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring.
Submitted Solution:
```
def helper(s):
ans=0
for i in range(len(s)-6):
temp=s[i:i+7]
if temp=="abacaba":
ans+=1
return ans
t=int(input())
for i in range(t):
n=int(input())
s=input()
if helper(s)>1:
print("NO")
elif helper(s)==1:
temp=""
for i in range(len(s)):
if s[i]=="?":
temp+="d"
else:
temp+=s[i]
print("YES")
print(temp)
else:
rec=0
t="abacaba"
for i in range(len(s)-6):
ans=1
temp=s[i:i+7]
for q in range(len(temp)):
if temp[q]!=t[q] and temp[q]!="?":
ans=0
if ans==1:
r=""
for p in range(i):
if s[p]=="?":
r+="d"
else:
r+=s[p]
r+="abacaba"
for p in range(i+7,len(s)):
r+=s[p]
if helper(r)==1:
print("YES")
print(r)
rec=1
break
if rec==0:
print("NO")
``` | instruction | 0 | 47,186 | 0 | 94,372 |
No | output | 1 | 47,186 | 0 | 94,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# 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")
# ------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
def toord(c): return ord(c)-ord('a')
mod = 998244353
INF = float('inf')
from math import factorial, sqrt, ceil, floor
from collections import Counter, defaultdict, deque
from heapq import heapify, heappop, heappush
# ------------------------------
# f = open('./input.txt')
# sys.stdin = f
def main():
for _ in range(N()):
n = N()
s = input()
cp = 'abacaba'
num = 0
for i in range(n-6):
if s[i: i+7]==cp:
num+=1
if num>1:
print('No')
else:
if num==1:
print('Yes')
print(s.replace('?', 'z'))
else:
tag = 0
for i in range(n-6):
for j in range(7):
if s[i+j]=='?' or s[i+j]==cp[j]:
continue
else:
break
else:
tag = 1
s = s[:i] + cp + s[i+8:]
break
if tag==0:
print("No")
else:
print("Yes")
print(s.replace('?', 'z'))
if __name__ == "__main__":
main()
``` | instruction | 0 | 47,187 | 0 | 94,374 |
No | output | 1 | 47,187 | 0 | 94,375 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acacius is studying strings theory. Today he came with the following problem.
You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?
Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".
Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 β€ i β€ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.
Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.
Input
First line of input contains an integer T (1 β€ T β€ 5000), number of test cases. T pairs of lines with test case descriptions follow.
The first line of a test case description contains a single integer n (7 β€ n β€ 50), length of a string s.
The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.
Output
For each test case output an answer for it.
In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".
Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.
You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).
Example
Input
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba
Output
Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No
Note
In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.
In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.
In sixth example there are two occurrences of a string "abacaba" as a substring.
Submitted Solution:
```
pat = 'abacaba'
def substring_count(string):
count = 0
start = 0
# Search through the string till
# we reach the end of it
while start < len(string):
# Check if a substring is present from
# 'start' position till the end
pos = string.find(pat, start)
if pos != -1:
# If a substring is present, move 'start' to
# the next position from start of the substring
start = pos + 1
# Increment the count
count += 1
else:
# If no further substring is present
break
# return the value of count
return count
def process(s, n):
i=0
while i != n - 7:
temp = s
flag = True
j = 0
while j != 7:
if temp[i + j] != '?' and temp[i + j] != pat[j]:
flag = False
break
else:
temp = temp[:i+j] + pat[j] + temp[i+j+1:]
j = j + 1
if flag and substring_count(temp) == 1:
return temp
else:
print('No')
return
i = i + 1
def solve():
n = int(input())
s = input()
if substring_count(s) == 1:
for i in range(len(s)):
if s[i] == '?':
s = s[:i] + 'x' + s[i+1:]
else:
s[i]
print('Yes')
print(s)
elif substring_count(s) > 1:
print('No')
elif substring_count(s) == 0:
process(s, n)
def main():
test_case = int(input())
while test_case != 0:
solve()
test_case = test_case - 1
if __name__ == '__main__':
main()
``` | instruction | 0 | 47,188 | 0 | 94,376 |
No | output | 1 | 47,188 | 0 | 94,377 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your classmate, whom you do not like because he is boring, but whom you respect for his intellect, has two strings: s of length n and t of length m.
