message
stringlengths 2
23.8k
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stringclasses 2
values | message_id
int64 0
1
| conversation_id
int64 97
109k
| cluster
float64 0
0
| __index_level_0__
int64 194
217k
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|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fedya has a string S, initially empty, and an array W, also initially empty.
There are n queries to process, one at a time. Query i consists of a lowercase English letter c_i and a nonnegative integer w_i. First, c_i must be appended to S, and w_i must be appended to W. The answer to the query is the sum of suspiciousnesses for all subsegments of W [L, \ R], (1 ≤ L ≤ R ≤ i).
We define the suspiciousness of a subsegment as follows: if the substring of S corresponding to this subsegment (that is, a string of consecutive characters from L-th to R-th, inclusive) matches the prefix of S of the same length (that is, a substring corresponding to the subsegment [1, \ R - L + 1]), then its suspiciousness is equal to the minimum in the array W on the [L, \ R] subsegment. Otherwise, in case the substring does not match the corresponding prefix, the suspiciousness is 0.
Help Fedya answer all the queries before the orderlies come for him!
Input
The first line contains an integer n (1 ≤ n ≤ 600 000) — the number of queries.
The i-th of the following n lines contains the query i: a lowercase letter of the Latin alphabet c_i and an integer w_i (0 ≤ w_i ≤ 2^{30} - 1).
All queries are given in an encrypted form. Let ans be the answer to the previous query (for the first query we set this value equal to 0). Then, in order to get the real query, you need to do the following: perform a cyclic shift of c_i in the alphabet forward by ans, and set w_i equal to w_i ⊕ (ans \ \& \ MASK), where ⊕ is the bitwise exclusive "or", \& is the bitwise "and", and MASK = 2^{30} - 1.
Output
Print n lines, i-th line should contain a single integer — the answer to the i-th query.
Examples
Input
7
a 1
a 0
y 3
y 5
v 4
u 6
r 8
Output
1
2
4
5
7
9
12
Input
4
a 2
y 2
z 0
y 2
Output
2
2
2
2
Input
5
a 7
u 5
t 3
s 10
s 11
Output
7
9
11
12
13
Note
For convenience, we will call "suspicious" those subsegments for which the corresponding lines are prefixes of S, that is, those whose suspiciousness may not be zero.
As a result of decryption in the first example, after all requests, the string S is equal to "abacaba", and all w_i = 1, that is, the suspiciousness of all suspicious sub-segments is simply equal to 1. Let's see how the answer is obtained after each request:
1. S = "a", the array W has a single subsegment — [1, \ 1], and the corresponding substring is "a", that is, the entire string S, thus it is a prefix of S, and the suspiciousness of the subsegment is 1.
2. S = "ab", suspicious subsegments: [1, \ 1] and [1, \ 2], total 2.
3. S = "aba", suspicious subsegments: [1, \ 1], [1, \ 2], [1, \ 3] and [3, \ 3], total 4.
4. S = "abac", suspicious subsegments: [1, \ 1], [1, \ 2], [1, \ 3], [1, \ 4] and [3, \ 3], total 5.
5. S = "abaca", suspicious subsegments: [1, \ 1], [1, \ 2], [1, \ 3], [1, \ 4] , [1, \ 5], [3, \ 3] and [5, \ 5], total 7.
6. S = "abacab", suspicious subsegments: [1, \ 1], [1, \ 2], [1, \ 3], [1, \ 4] , [1, \ 5], [1, \ 6], [3, \ 3], [5, \ 5] and [5, \ 6], total 9.
7. S = "abacaba", suspicious subsegments: [1, \ 1], [1, \ 2], [1, \ 3], [1, \ 4] , [1, \ 5], [1, \ 6], [1, \ 7], [3, \ 3], [5, \ 5], [5, \ 6], [5, \ 7] and [7, \ 7], total 12.
In the second example, after all requests S = "aaba", W = [2, 0, 2, 0].
1. S = "a", suspicious subsegments: [1, \ 1] (suspiciousness 2), totaling 2.
2. S = "aa", suspicious subsegments: [1, \ 1] (2), [1, \ 2] (0), [2, \ 2] ( 0), totaling 2.
3. S = "aab", suspicious subsegments: [1, \ 1] (2), [1, \ 2] (0), [1, \ 3] ( 0), [2, \ 2] (0), totaling 2.
4. S = "aaba", suspicious subsegments: [1, \ 1] (2), [1, \ 2] (0), [1, \ 3] ( 0), [1, \ 4] (0), [2, \ 2] (0), [4, \ 4] (0), totaling 2.
In the third example, from the condition after all requests S = "abcde", W = [7, 2, 10, 1, 7].
1. S = "a", suspicious subsegments: [1, \ 1] (7), totaling 7.
2. S = "ab", suspicious subsegments: [1, \ 1] (7), [1, \ 2] (2), totaling 9.
3. S = "abc", suspicious subsegments: [1, \ 1] (7), [1, \ 2] (2), [1, \ 3] ( 2), totaling 11.
4. S = "abcd", suspicious subsegments: [1, \ 1] (7), [1, \ 2] (2), [1, \ 3] ( 2), [1, \ 4] (1), totaling 12.
5. S = "abcde", suspicious subsegments: [1, \ 1] (7), [1, \ 2] (2), [1, \ 3] ( 2), [1, \ 4] (1), [1, \ 5] (1), totaling 13.
Submitted Solution:
```
n = int(input()) # number of input
s = []
q = []
ans = 0
def susp(string, queries):
l = len(string)
val = 0
for i in range(l):
for j in range(i, l):
# print(i, j)
if string[i: j + 1] == string[:j - i + 1]:
val += min(queries[i: j + 1])
print(val)
return val
for i in range(n):
a = input().split()
if ord(a[0]) + ans <= 122:
req = chr(ord(a[0]) + ans)
else:
req = chr(ord(a[0]) + ans - 123 + 97)
s.append(req)
# appending in q has to be done very carefully
q.append(int(a[1]) ^ (ans & (2**30 - 1)))
# print(s, q)
ans = susp(s, q)
del a
```
|
instruction
| 0
| 11,756
| 0
| 23,512
|
No
|
output
| 1
| 11,756
| 0
| 23,513
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fedya has a string S, initially empty, and an array W, also initially empty.
There are n queries to process, one at a time. Query i consists of a lowercase English letter c_i and a nonnegative integer w_i. First, c_i must be appended to S, and w_i must be appended to W. The answer to the query is the sum of suspiciousnesses for all subsegments of W [L, \ R], (1 ≤ L ≤ R ≤ i).
We define the suspiciousness of a subsegment as follows: if the substring of S corresponding to this subsegment (that is, a string of consecutive characters from L-th to R-th, inclusive) matches the prefix of S of the same length (that is, a substring corresponding to the subsegment [1, \ R - L + 1]), then its suspiciousness is equal to the minimum in the array W on the [L, \ R] subsegment. Otherwise, in case the substring does not match the corresponding prefix, the suspiciousness is 0.
Help Fedya answer all the queries before the orderlies come for him!
Input
The first line contains an integer n (1 ≤ n ≤ 600 000) — the number of queries.
The i-th of the following n lines contains the query i: a lowercase letter of the Latin alphabet c_i and an integer w_i (0 ≤ w_i ≤ 2^{30} - 1).