A sequence p_1, p_2, β¦, p_m, where 1 β€ p_1 < p_2 < β¦ < p_m β€ n, is called beautiful, if s_{p_i} = t_i for all i from 1 to m. The width of a sequence is defined as max_{1 β€ i < m} \left(p_{i + 1} - p_i\right).
Please help your classmate to identify the beautiful sequence with the maximum width. Your classmate promised you that for the given strings s and t there is at least one beautiful sequence.
Input
The first input line contains two integers n and m (2 β€ m β€ n β€ 2 β
10^5) β the lengths of the strings s and t.
The following line contains a single string s of length n, consisting of lowercase letters of the Latin alphabet.
The last line contains a single string t of length m, consisting of lowercase letters of the Latin alphabet.
It is guaranteed that there is at least one beautiful sequence for the given strings.
Output
Output one integer β the maximum width of a beautiful sequence.
Examples
Input
5 3
abbbc
abc
Output
3
Input
5 2
aaaaa
aa
Output
4
Input
5 5
abcdf
abcdf
Output
1
Input
2 2
ab
ab
Output
1
Note
In the first example there are two beautiful sequences of width 3: they are \{1, 2, 5\} and \{1, 4, 5\}.
In the second example the beautiful sequence with the maximum width is \{1, 5\}.
In the third example there is exactly one beautiful sequence β it is \{1, 2, 3, 4, 5\}.
In the fourth example there is exactly one beautiful sequence β it is \{1, 2\}. | instruction | 0 | 47,237 | 0 | 94,474 |
Tags: binary search, data structures, dp, greedy, two pointers
Correct Solution:
```
def solve(n, m, s, t):
idx = [0] * m
j = 0
for i in range(n):
if j < m and t[j] == s[i]:
idx[j] = i
j += 1
res = 0
j = m - 1
for i in range(n - 1, -1, -1):
if j >= 0 and t[j] == s[i]:
res = max(res, i - idx[j - 1])
j -= 1
return res
n, m = map(int, input().split())
s = input()
t = input()
print(solve(n, m, s, t))
``` | output | 1 | 47,237 | 0 | 94,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your classmate, whom you do not like because he is boring, but whom you respect for his intellect, has two strings: s of length n and t of length m.
A sequence p_1, p_2, β¦, p_m, where 1 β€ p_1 < p_2 < β¦ < p_m β€ n, is called beautiful, if s_{p_i} = t_i for all i from 1 to m. The width of a sequence is defined as max_{1 β€ i < m} \left(p_{i + 1} - p_i\right).
Please help your classmate to identify the beautiful sequence with the maximum width. Your classmate promised you that for the given strings s and t there is at least one beautiful sequence.
Input
The first input line contains two integers n and m (2 β€ m β€ n β€ 2 β
10^5) β the lengths of the strings s and t.
The following line contains a single string s of length n, consisting of lowercase letters of the Latin alphabet.
The last line contains a single string t of length m, consisting of lowercase letters of the Latin alphabet.
It is guaranteed that there is at least one beautiful sequence for the given strings.
Output
Output one integer β the maximum width of a beautiful sequence.
Examples
Input
5 3
abbbc
abc
Output
3
Input
5 2
aaaaa
aa
Output
4
Input
5 5
abcdf
abcdf
Output
1
Input
2 2
ab
ab
Output
1
Note
In the first example there are two beautiful sequences of width 3: they are \{1, 2, 5\} and \{1, 4, 5\}.
In the second example the beautiful sequence with the maximum width is \{1, 5\}.
In the third example there is exactly one beautiful sequence β it is \{1, 2, 3, 4, 5\}.
In the fourth example there is exactly one beautiful sequence β it is \{1, 2\}. | instruction | 0 | 47,238 | 0 | 94,476 |
Tags: binary search, data structures, dp, greedy, two pointers
Correct Solution:
```
n,m=map(int,input().split())
a,b,q,w,e=input(),input(),[],[],0
for i in range(n):
if b[e]==a[i]:q+=[i];e+=1
if e==m:break
e-=1
for i in range(n-1,-1,-1):
if b[e]==a[i]:w+=[i];e-=1
if e==-1:break
w=w[::-1]
print(max(w[i+1]-q[i] for i in range(m-1)))
``` | output | 1 | 47,238 | 0 | 94,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your classmate, whom you do not like because he is boring, but whom you respect for his intellect, has two strings: s of length n and t of length m.
A sequence p_1, p_2, β¦, p_m, where 1 β€ p_1 < p_2 < β¦ < p_m β€ n, is called beautiful, if s_{p_i} = t_i for all i from 1 to m. The width of a sequence is defined as max_{1 β€ i < m} \left(p_{i + 1} - p_i\right).
Please help your classmate to identify the beautiful sequence with the maximum width. Your classmate promised you that for the given strings s and t there is at least one beautiful sequence.
Input
The first input line contains two integers n and m (2 β€ m β€ n β€ 2 β
10^5) β the lengths of the strings s and t.
The following line contains a single string s of length n, consisting of lowercase letters of the Latin alphabet.
The last line contains a single string t of length m, consisting of lowercase letters of the Latin alphabet.
It is guaranteed that there is at least one beautiful sequence for the given strings.
Output
Output one integer β the maximum width of a beautiful sequence.
Examples
Input
5 3
abbbc
abc
Output
3
Input
5 2
aaaaa
aa
Output
4
Input
5 5
abcdf
abcdf
Output
1
Input
2 2
ab
ab
Output
1
Note
In the first example there are two beautiful sequences of width 3: they are \{1, 2, 5\} and \{1, 4, 5\}.
In the second example the beautiful sequence with the maximum width is \{1, 5\}.
In the third example there is exactly one beautiful sequence β it is \{1, 2, 3, 4, 5\}.
In the fourth example there is exactly one beautiful sequence β it is \{1, 2\}. | instruction | 0 | 47,239 | 0 | 94,478 |
Tags: binary search, data structures, dp, greedy, two pointers
Correct Solution:
```
#!/usr/bin/env python3.9
def findmax(s, t):
l_indexies = []
start = 0
for c in t:
idx = s.find(c, start)
l_indexies.append(idx)
start = idx + 1
r_indexies = []
end = len(s)
for c in reversed(t):
idx = s.rfind(c, 0, end)
r_indexies.append(idx)
end = idx
r_indexies = r_indexies[::-1]
diff1 = (r - l for l, r in zip(l_indexies[:-1], l_indexies[1:]))
diff2 = (r - l for l, r in zip(l_indexies[:-1], r_indexies[1:]))
maxwidth = max(*diff1, *diff2)
return maxwidth
n, m = list(map(int, input().split(' ')))
s = input()
t = input()
# start = s.find(t[0])
# end = s.rfind(t[-1])
# s = s[start:end+1]
print(findmax(s, t))
``` | output | 1 | 47,239 | 0 | 94,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your classmate, whom you do not like because he is boring, but whom you respect for his intellect, has two strings: s of length n and t of length m.
A sequence p_1, p_2, β¦, p_m, where 1 β€ p_1 < p_2 < β¦ < p_m β€ n, is called beautiful, if s_{p_i} = t_i for all i from 1 to m. The width of a sequence is defined as max_{1 β€ i < m} \left(p_{i + 1} - p_i\right).
Please help your classmate to identify the beautiful sequence with the maximum width. Your classmate promised you that for the given strings s and t there is at least one beautiful sequence.
Input
The first input line contains two integers n and m (2 β€ m β€ n β€ 2 β
10^5) β the lengths of the strings s and t.
The following line contains a single string s of length n, consisting of lowercase letters of the Latin alphabet.
The last line contains a single string t of length m, consisting of lowercase letters of the Latin alphabet.
It is guaranteed that there is at least one beautiful sequence for the given strings.
Output
Output one integer β the maximum width of a beautiful sequence.
Examples
Input
5 3
abbbc
abc
Output
3
Input
5 2
aaaaa
aa
Output
4
Input
5 5
abcdf
abcdf
Output
1
Input
2 2
ab
ab
Output
1
Note
In the first example there are two beautiful sequences of width 3: they are \{1, 2, 5\} and \{1, 4, 5\}.