All queries are given in an encrypted form. Let ans be the answer to the previous query (for the first query we set this value equal to 0). Then, in order to get the real query, you need to do the following: perform a cyclic shift of c_i in the alphabet forward by ans, and set w_i equal to w_i ⊕ (ans \ \& \ MASK), where ⊕ is the bitwise exclusive "or", \& is the bitwise "and", and MASK = 2^{30} - 1.
Output
Print n lines, i-th line should contain a single integer — the answer to the i-th query.
Examples
Input
7
a 1
a 0
y 3
y 5
v 4
u 6
r 8
Output
1
2
4
5
7
9
12
Input
4
a 2
y 2
z 0
y 2
Output
2
2
2
2
Input
5
a 7
u 5
t 3
s 10
s 11
Output
7
9
11
12
13
Note
For convenience, we will call "suspicious" those subsegments for which the corresponding lines are prefixes of S, that is, those whose suspiciousness may not be zero.
As a result of decryption in the first example, after all requests, the string S is equal to "abacaba", and all w_i = 1, that is, the suspiciousness of all suspicious sub-segments is simply equal to 1. Let's see how the answer is obtained after each request:
1. S = "a", the array W has a single subsegment — [1, \ 1], and the corresponding substring is "a", that is, the entire string S, thus it is a prefix of S, and the suspiciousness of the subsegment is 1.
2. S = "ab", suspicious subsegments: [1, \ 1] and [1, \ 2], total 2.
3. S = "aba", suspicious subsegments: [1, \ 1], [1, \ 2], [1, \ 3] and [3, \ 3], total 4.
4. S = "abac", suspicious subsegments: [1, \ 1], [1, \ 2], [1, \ 3], [1, \ 4] and [3, \ 3], total 5.
5. S = "abaca", suspicious subsegments: [1, \ 1], [1, \ 2], [1, \ 3], [1, \ 4] , [1, \ 5], [3, \ 3] and [5, \ 5], total 7.
6. S = "abacab", suspicious subsegments: [1, \ 1], [1, \ 2], [1, \ 3], [1, \ 4] , [1, \ 5], [1, \ 6], [3, \ 3], [5, \ 5] and [5, \ 6], total 9.
7. S = "abacaba", suspicious subsegments: [1, \ 1], [1, \ 2], [1, \ 3], [1, \ 4] , [1, \ 5], [1, \ 6], [1, \ 7], [3, \ 3], [5, \ 5], [5, \ 6], [5, \ 7] and [7, \ 7], total 12.
In the second example, after all requests S = "aaba", W = [2, 0, 2, 0].
1. S = "a", suspicious subsegments: [1, \ 1] (suspiciousness 2), totaling 2.
2. S = "aa", suspicious subsegments: [1, \ 1] (2), [1, \ 2] (0), [2, \ 2] ( 0), totaling 2.
3. S = "aab", suspicious subsegments: [1, \ 1] (2), [1, \ 2] (0), [1, \ 3] ( 0), [2, \ 2] (0), totaling 2.
4. S = "aaba", suspicious subsegments: [1, \ 1] (2), [1, \ 2] (0), [1, \ 3] ( 0), [1, \ 4] (0), [2, \ 2] (0), [4, \ 4] (0), totaling 2.
In the third example, from the condition after all requests S = "abcde", W = [7, 2, 10, 1, 7].
1. S = "a", suspicious subsegments: [1, \ 1] (7), totaling 7.
2. S = "ab", suspicious subsegments: [1, \ 1] (7), [1, \ 2] (2), totaling 9.
3. S = "abc", suspicious subsegments: [1, \ 1] (7), [1, \ 2] (2), [1, \ 3] ( 2), totaling 11.
4. S = "abcd", suspicious subsegments: [1, \ 1] (7), [1, \ 2] (2), [1, \ 3] ( 2), [1, \ 4] (1), totaling 12.
5. S = "abcde", suspicious subsegments: [1, \ 1] (7), [1, \ 2] (2), [1, \ 3] ( 2), [1, \ 4] (1), [1, \ 5] (1), totaling 13.
Submitted Solution:
```
n = int(input()) # number of input
s = []
q = []
ans = 0
def susp(string, queries):
l = len(string)
val = 0
for i in range(l):
for j in range(i, l):
# print(i, j)
if string[i: j + 1] == string[:j - i + 1]:
val += min(queries[i: j + 1])
print(val)
return val
for i in range(n):
a = input().split()
req = chr(97 + (ord(a[0]) + ans - 97) % 26)
s.append(req)
# appending in q has to be done very carefully
q.append(int(a[1]) ^ (ans & (2**30 - 1)))
print(s, q)
ans = susp(s, q)
del a
```
|
instruction
| 0
| 11,757
| 0
| 23,514
|
No
|
output
| 1
| 11,757
| 0
| 23,515
|
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
|
instruction
| 0
| 11,860
| 0
| 23,720
|
Tags: greedy
Correct Solution:
```
def cal(s):
m=0;t=0
boo = True
x=n;j=0
for i in s:
if(i == 'T'):
t+=1
else:
m+=1
if t<m:
boo=False
return (t == 2*m and boo)
for _ in range(int(input())):
n = int(input())
s = input()
ans = "YES" if(cal(s) and cal(s[::-1])) else "NO"
print(ans)
"""
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
"""
```
|
output
| 1
| 11,860
| 0
| 23,721
|
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
|
instruction
| 0
| 11,861
| 0
| 23,722
|
Tags: greedy
Correct Solution:
```
def solve(s,n):
t,m=0,0
if True:
for i in range(n):
if s[i]=='T':
t+=1
if s[i]=='M':
m+=1
if t<m:
print("NO")
return
t,m=0,0
for i in range(n-1,-1,-1):
if s[i]=='T':
t+=1
if s[i]=='M':
m+=1
if t<m:
print("NO")
return
print("YES")
for _ in range(int(input())):
n=int(input())
s=input()
if s.count("T")==s.count("M")*2:
solve(s,n)
else:
print("NO")
```
|
output
| 1
| 11,861
| 0
| 23,723
|
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
|
instruction
| 0
| 11,862
| 0
| 23,724
|
Tags: greedy
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
from bisect import bisect_right
from math import gcd,log
from collections import Counter,defaultdict
from pprint import pprint
from itertools import permutations
from bisect import bisect_right
def main():
n=int(input())
s=input()
bol=True
mc=s.count('M')
tc=0
m=0
for i in range(n):
if s[i]=='T':
if mc:
mc-=1
tc+=1
else:
m-=1
if s[i]=='M':
tc-=1
m+=1
if m<0 or tc<0 :
print('NO')
return
if m!=0 or tc!=0:
print('NO')
return
print('YES')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
for _ in range(int(input())):
main()
```
|
output
| 1
| 11,862
| 0
| 23,725
|
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
|
instruction
| 0
| 11,863
| 0
| 23,726
|
Tags: greedy
Correct Solution:
```
from collections import defaultdict, deque
import sys
from math import gcd
#import heapq
input = sys.stdin.readline
t = int(input().rstrip())
maxn = 2000005
Jc = [0 for i in range(maxn)];
mod = 10**9+7
for _ in range(t):
n = int(input().rstrip())