In the second example the beautiful sequence with the maximum width is \{1, 5\}.
In the third example there is exactly one beautiful sequence β it is \{1, 2, 3, 4, 5\}.
In the fourth example there is exactly one beautiful sequence β it is \{1, 2\}. | instruction | 0 | 47,240 | 0 | 94,480 |
Tags: binary search, data structures, dp, greedy, two pointers
Correct Solution:
```
import math
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
n,m=map(int,input().split())
s=input()
k=input()
a=[]
b=[]
for j in s:
a.append(j)
for j in k:
b.append(j)
ind=defaultdict(lambda:[])
for j in range(n):
ind[a[j]].append(j)
val = [0]*(m)
last = -1
j = 0
while (j < m):
val[j] = ind[b[j]][bisect_right(ind[b[j]],last)]
last=val[j]
j+=1
val1=[0]*(m)
last=float("inf")
j = m-1
while (j >=0):
val1[j] = ind[b[j]][bisect_left(ind[b[j]],last)-1]
last=val1[j]
j+=-1
j=m-1
ans=1
while(j>=1):
curr=b[j]
ans=max(ans,val1[j]-val[j-1])
ind[curr].pop()
j+=-1
print(ans)
``` | output | 1 | 47,240 | 0 | 94,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your classmate, whom you do not like because he is boring, but whom you respect for his intellect, has two strings: s of length n and t of length m.
A sequence p_1, p_2, β¦, p_m, where 1 β€ p_1 < p_2 < β¦ < p_m β€ n, is called beautiful, if s_{p_i} = t_i for all i from 1 to m. The width of a sequence is defined as max_{1 β€ i < m} \left(p_{i + 1} - p_i\right).
Please help your classmate to identify the beautiful sequence with the maximum width. Your classmate promised you that for the given strings s and t there is at least one beautiful sequence.
Input
The first input line contains two integers n and m (2 β€ m β€ n β€ 2 β
10^5) β the lengths of the strings s and t.
The following line contains a single string s of length n, consisting of lowercase letters of the Latin alphabet.
The last line contains a single string t of length m, consisting of lowercase letters of the Latin alphabet.
It is guaranteed that there is at least one beautiful sequence for the given strings.
Output
Output one integer β the maximum width of a beautiful sequence.
Examples
Input
5 3
abbbc
abc
Output
3
Input
5 2
aaaaa
aa
Output
4
Input
5 5
abcdf
abcdf
Output
1
Input
2 2
ab
ab
Output
1
Note
In the first example there are two beautiful sequences of width 3: they are \{1, 2, 5\} and \{1, 4, 5\}.
In the second example the beautiful sequence with the maximum width is \{1, 5\}.
In the third example there is exactly one beautiful sequence β it is \{1, 2, 3, 4, 5\}.
In the fourth example there is exactly one beautiful sequence β it is \{1, 2\}. | instruction | 0 | 47,241 | 0 | 94,482 |
Tags: binary search, data structures, dp, greedy, two pointers
Correct Solution:
```
def main():
n, m = map(int, input().split())
s = str(input())
t = str(input())
from collections import deque, defaultdict
import bisect
X = [[] for i in range(26)]
for i, c in enumerate(s):
c = ord(c)-ord('a')
X[c].append(i)
A = []
cur = -1
for i, c in enumerate(t):
c = ord(c)-ord('a')
j = bisect.bisect_right(X[c], cur)
A.append(X[c][j])
cur = X[c][j]
#print(A)
B = []
cur = n
for i in reversed(range(m)):
c = t[i]
c = ord(c)-ord('a')
j = bisect.bisect_left(X[c], cur)
B.append(X[c][j-1])
cur = X[c][j-1]
B.reverse()
#print(B)
ans = 0
for i in range(m-1):
ans = max(ans, A[i+1]-A[i])
ans = max(ans, B[i+1]-A[i])
ans = max(ans, B[i+1]-B[i])
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 47,241 | 0 | 94,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your classmate, whom you do not like because he is boring, but whom you respect for his intellect, has two strings: s of length n and t of length m.
A sequence p_1, p_2, β¦, p_m, where 1 β€ p_1 < p_2 < β¦ < p_m β€ n, is called beautiful, if s_{p_i} = t_i for all i from 1 to m. The width of a sequence is defined as max_{1 β€ i < m} \left(p_{i + 1} - p_i\right).
Please help your classmate to identify the beautiful sequence with the maximum width. Your classmate promised you that for the given strings s and t there is at least one beautiful sequence.
Input
The first input line contains two integers n and m (2 β€ m β€ n β€ 2 β
10^5) β the lengths of the strings s and t.
The following line contains a single string s of length n, consisting of lowercase letters of the Latin alphabet.
The last line contains a single string t of length m, consisting of lowercase letters of the Latin alphabet.
It is guaranteed that there is at least one beautiful sequence for the given strings.
Output
Output one integer β the maximum width of a beautiful sequence.
Examples
Input
5 3
abbbc
abc
Output
3
Input
5 2
aaaaa
aa
Output
4
Input
5 5
abcdf
abcdf
Output
1
Input
2 2
ab
ab
Output
1
Note
In the first example there are two beautiful sequences of width 3: they are \{1, 2, 5\} and \{1, 4, 5\}.
In the second example the beautiful sequence with the maximum width is \{1, 5\}.
In the third example there is exactly one beautiful sequence β it is \{1, 2, 3, 4, 5\}.
In the fourth example there is exactly one beautiful sequence β it is \{1, 2\}. | instruction | 0 | 47,242 | 0 | 94,484 |
Tags: binary search, data structures, dp, greedy, two pointers
Correct Solution:
```
for u in range(1):
n, m = map(int, input().split())
s = input()
t = input()
left = []
right = []
p = 0
for i in range(n):
if(s[i] == t[p]):
left.append(i)
p += 1
if(p == m):
break
p = m-1
for i in range(n-1, -1, -1):
if(s[i] == t[p]):
right.append(i)
p -= 1
if(p == -1):
break
right = right[::-1]
ans = -1
for i in range(1, m):
ans = max(ans, right[i] - left[i-1])
print(ans)
``` | output | 1 | 47,242 | 0 | 94,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your classmate, whom you do not like because he is boring, but whom you respect for his intellect, has two strings: s of length n and t of length m.
A sequence p_1, p_2, β¦, p_m, where 1 β€ p_1 < p_2 < β¦ < p_m β€ n, is called beautiful, if s_{p_i} = t_i for all i from 1 to m. The width of a sequence is defined as max_{1 β€ i < m} \left(p_{i + 1} - p_i\right).
Please help your classmate to identify the beautiful sequence with the maximum width. Your classmate promised you that for the given strings s and t there is at least one beautiful sequence.
Input
The first input line contains two integers n and m (2 β€ m β€ n β€ 2 β
10^5) β the lengths of the strings s and t.
The following line contains a single string s of length n, consisting of lowercase letters of the Latin alphabet.
The last line contains a single string t of length m, consisting of lowercase letters of the Latin alphabet.
It is guaranteed that there is at least one beautiful sequence for the given strings.
Output
Output one integer β the maximum width of a beautiful sequence.
Examples
Input
5 3
abbbc
abc
Output
3
Input
5 2
aaaaa
aa
Output
4
Input
5 5
abcdf
abcdf
Output
1
Input
2 2
ab
ab
Output
1
Note
In the first example there are two beautiful sequences of width 3: they are \{1, 2, 5\} and \{1, 4, 5\}.
In the second example the beautiful sequence with the maximum width is \{1, 5\}.
In the third example there is exactly one beautiful sequence β it is \{1, 2, 3, 4, 5\}.