#n,k = map(int,input().rstrip().rsplit())
#arr = [int(i) for i in input().rstrip().split()]
s = input().rstrip()
cnt = defaultdict(int)
for i in s:
cnt[i] += 1
if cnt['T'] != n//3 * 2:
print('NO')
continue
has = True
cnt = defaultdict(int)
for i in s:
cnt[i] += 1
if i == 'M':
if cnt['M'] > cnt['T']:
has = False
break
cnt = defaultdict(int)
for i in range(len(s) - 1,-1,-1):
cnt[s[i]] += 1
if s[i] == 'M':
if cnt['M'] > cnt['T']:
has = False
break
if has:
print('YES')
else:
print('NO')
#print(' '.join(str(i) for i in ans))
```
|
output
| 1
| 11,863
| 0
| 23,727
|
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
|
instruction
| 0
| 11,864
| 0
| 23,728
|
Tags: greedy
Correct Solution:
```
def solve(n, s):
t, m = [], []
for i in range(n):
if s[i] == 'T':
t.append(i)
else:
m.append(i)
if len(t) != 2*len(m):
return False
for i in range(len(m)):
if m[i] < t[i] or m[i] > t[i + len(m)]:
return False
return True
for __ in range(int(input())):
if(solve(int(input()), input())):
print("YES")
else:
print("NO")
```
|
output
| 1
| 11,864
| 0
| 23,729
|
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
|
instruction
| 0
| 11,865
| 0
| 23,730
|
Tags: greedy
Correct Solution:
```
t = int(input())
for i in range(t):
n = int(input())
s = str(input())
m = 0
for x in s:
if x == 'M':
m += 1
if m != n / 3:
print("NO")
continue
cur = 0
f = True
for x in s:
if x == 'T':
cur += 1
else:
cur -= 1
if cur < 0:
f = False
break
cur = 0
for x in reversed(s):
if x == 'T':
cur += 1
else:
cur -= 1
if cur < 0:
f = False
break
if f:
print("YES")
else:
print("NO")
```
|
output
| 1
| 11,865
| 0
| 23,731
|
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
|
instruction
| 0
| 11,866
| 0
| 23,732
|
Tags: greedy
Correct Solution:
```
n=int(input())
for g in range(n):
no=int(input())
s=input()
t=[]
m=[]
for i in range(len(s)):
if s[i]=='T':
t.append(i)
else:
m.append(i)
if len(t)!=2*len(m):
print('NO')
else:
for i in range(len(m)):
if (t[i]>m[i]) or (m[i]>t[len(m)+i]):
print('NO')
break
elif i==len(m)-1:
print('YES')
```
|
output
| 1
| 11,866
| 0
| 23,733
|
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
|
instruction
| 0
| 11,867
| 0
| 23,734
|
Tags: greedy
Correct Solution:
```
import sys
from io import BytesIO, IOBase
import heapq as h
import bisect
import math as mt
from types import GeneratorType
BUFSIZE = 8192
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index+1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
import sys
from math import ceil,log2
INT_MAX = sys.maxsize
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
import collections as col
import math, string
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
def minVal(x, y) :
return x if (x < y) else y;
# A utility function to get the
# middle index from corner indexes.
def getMid(s, e) :
return s + (e - s) // 2;
""" A recursive function to get the
minimum value in a given range
of array indexes. The following
are parameters for this function.
st --> Pointer to segment tree
index --> Index of current node in the
segment tree. Initially 0 is
passed as root is always at index 0
ss & se --> Starting and ending indexes
of the segment represented
by current node, i.e., st[index]
qs & qe --> Starting and ending indexes of query range """
def RMQUtil( st, ss, se, qs, qe, index) :
# If segment of this node is a part
# of given range, then return
# the min of the segment
if (qs <= ss and qe >= se) :
return st[index];
# If segment of this node
# is outside the given range
if (se < qs or ss > qe) :
return INT_MAX;
# If a part of this segment
# overlaps with the given range
mid = getMid(ss, se);
return minVal(RMQUtil(st, ss, mid, qs,
qe, 2 * index + 1),
RMQUtil(st, mid + 1, se,
qs, qe, 2 * index + 2));
# Return minimum of elements in range
# from index qs (query start) to
# qe (query end). It mainly uses RMQUtil()
def RMQ( st, n, qs, qe) :
# Check for erroneous input values
if (qs < 0 or qe > n - 1 or qs > qe) :
print("Invalid Input");
return -1;
return RMQUtil(st, 0, n - 1, qs, qe, 0);
# A recursive function that constructs
# Segment Tree for array[ss..se].
# si is index of current node in segment tree st
def constructSTUtil(arr, ss, se, st, si) :
# If there is one element in array,
# store it in current node of
# segment tree and return
if (ss == se) :
st[si] = arr[ss];
return arr[ss];
# If there are more than one elements,
# then recur for left and right subtrees
# and store the minimum of two values in this node
mid = getMid(ss, se);
st[si] = minVal(constructSTUtil(arr, ss, mid,
st, si * 2 + 1),
constructSTUtil(arr, mid + 1, se,
st, si * 2 + 2));
return st[si];
"""Function to construct segment tree
from given array. This function allocates
memory for segment tree and calls constructSTUtil()
to fill the allocated memory """
def constructST( arr, n) :
# Allocate memory for segment tree
# Height of segment tree
x = (int)(ceil(log2(n)));
# Maximum size of segment tree
max_size = 2 * (int)(2**x) - 1;
st = [0] * (max_size);
# Fill the allocated memory st
constructSTUtil(arr, 0, n - 1, st, 0);
# Return the constructed segment tree
return st;
MOD = 10**9+7
mod=10**9+7
#t=1
p=10**9+7
def ncr_util():
inv[0]=inv[1]=1
fact[0]=fact[1]=1
for i in range(2,300001):
inv[i]=(inv[i%p]*(p-p//i))%p
for i in range(1,300001):
inv[i]=(inv[i-1]*inv[i])%p
fact[i]=(fact[i-1]*i)%p
def z_array(s1):
n = len(s1)
z=[0]*(n)
l, r, k = 0, 0, 0
for i in range(1,n):
# if i>R nothing matches so we will calculate.
# Z[i] using naive way.
if i > r:
l, r = i, i
# R-L = 0 in starting, so it will start
# checking from 0'th index. For example,
# for "ababab" and i = 1, the value of R
# remains 0 and Z[i] becomes 0. For string
# "aaaaaa" and i = 1, Z[i] and R become 5
while r < n and s1[r - l] == s1[r]:
r += 1
z[i] = r - l
r -= 1
else:
# k = i-L so k corresponds to number which
# matches in [L,R] interval.
k = i - l
# if Z[k] is less than remaining interval
# then Z[i] will be equal to Z[k].
# For example, str = "ababab", i = 3, R = 5
# and L = 2
if z[k] < r - i + 1:
z[i] = z[k]
# For example str = "aaaaaa" and i = 2,
# R is 5, L is 0
else:
# else start from R and check manually
l = i
while r < n and s1[r - l] == s1[r]:
r += 1
z[i] = r - l
r -= 1
return z
MAXN1=10000001
#spf=[0]*MAXN1
def sieve():
d1={}
d2=[0]*(MAXN1)
for i in range(1,MAXN1):
for j in range(i,MAXN1,i):
d2[j]+=i
if d2[i]<MAXN1 and d1.get(d2[i],-1)==-1:
d1[d2[i]]=i
return d1
#d1=sieve()
def factor(x):
d1={}
x1=x
while x!=1:
d1[spf[x]]=d1.get(spf[x],0)+1
x//=spf[x]
def primeFactors(n):
d1={}
while n % 2 == 0:
d1[2]=d1.get(2,0)+1
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
# while i divides n , print i ad divide n
while n % i== 0:
d1[i]=d1.get(i,0)+1
n = n // i
if n > 2:
d1[n]=d1.get(n,0)+1
return d1
def fs(x):
if x<2:
return 0
return (x*(x-1))//2
t=int(input())
#t=1
def solve():
ind=[]
for i in range(n):
if s[i]=='T':
ind.append(i)
l=0
d={}
for i in range(n):
if s[i]=='M':
if l>=len(ind) :
return "NO"
if ind[l]>i:
return "NO"
d[ind[l]]=1
l+=1
l=0
#print(d)
for i in range(n):
if s[i]=='M':
i1=-1
while l<len(ind) :
if ind[l]>i and d.get(ind[l],-1)==-1:
i1=l
d[ind[l]]=1
break
l+=1
if i1==-1:
return "NO"
if len(d)!=len(ind):
return "NO"
return "YES"
for _ in range(t):
#d={}
#n,l,r,s=map(int,input().split())
n=int(input())
s=input()
#l=list(map(int,input().split()))
#l2=list(map(int,input().split()))
# a=list(map(int,input().split()))
# b=list(map(int,input().split()))
#l.sort()
#l2=list(map(int,input().split()))
#l.sort()
#l.sort(revrese=True)
#x,y=(map(int,input().split()))
#l=str(n)
#l.sort(reverse=True)
#l2.sort(reverse=True)
#l1.sort(reverse=True)
print(solve())
#print()
#print(n,u,r,d,l)
```
|
output
| 1
| 11,867
| 0
| 23,735
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
tm = input()
mt = tm[::-1]
t = 0
m = 0
tt = 0
mm = 0
fail = 0
for j in range(n):
if tm[j] == "T":
t += 1
else:
m += 1
if mt[j] == "T":
tt += 1
else:
mm += 1
if m > t or mm > tt:
fail = 1
break
if t == 2 * m and fail == 0:
print("YES")
else:
print("NO")
```
|
instruction
| 0
| 11,868
| 0
| 23,736
|
Yes
|
output
| 1
| 11,868
| 0
| 23,737
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
Submitted Solution:
```
x=int(input())
for i in range(x):
n=int(input())
a =input()
score=0
flag=True
for j in a:
if j=="T":
score+=1
else :
score-=1
if score<0 or score>n//3:
flag=False
break
if (flag and score==(n-2*(n//3))):
print("YES")
else :
print("NO")
```
|
instruction
| 0
| 11,869
| 0
| 23,738
|
Yes
|
output
| 1
| 11,869
| 0
| 23,739
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
Submitted Solution:
```
def check(s):
c=0
for i in s:
if i=='T':
c+=1
else:
c-=1
if (c<0):
return False
c=0
s=s[::-1]
for i in s:
if i=='T':
c+=1
else:
c-=1
if (c<0):
return False
return True
for _ in range(int(input())):
n=int(input())
s=input()
if s.count('T')!=2*s.count('M'):
print("NO")
else:
if check(s):
print("YES")
else:
print("NO")
```
|
instruction
| 0
| 11,870
| 0
| 23,740
|
Yes
|
output
| 1
| 11,870
| 0
| 23,741
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
Submitted Solution:
```
# import math as mt
# from collections import defaultdict
# from collections import Counter, deque
# from itertools import permutations
# from functools import reduce
# from heapq import heapify, heappop, heappush, heapreplace
def getInput(): return sys.stdin.readline().strip()
def getInt(): return int(getInput())
def getInts(): return map(int, getInput().split())
def getArray(): return list(getInts())
# sys.setrecursionlimit(10**7)
# INF = float('inf')
# MOD1, MOD2 = 10**9+7, 998244353
# def def_value():
# return 0
# Defining the dict
for _ in range(int(input())):
n = int(input())
st = input()
flag1= True
flag2= True
check = 0
for i in st:
if(i=="T"):
check+=1
elif(i=="M"):
check-=1
if(check<0):
flag1 = False
break
check =0
for i in st[::-1]:
if(i=="T"):
check+=1
elif(i=="M"):
check-=1
if(check<0):
flag2 = False
break
# print(flag1,flag2,st.count("T"),st.count("M"))
if(flag1 == True and flag2 == True and st.count("T") == st.count("M")*2):
print("YES")
else:
print("NO")
```
|
instruction
| 0
| 11,871
| 0
| 23,742
|
Yes
|
output
| 1
| 11,871
| 0
| 23,743
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
Submitted Solution:
```
def STR(): return list(input())
def INT(): return int(input())
def MAP(): return map(int, input().split())
def MAP2():return map(float,input().split())
def LIST(): return list(map(int, input().split()))
def STRING(): return input()
import string
import sys
from heapq import heappop , heappush
from bisect import *
from collections import deque , Counter , defaultdict
from math import *
from itertools import permutations , accumulate
dx = [-1 , 1 , 0 , 0 ]
dy = [0 , 0 , 1 , - 1]
#visited = [[False for i in range(m)] for j in range(n)]
# primes = [2,11,101,1009,10007,100003,1000003,10000019,102345689]
#sys.stdin = open(r'input.txt' , 'r')
#sys.stdout = open(r'output.txt' , 'w')
#for tt in range(INT()):
#Code
def solve(s):
cnt = 0
flag = True
for i in s :
if i == 'T':
cnt+=1
else:
cnt-=1
if cnt < 0 :
flag = False
break
if flag:
return True
return False
for tt in range(INT()):
n = INT()
s = STR()
cntM = s.count('M')
cntT = s.count('T')
if cntT == 0 or cntM == 0 or s[0] == 'M' or s[-1] == 'M' or 2 * cntM != cntT:
print('NO')
else:
if solve(s):
print('YES')
else:
print('NO')
```
|
instruction
| 0
| 11,872
| 0
| 23,744
|
No
|
output
| 1
| 11,872
| 0
| 23,745
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
Submitted Solution:
```
t=int(input())
for i in range(t):
n=int(input())
string=input()
i=0
j=n-1
string=list(string)
print(string)
while string:
if string[0]=='M' or string[len(string)-1]=='M':
print("NO")
break
string.pop(0)
string.pop()
if 'M' in string:
string.remove("M")
else:
print("NO")
break
#print(string)
else:
print("YES")
```
|
instruction
| 0
| 11,873
| 0
| 23,746
|
No
|
output
| 1
| 11,873
| 0
| 23,747
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
Submitted Solution:
```
import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools,itertools,operator,bisect,fractions,statistics
from collections import deque,defaultdict,OrderedDict,Counter
from fractions import Fraction
from decimal import Decimal
from sys import stdout
from heapq import heappush, heappop, heapify ,_heapify_max,_heappop_max,nsmallest,nlargest
# sys.setrecursionlimit(111111)
INF=999999999999999999999999
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
def main():
mod=1000000007
# InverseofNumber(mod)
# InverseofFactorial(mod)
# factorial(mod)
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
###CODE
tc = ri()
for _ in range(tc):
n=ri()
s=rs()
lp=n-1
fp=0
mp=1
count=0
while fp<mp<lp:
while fp<mp<lp and s[mp]!='M':
mp+=1
if not (fp<mp<lp):
break
else:
while fp<mp<lp and s[lp]!='T':
lp-=1
if not (fp<mp<lp):
break
else:
count+=1
lp-=1
fp+=1
mp+=1
if count*3==n:
ws("YES")
else:
ws("NO")
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
```
|
instruction
| 0
| 11,874
| 0
| 23,748
|
No
|
output
| 1
| 11,874
| 0
| 23,749
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council has a shared document file. Every day, some members of the student council write the sequence TMT (short for Towa Maji Tenshi) in it.