In the fourth example there is exactly one beautiful sequence β it is \{1, 2\}. | instruction | 0 | 47,243 | 0 | 94,486 |
Tags: binary search, data structures, dp, greedy, two pointers
Correct Solution:
```
n,m = map(int,input().split(' '))
s = [w for w in input()]
t = [w for w in input()]
left = []
j = 0
for i in range(n):
if s[i] == t[j]:
left.append(i)
j = j + 1
if j>=m:
break
right = []
i = n - 1
j = m - 1
while i>=0:
if s[i] == t[j]:
right.append(i)
j = j - 1
if j<0:
break
i = i - 1
right.reverse()
#print(left)
#print(right)
if m>2:
ans = 0
prev = left[0]
for i in range(1,m-1):
ans = max(ans,right[i] - prev)
prev = left[i]
ans = max(ans,right[-1] - left[-2] )
else:
ans = right[-1] - left[0]
print(ans)
``` | output | 1 | 47,243 | 0 | 94,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your classmate, whom you do not like because he is boring, but whom you respect for his intellect, has two strings: s of length n and t of length m.
A sequence p_1, p_2, β¦, p_m, where 1 β€ p_1 < p_2 < β¦ < p_m β€ n, is called beautiful, if s_{p_i} = t_i for all i from 1 to m. The width of a sequence is defined as max_{1 β€ i < m} \left(p_{i + 1} - p_i\right).
Please help your classmate to identify the beautiful sequence with the maximum width. Your classmate promised you that for the given strings s and t there is at least one beautiful sequence.
Input
The first input line contains two integers n and m (2 β€ m β€ n β€ 2 β
10^5) β the lengths of the strings s and t.
The following line contains a single string s of length n, consisting of lowercase letters of the Latin alphabet.
The last line contains a single string t of length m, consisting of lowercase letters of the Latin alphabet.
It is guaranteed that there is at least one beautiful sequence for the given strings.
Output
Output one integer β the maximum width of a beautiful sequence.
Examples
Input
5 3
abbbc
abc
Output
3
Input
5 2
aaaaa
aa
Output
4
Input
5 5
abcdf
abcdf
Output
1
Input
2 2
ab
ab
Output
1
Note
In the first example there are two beautiful sequences of width 3: they are \{1, 2, 5\} and \{1, 4, 5\}.
In the second example the beautiful sequence with the maximum width is \{1, 5\}.
In the third example there is exactly one beautiful sequence β it is \{1, 2, 3, 4, 5\}.
In the fourth example there is exactly one beautiful sequence β it is \{1, 2\}. | instruction | 0 | 47,244 | 0 | 94,488 |
Tags: binary search, data structures, dp, greedy, two pointers
Correct Solution:
```
a, b = map(int, input().split())
s = input()
t = input()
l1 = len(s)
l2 = len(t)
p1 = []
pos1 = 0
pos2 = 0
while(pos1 < l1 and pos2 < l2):
if(s[pos1] == t[pos2]):
p1.append(pos1)
pos2 += 1
pos1 += 1
p2 = []
pos1 = l1-1
pos2 = l2-1
while(pos1 >= 0 and pos2 >= 0):
if(s[pos1] == t[pos2]):
p2.append(pos1)
pos2 -= 1
pos1 -= 1
p2.reverse()
ans = 0
for i in range(1, l2):
ans = max(ans, p2[i]-p1[i-1])
print(ans)
``` | output | 1 | 47,244 | 0 | 94,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your classmate, whom you do not like because he is boring, but whom you respect for his intellect, has two strings: s of length n and t of length m.
A sequence p_1, p_2, β¦, p_m, where 1 β€ p_1 < p_2 < β¦ < p_m β€ n, is called beautiful, if s_{p_i} = t_i for all i from 1 to m. The width of a sequence is defined as max_{1 β€ i < m} \left(p_{i + 1} - p_i\right).
Please help your classmate to identify the beautiful sequence with the maximum width. Your classmate promised you that for the given strings s and t there is at least one beautiful sequence.
Input
The first input line contains two integers n and m (2 β€ m β€ n β€ 2 β
10^5) β the lengths of the strings s and t.
The following line contains a single string s of length n, consisting of lowercase letters of the Latin alphabet.
The last line contains a single string t of length m, consisting of lowercase letters of the Latin alphabet.
It is guaranteed that there is at least one beautiful sequence for the given strings.
Output
Output one integer β the maximum width of a beautiful sequence.