However, one day, the members somehow entered the sequence into the document at the same time, creating a jumbled mess. Therefore, it is Suguru Doujima's task to figure out whether the document has malfunctioned. Specifically, he is given a string of length n whose characters are all either T or M, and he wants to figure out if it is possible to partition it into some number of disjoint subsequences, all of which are equal to TMT. That is, each character of the string should belong to exactly one of the subsequences.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.
Input
The first line contains an integer t (1 ≤ t ≤ 5000) — the number of test cases.
The first line of each test case contains an integer n (3 ≤ n < 10^5), the number of characters in the string entered in the document. It is guaranteed that n is divisible by 3.
The second line of each test case contains a string of length n consisting of only the characters T and M.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, print a single line containing YES if the described partition exists, and a single line containing NO otherwise.
Example
Input
5
3
TMT
3
MTT
6
TMTMTT
6
TMTTTT
6
TTMMTT
Output
YES
NO
YES
NO
YES
Note
In the first test case, the string itself is already a sequence equal to TMT.
In the third test case, we may partition the string into the subsequences TMTMTT. Both the bolded and the non-bolded subsequences are equal to TMT.
Submitted Solution:
```
t=int(input())
def TMT(s,n):
t,m,f1,f2=0,0,1,1
i=0
while(i<n):
if i<n and s[i]=="T":
t=t+1
elif i<n and s[i]=="M":
f1=0
m=m+1
else:
f2=0
if m>t:
return True
else:
i=i-1
t,m,f1,f2=0,0,1,1
i=i+1
return False
for _ in range(t):
n=int(input())
s=input()
t=s.count("T")
m=s.count("M")
if (t!=(n//3)*2) or (m!=(n//3)) or (s[0]=="M") or (s[n-1]=="M") or TMT(s,n) or TMT(s[::-1],n):
print("NO")
else:
print("YES")
```
|
instruction
| 0
| 11,875
| 0
| 23,750
|
No
|
output
| 1
| 11,875
| 0
| 23,751
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
|
instruction
| 0
| 11,876
| 0
| 23,752
|
Tags: binary search, dp, greedy, implementation, strings, two pointers
Correct Solution:
```
def main():
t=int(input())
allans=[]
for _ in range(t):
s=input()
n=len(s)
dp0=[0]*n # max len of beautiful ending with 0 here
dp1=[0]*n # max len of beautiful ending with 1 here
# dpq0=[0]*n # max len of beautiful ending with ? here, if made to be 0
for i,x in enumerate(s):
if s[i]=='?':
if i-1>=0:
dp0[i]=1+dp1[i-1]
dp1[i]=1+dp0[i-1]
else:
dp0[i]=1
dp1[i]=1
elif s[i]=='0':
if i-1>=0:
dp0[i]=1+dp1[i-1]
else:
dp0[i]=1
else:
if i-1>=0:
dp1[i]=1+dp0[i-1]
else:
dp1[i]=1
cnts=0
for i in range(n):
if s[i]=='?':
cnts+=max(dp0[i],dp1[i])
else:
cnts+=dp0[i]+dp1[i]
# print(dp0)#
# print(dp1)#
allans.append(cnts)
multiLineArrayPrint(allans)
return
import sys
# input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
# def readFloatArr():
# return [float(x) for x in input().split()]
def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
dv=defaultValFactory;da=dimensionArr
if len(da)==1:return [dv() for _ in range(da[0])]
else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
def queryInteractive(i,j):
print('? {} {}'.format(i,j))
sys.stdout.flush()
return int(input())
def answerInteractive(ans):
print('! {}'.format(' '.join([str(x) for x in ans])))
sys.stdout.flush()
inf=float('inf')
MOD=10**9+7
# MOD=998244353
for _abc in range(1):
main()
```
|
output
| 1
| 11,876
| 0
| 23,753
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
|
instruction
| 0
| 11,877
| 0
| 23,754
|
Tags: binary search, dp, greedy, implementation, strings, two pointers
Correct Solution:
```
# Fast IO Region
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# Get out of main function
def main():
pass
# decimal to binary
def binary(n):
return (bin(n).replace("0b", ""))
# binary to decimal
def decimal(s):
return (int(s, 2))
# power of a number base 2
def pow2(n):
p = 0
while n > 1:
n //= 2
p += 1
return (p)
# if number is prime in √n time
def isPrime(n):
if (n == 1):
return (False)
else:
root = int(n ** 0.5)
root += 1
for i in range(2, root):
if (n % i == 0):
return (False)
return (True)
# list to string ,no spaces
def lts(l):
s = ''.join(map(str, l))
return s
# String to list
def stl(s):
# for each character in string to list with no spaces -->
l = list(s)
# for space in string -->
# l=list(s.split(" "))
return l
# Returns list of numbers with a particular sum
def sq(a, target, arr=[]):
s = sum(arr)
if (s == target):
return arr
if (s >= target):
return
for i in range(len(a)):
n = a[i]
remaining = a[i + 1:]
ans = sq(remaining, target, arr + [n])
if (ans):
return ans
# Sieve for prime numbers in a range
def SieveOfEratosthenes(n):
cnt = 0
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n + 1, p):
prime[i] = False
p += 1
for p in range(2, n + 1):
if prime[p]:
cnt += 1
# print(p)
return (cnt)
# for positive integerse only
def nCr(n, r):
f = math.factorial
return f(n) // f(r) // f(n - r)
# 1000000007
mod = int(1e9) + 7
def ssinp(): return input()
# s=input()
def iinp(): return int(input())
# n=int(input())
def nninp(): return map(int, input().split())
# a,b,c=map(int,input().split())
def llinp(): return list(map(int, input().split()))
# a=list(map(int,input().split()))
def p(xyz): print(xyz)
def p2(a, b): print(a, b)
import math
# import random
# sys.setrecursionlimit(300000)
# from fractions import Fraction
# from collections import OrderedDict
# from collections import deque
######################## mat=[[0 for i in range(n)] for j in range(m)] ########################
######################## list.sort(key=lambda x:x[1]) for sorting a list according to second element in sublist ########################
######################## Speed: STRING < LIST < SET,DICTIONARY ########################
######################## from collections import deque ########################
######################## ASCII of A-Z= 65-90 ########################
######################## ASCII of a-z= 97-122 ########################
######################## d1.setdefault(key, []).append(value) ########################
for __ in range(iinp()):
s=ssinp()
l=list(s)
n=len(l)
ans =ans1=0
curr = l[0]
cnt=1
ques=0
if(curr=="?"):
ques+=1
for i in range(1,n):
if(curr=="?"):
if(l[i]=="1"):
curr="0"
elif(l[i]=="0"):
curr="1"
else:
cnt+=1
ques+=1
if(curr=="1"):
if(l[i]=="0" or l[i]=="?"):
cnt+=1
curr="0"
if(l[i]=="?"):
ques+=1
else:
ques=0
else:
ans+=(cnt*(cnt+1))//2
cnt=1+ques
ans1+=(ques*(ques+1))//2
curr="1"
ques=0
elif(curr=="0"):
if(l[i]=="1" or l[i]=="?"):
cnt+=1
curr="1"
if (l[i] == "?"):
ques += 1
else:
ques = 0
else:
ans+=(cnt*(cnt+1))//2
cnt = 1 + ques
ans1 += (ques * (ques + 1)) // 2
curr="0"
ques=0
ans+=(cnt*(cnt+1))//2
print(ans-ans1)
```
|
output
| 1
| 11,877
| 0
| 23,755
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
|
instruction
| 0
| 11,878
| 0
| 23,756
|
Tags: binary search, dp, greedy, implementation, strings, two pointers
Correct Solution:
```
import sys
read=sys.stdin.buffer.read;readline=sys.stdin.buffer.readline;input=lambda:sys.stdin.readline().rstrip()
import bisect,string,math,time,functools,random,fractions
from bisect import*
from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations,groupby
rep=range;R=range
def I():return int(input())
def LI():return [int(i) for i in input().split()]
def LI_():return [int(i)-1 for i in input().split()]
def AI():return map(int,open(0).read().split())
def S_():return input()
def IS():return input().split()
def LS():return [i for i in input().split()]
def NI(n):return [int(input()) for i in range(n)]
def NI_(n):return [int(input())-1 for i in range(n)]
def NLI(n):return [[int(i) for i in input().split()] for i in range(n)]
def NLI_(n):return [[int(i)-1 for i in input().split()] for i in range(n)]
def StoLI():return [ord(i)-97 for i in input()]
def ItoS(n):return chr(n+97)
def LtoS(ls):return ''.join([chr(i+97) for i in ls])
def RLI(n=8,a=1,b=10):return [random.randint(a,b)for i in range(n)]
def RI(a=1,b=10):return random.randint(a,b)
def INP():
N=10
n=random.randint(1,N)
a=RLI(n,0,n-1)
return n,a
def Rtest(T):
case,err=0,0
for i in range(T):
inp=INP()
a1=naive(*inp)
a2=solve(*inp)
if a1!=a2:
print(inp)
print('naive',a1)
print('solve',a2)
err+=1
case+=1
print('Tested',case,'case with',err,'errors')
def GI(V,E,ls=None,Directed=False,index=1):
org_inp=[];g=[[] for i in range(V)]
FromStdin=True if ls==None else False
for i in range(E):
if FromStdin:
inp=LI()
org_inp.append(inp)
else:
inp=ls[i]
if len(inp)==2:a,b=inp;c=1
else:a,b,c=inp
if index==1:a-=1;b-=1
aa=(a,c);bb=(b,c);g[a].append(bb)
if not Directed:g[b].append(aa)
return g,org_inp
def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1):
#h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1) # sample usage
mp=[boundary]*(w+2);found={}
for i in R(h):
s=input()
for char in search:
if char in s:
found[char]=((i+1)*(w+2)+s.index(char)+1)
mp_def[char]=mp_def[replacement_of_found]
mp+=[boundary]+[mp_def[j] for j in s]+[boundary]
mp+=[boundary]*(w+2)
return h+2,w+2,mp,found
def TI(n):return GI(n,n-1)
def accum(ls):
rt=[0]
for i in ls:rt+=[rt[-1]+i]
return rt
def bit_combination(n,base=2):
rt=[]
for tb in R(base**n):s=[tb//(base**bt)%base for bt in R(n)];rt+=[s]
return rt
def gcd(x,y):
if y==0:return x
if x%y==0:return y
while x%y!=0:x,y=y,x%y
return y
def YN(x):print(['NO','YES'][x])
def Yn(x):print(['No','Yes'][x])
def show(*inp,end='\n'):
if show_flg:print(*inp,end=end)
mo=10**9+7
#mo=998244353
inf=float('inf')
FourNb=[(-1,0),(1,0),(0,1),(0,-1)];EightNb=[(-1,0),(1,0),(0,1),(0,-1),(1,1),(-1,-1),(1,-1),(-1,1)];compas=dict(zip('WENS',FourNb));cursol=dict(zip('LRUD',FourNb))
alp=[chr(ord('a')+i)for i in range(26)]
#sys.setrecursionlimit(10**7)
def gcj(c,x):
print("Case #{0}:".format(c+1),x)
show_flg=False
show_flg=True
import sys
read=sys.stdin.buffer.read;readline=sys.stdin.buffer.readline;input=lambda:sys.stdin.readline().rstrip()
def gcd(x,y):
if y==0:return x
if x%y==0:return y
while x%y!=0:x,y=y,x%y
return y
for i in range(int(input())):
s=list(input())
n=len(s)
s+=' ',
s0=[0]
for i in range(n):
s0+=1-s0[-1],
s1=[1]
for i in range(n):
s1+=1-s1[-1],
x=[]
l=0
for i in range(n+1):
c=s[i]
if c=='?' or c==str(s0[i]):
l+=1
else:
if l==0:
continue
x+=l,
l=0
y=[]
z=[]
l=0
r=0
for i in range(n+1):
c=s[i]
if c=='?' or c==str(s1[i]):
l+=1
else:
if l==0:
continue
y+=l,
l=0
if c=='?':
r+=1
else:
if r==0:
continue
z+=r,
r=0
ans=0
ans+=sum(i*(i+1)//2for i in x)
ans+=sum(i*(i+1)//2for i in y)
ans-=sum(i*(i+1)//2for i in z)
#show(x,y,z)
print(ans)
```
|
output
| 1
| 11,878
| 0
| 23,757
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
|
instruction
| 0
| 11,879
| 0
| 23,758
|
Tags: binary search, dp, greedy, implementation, strings, two pointers
Correct Solution:
```
import bisect
import copy
import decimal
import fractions
import heapq
import itertools
import math
import random
import sys
from collections import Counter, deque,defaultdict
from functools import lru_cache,reduce
from heapq import heappush,heappop,heapify,_heappop_max,_heapify_max
def _heappush_max(heap,item):
heap.append(item)
heapq._siftdown_max(heap,0,len(heap)-1)
from math import gcd as Gcd
read=sys.stdin.read
readline=sys.stdin.readline
readlines=sys.stdin.readlines
t=int(readline())
for _ in range(t):
S=readline().rstrip()
N=len(S)
idx=[None]*N
bound=[]
for i in range(1,N):
if S[i-1]!='?':
idx[i]=i-1
else:
idx[i]=idx[i-1]
for i in range(N):
if idx[i]==None:
continue
if S[i]=='?':
continue
if (i%2==idx[i]%2)!=(S[i]==S[idx[i]]):
bound.append((idx[i],i+1))
L,R=[0],[]
for i,j in bound:
L.append(i+1)
R.append(j-1)
R.append(N)
ans=0
for l,r in zip(L,R):
d=r-l+1
ans+=d*(d-1)//2
for i,j in bound:
d=j-i-1
ans-=d*(d-1)//2
print(ans)
```
|
output
| 1
| 11,879
| 0
| 23,759
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
|
instruction
| 0
| 11,880
| 0
| 23,760
|
Tags: binary search, dp, greedy, implementation, strings, two pointers
Correct Solution:
```
import math
t=int(input())
for _ in range(t):
s=input()
b=[]
for j in s:
b.append(j)
n=len(b)
s=[]
ans=1
dp=[0 for i in range(n)]
la=0
dp[0]=1
for j in range(1,n):
if b[la]=="?":
dp[j]=j-la+1
else:
if b[j]=="?":
dp[j]=j-la+dp[la]
else:
if (j-la)%2:
if b[j]!=b[la]:
dp[j]=dp[la]+(j-la)
else:
dp[j]=j-la
else:
if b[j] == b[la]:
dp[j] = dp[la] + (j - la)
else:
dp[j] = j - la
if b[j]!="?":
la=j