Examples
Input
5 3
abbbc
abc
Output
3
Input
5 2
aaaaa
aa
Output
4
Input
5 5
abcdf
abcdf
Output
1
Input
2 2
ab
ab
Output
1
Note
In the first example there are two beautiful sequences of width 3: they are \{1, 2, 5\} and \{1, 4, 5\}.
In the second example the beautiful sequence with the maximum width is \{1, 5\}.
In the third example there is exactly one beautiful sequence β it is \{1, 2, 3, 4, 5\}.
In the fourth example there is exactly one beautiful sequence β it is \{1, 2\}.
Submitted Solution:
```
n,m=map(int,input().split())
s=input()
t=input()
l=[0]*m
r=[0]*m
i,j=0,0
while j<m:
if s[i]==t[j]:
l[j]=i
j+=1
i+=1
i=n-1
j=m-1
while j>0:
if s[i]==t[j]:
r[j]=i
j-=1
i-=1
print(max(r[i+1]-l[i] for i in range(m-1)))
``` | instruction | 0 | 47,245 | 0 | 94,490 |
Yes | output | 1 | 47,245 | 0 | 94,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your classmate, whom you do not like because he is boring, but whom you respect for his intellect, has two strings: s of length n and t of length m.
A sequence p_1, p_2, β¦, p_m, where 1 β€ p_1 < p_2 < β¦ < p_m β€ n, is called beautiful, if s_{p_i} = t_i for all i from 1 to m. The width of a sequence is defined as max_{1 β€ i < m} \left(p_{i + 1} - p_i\right).
Please help your classmate to identify the beautiful sequence with the maximum width. Your classmate promised you that for the given strings s and t there is at least one beautiful sequence.
Input
The first input line contains two integers n and m (2 β€ m β€ n β€ 2 β
10^5) β the lengths of the strings s and t.
The following line contains a single string s of length n, consisting of lowercase letters of the Latin alphabet.
The last line contains a single string t of length m, consisting of lowercase letters of the Latin alphabet.
It is guaranteed that there is at least one beautiful sequence for the given strings.
Output
Output one integer β the maximum width of a beautiful sequence.
Examples
Input
5 3
abbbc
abc
Output
3
Input
5 2
aaaaa
aa
Output
4
Input
5 5
abcdf
abcdf
Output
1
Input
2 2
ab
ab
Output
1
Note
In the first example there are two beautiful sequences of width 3: they are \{1, 2, 5\} and \{1, 4, 5\}.
In the second example the beautiful sequence with the maximum width is \{1, 5\}.
In the third example there is exactly one beautiful sequence β it is \{1, 2, 3, 4, 5\}.
In the fourth example there is exactly one beautiful sequence β it is \{1, 2\}.
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
"""Codeforces Round #704 (Div. 2)
Problem C. Maximum width
:author: Kitchen Tong
:mail: kctong529@gmail.com
Please feel free to contact me if you have any question
regarding the implementation below.
"""
__version__ = '3.0'
__date__ = '2021-03-08'
import sys
def solve(n, m, s, t):
'''
All characters of t can be found in the same order in s.
Use two lists of size m to denote the position of leftmost and rightmost
occurrence of each t character in s.
The positions found in each list should be valid on its own, i.e.
l[i] < l[j] where i < j, for all elements in the list
Therefore, it is still valid when we combine the two lists,
provided that the position of the element replaced in the junction
does not violate the above rule, i.e.
left[k] <= right[k] where k is the element to be replaced
'''
left_pos = [-1] * m
right_pos = [-1] * m
curr = 0
for pos in range(n):
if curr == m:
break
if s[pos] == t[curr]:
left_pos[curr] = pos
curr += 1
curr = m - 1
for pos in range(n-1, -1, -1):
if curr == -1:
break
if s[pos] == t[curr]:
right_pos[curr] = pos
curr -= 1
answer = 1
for i in range(m-1):
answer = max(answer, right_pos[i+1] - left_pos[i])
return answer
def main(argv=None):
n, m = map(int, input().split())
s = input()
t = input()
print(solve(n, m, s, t))
return 0
if __name__ == "__main__":
STATUS = main()
sys.exit(STATUS)
``` | instruction | 0 | 47,246 | 0 | 94,492 |
Yes | output | 1 | 47,246 | 0 | 94,493 |
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