print(sum(dp))
```
|
output
| 1
| 11,880
| 0
| 23,761
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
|
instruction
| 0
| 11,881
| 0
| 23,762
|
Tags: binary search, dp, greedy, implementation, strings, two pointers
Correct Solution:
```
from sys import stdin
from math import gcd
input=lambda:stdin.readline().strip()
for _ in range(int(input())):
s=input()
n=len(s)
ans=-1
i=0
nd=0
sum1=0
while i<n:
if s[i]=='?':
if ans!=-1:
ans=ans^1
nd+=1
i+=1
else:
nd+=1
i+=1
else:
if ans==-1:
ans=int(s[i])^1
i+=1
nd+=1
else:
if ans!=int(s[i]):
#print("nd ",nd,i,ans,s[i])
sum1+=((nd*(nd+1))//2)
nd=0
nk=0
while i>0:
if s[i-1]=='?':
i-=1
nk+=1
else:
break
sum1-=((nk*(nk+1))//2)
ans=-1
else:
ans=ans^1
nd+=1
i+=1
#print(i)
if nd!=0:
sum1+=((nd*(nd+1))//2)
print(sum1)
```
|
output
| 1
| 11,881
| 0
| 23,763
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
|
instruction
| 0
| 11,882
| 0
| 23,764
|
Tags: binary search, dp, greedy, implementation, strings, two pointers
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import math
from collections import Counter
def func(array):
num = 0
last_question = None
last_number = None
start = None
first = None
first_index = None
for i, val in enumerate(array):
# print("Start", i, num, first_index, start, last_question)
if first is None:
if val != "?":
first = val
first_index = i
last_question = None
last_number = i
if start is None:
start = i
elif last_question is None:
last_question = i
start = i
num += i - start + 1
else:
if val == "?":
num += i - start + 1
if last_question is None or last_question < last_number:
last_question = i
elif (val == first and (i - first_index) % 2 == 0) or (
val != first and (i - first_index) % 2 != 0
):
num += i - start + 1
last_number = i
last_question = None
else:
# num += ((i - start) * (i - start - 1)) // 2
# print("last")
if last_question is not None and last_question > first_index:
# print("Here")
start = last_question
else:
start = i
last_question = None
first_index = i
first = val
last_number = i
num += i - start + 1
# print(i, val, num, first_index, start, last_question)
# num += ((len(array) - start) * (len(array) - 1 - start - 1)) // 2
return num
def main():
num_test = int(parse_input())
result = []
for _ in range(num_test):
array = parse_input()
result.append(func(array))
print("\n".join(map(str, result)))
# 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)
parse_input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
|
output
| 1
| 11,882
| 0
| 23,765
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting of the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
|
instruction
| 0
| 11,883
| 0
| 23,766
|
Tags: binary search, dp, greedy, implementation, strings, two pointers
Correct Solution:
```
from collections import Counter
def unstable(s):
P = [None if c == '?' else ((c == '1') == (i % 2))
for (i, c) in enumerate(s)]
q = 0
k = 0
r = None
u = 0
for p in P:
if p is None:
q += 1
k += 1
elif p == r:
q = 0
k += 1
else:
u += k * (k+1) // 2
u -= q * (q+1) // 2
k = q + 1
q = 0
r = p
u += k * (k + 1) // 2
return u
def main():
t = readint()
print('\n'.join(map(str, (unstable(input()) for _ in range(t)))))
##########
import sys
def readint():
return int(input())
def readinti():
return map(int, input().split())
def readintt():
return tuple(readinti())
def readintl():
return list(readinti())
def readinttl(k):
return [readintt() for _ in range(k)]
def readintll(k):
return [readintl() for _ in range(k)]
def log(*args, **kwargs):
print(*args, **kwargs, file=sys.stderr)
if __name__ == '__main__':
main()
```
|
output
| 1
| 11,883
| 0
| 23,767
|
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 the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
Submitted Solution:
```
bu=int(input())
for j in range(bu):
ap=input()
i=len(ap)
for az in range(len(ap)):
if ap[az]!='?':
i=az+1
digit=ap[az]
count=az+1
lastquestionmark=0
break
else:pass
if i==len(ap):
print((len(ap)*(len(ap)+1))//2)
continue
ans=len(ap)
while i<len(ap):
if ap[i]=='?':
if ap[i-1]!='?':
lastquestionmark=i
count+=1
i+=1
if digit=='1':
digit='0'
else :
digit='1'
else:
count+=1
i+=1
if digit=='1':
digit='0'
else :
digit='1'
elif digit==ap[i]:
ans=ans+((count*(count-1))//2)
if ap[i-1]=='?' and lastquestionmark!=i-1:
i+=1
count=i-lastquestionmark
ans-=((count-1)*(count-2))//2
elif ap[i-1]=='?' and lastquestionmark==i-1:
count=2
i+=1
else:
i+=1
count=1
else:
i+=1
count+=1
if digit=='1':
digit='0'
else :digit='1'
ans+=(count*(count-1))//2
print(ans)
```
|
instruction
| 0
| 11,884
| 0
| 23,768
|
Yes
|
output
| 1
| 11,884
| 0
| 23,769
|
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 the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
Submitted Solution:
```
import os, sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
for _ in range(int(input())):
a = [i for i in input()]
n = len(a)
i = 0
j = 0
ans = 0
ct = 0
while i < n:
while j < n:
if i == j:
if a[j] == '?':
ct += 1
j += 1
else:
if a[j] == '?':
ct += 1
j += 1
else:
if ct == j - i:
j += 1
else:
if a[j - ct - 1] == a[j]:
if ct % 2:
j += 1
else:
while i != j - ct:
ans += j - i
i += 1
else:
if ct % 2:
while i != j - ct:
ans += j - i
i += 1
else:
j += 1
ct = 0
break
while i != j - 1:
ans += j - i
i += 1
print(ans + 1)
```
|
instruction
| 0
| 11,885
| 0
| 23,770
|
Yes
|
output
| 1
| 11,885
| 0
| 23,771
|
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 the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
Submitted Solution:
```
t=int(input())
for test in range(t):
s=input()
n=len(s)
ans=0
l=0
r=0
pr=0
while r<n:
p=True
zp=-1
z=-1
op=-2
o=-2
while r<n and p:
if s[r]=='1':
zp=r%2
if zp==op:
#r-=1
break
z=r
op=1-zp
elif s[r]=='0':
op=r%2
if zp==op:
#r-=1
break
o=r
zp=1-op
r+=1
b=r-l
c=pr-l
#b-=1
ans+=b*(b+1)//2
ans-=c*(c+1)//2
#print(*[l,r])
l=max(z,o)+1
pr=r
print(ans)
```
|
instruction
| 0
| 11,886
| 0
| 23,772
|
Yes
|
output
| 1
| 11,886
| 0
| 23,773
|
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 the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
Submitted Solution:
```
for _ in range(int(input())):
s = list(input())
ret = 0
dp = [[0, 0] for _ in range(len(s) + 1)]
for i in range(1, len(s) + 1):
if s[i-1] != '0':
dp[i][0] = dp[i - 1][1] + 1
if s[i-1] != '1':
dp[i][1] = dp[i - 1][0] + 1
ret += max(dp[i][0], dp[i][1])
print(ret)
```
|
instruction
| 0
| 11,887
| 0
| 23,774
|
Yes
|
output
| 1
| 11,887
| 0
| 23,775
|
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 the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
Submitted Solution:
```
def Convert(string):
list1=[]
list1[:0]=string
return list1
for _ in range(int(input())):
s = Convert(input())
n = len(s)
count = 0
for i in range(n):
start = s[i]
qcount = 0
found = False
temp = count
for j in range(i,n):
# print("j is " ,j)
if (i == j):
count += 1
else:
if (s[j] != '?'):
if (s[j] == s[j-1]):
break
else:
if (found == False):
count += 1
else:
# print("i is ",i," qcount is ",qcount," start is ",start," s[j] is ",s[j]," count is ",count)
found = False
if (start == s[j]):
if (qcount%2 == 1):
count += 1
else:
break
else:
if (qcount%2 == 0):
count += 1
else:
# print("break")
break
qcount = 0
else:
if (found == False):
found = True
start = s[j-1]
qcount += 1
count += 1
# print(count-temp)
print(count)
```
|
instruction
| 0
| 11,888
| 0
| 23,776
|
No
|
output
| 1
| 11,888
| 0
| 23,777
|
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 the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
Submitted Solution:
```
def compliment(a):
if a == "1":
return "0"
else:
return "1"
def calc(n):
return n*(n+1)//2
for _ in range(int(input())):
s = input()
n = len(s)
i = 0
j = 0
substring = [(0,0)]
while j < n and s[j] == "?":
j += 1
if j == n:
print(n * (n+1)//2)
else:
curr = s[j]
j += 1
while j < n:
if s[j] == "?":
curr = compliment(curr)
j+=1
else:
if s[j] != curr:
curr = compliment(curr)
j += 1
else:
substring.append((i,j-1))
if s[j-1] == "?":
i = j-1
curr = s[j]
else:
i = j
curr = s[j]
j += 1
substring.append((i,j-1))
#print(substring)
substring = substring[::-1]
ans = 0
for i in range(len(substring)):
ans += calc((substring[i][1] - substring[i][0])+1)
if substring[i][0] == 0:
break
if substring[i][0] == substring[i+1][1]:
ans -= 1
print(ans)
```
|
instruction
| 0
| 11,889
| 0
| 23,778
|
No
|
output
| 1
| 11,889
| 0
| 23,779
|
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 the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
Submitted Solution:
```
from math import gcd
for _ in range(int(input())):
s = input()
n = len(s)
ans = 0
curr = 0
prevans = 0
prev = None
dic = {'0':'1', '1':'0'}
for i in s:
addn = 0
if i != prev and prev!=None:
ans += prevans+1
addn+=prevans+1
else:
ans += curr+1
addn += curr+1
prevans = addn
if i == '?':
curr += 1
else:
curr = 0
if i != '?':
prev = i
elif prev:
prev = dic[prev]
print(i, ans, prev, prevans)
print(ans)
```
|
instruction
| 0
| 11,890
| 0
| 23,780
|
No
|
output
| 1
| 11,890
| 0
| 23,781
|
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 the characters 0, 1, and ?.
Let's call a string unstable if it consists of the characters 0 and 1 and any two adjacent characters are different (i. e. it has the form 010101... or 101010...).
Let's call a string beautiful if it consists of the characters 0, 1, and ?, and you can replace the characters ? to 0 or 1 (for each character, the choice is independent), so that the string becomes unstable.
For example, the strings 0??10, 0, and ??? are beautiful, and the strings 00 and ?1??1 are not.
Calculate the number of beautiful contiguous substrings of the string s.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
The first and only line of each test case contains the string s (1 ≤ |s| ≤ 2 ⋅ 10^5) consisting of characters 0, 1, and ?.
It is guaranteed that the sum of the string lengths over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, output a single integer — the number of beautiful substrings of the string s.
Example
Input
3
0?10
???
?10??1100
Output
8
6
25
Submitted Solution:
```
def flip(a,x="?"):
if a=="1":
return("0")
elif a=="0":
return("1")
elif x=="?":
return "?"
else:
return x
for i in range(int(input())):
s=input()
ans=0
qlen=0
len=1
if s[0]=="?":
qlen+=1
ch=s[0]
for i in s[1:]:
if i=="?":
qlen+=1
len+=1
ch=flip(ch,i)
if i=="0":
if ch=="0":
ch=i
ans+=len*(len+1)/2-qlen*(qlen+1)/2
len=1
qlen=0
else:
qlen=0
len+=1
ch=flip(ch,i)
if i=="1":
if ch=="1":
ch=i
ans+=len*(len+1)/2-qlen*(qlen+1)/2
len=1
qlen=0
else:
qlen=0
len+=1
ch=flip(ch,i)
print(ans)
```
|
instruction
| 0
| 11,891
| 0
| 23,782
|
No
|
output
| 1
| 11,891
| 0
| 23,783
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
|
instruction
| 0
| 11,928
| 0
| 23,856
|
Tags: implementation
Correct Solution:
```
from math import *
from collections import deque
from copy import deepcopy
import sys
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def multi(): return map(int,input().split())
def strmulti(): return map(str, inp().split())
def lis(): return list(map(int, inp().split()))
def lcm(a,b): return (a*b)//gcd(a,b)
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def stringlis(): return list(map(str, inp().split()))
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def printlist(a) :
print(' '.join(str(a[i]) for i in range(len(a))))
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
#copied functions end
#start coding
s=str(inp())
if(s.count('x')>s.count('y')):
print('x'*(abs(s.count('x')-s.count('y'))))
else:
print('y' * (abs(s.count('x') - s.count('y'))))
```
|
output
| 1
| 11,928
| 0
| 23,857
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
|
instruction
| 0
| 11,929
| 0
| 23,858
|
Tags: implementation
Correct Solution:
```
from sys import stdin
from collections import defaultdict
input = stdin.readline
# ~ T = int(input())
T = 1
for t in range(1,T + 1):
s = input()
x = 0; y = 0
for i in s:
x += (i == 'x')
y += (i == 'y')
if x > y:
for i in range((x - y)):
print('x',end = "")
else:
for i in range((y - x)):
print('y',end = "")
```
|
output
| 1
| 11,929
| 0
| 23,859
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
|
instruction
| 0
| 11,930
| 0
| 23,860
|
Tags: implementation
Correct Solution:
```
#"from dust i have come, dust i will be"
s=input()
x=0;y=0
for i in range(len(s)):
if(s[i]=='x'):
x+=1
else:
y+=1
t=""
if(x>y):
for i in range(x-y):
t+="x"
else:
for i in range(y-x):
t+="y"
print(t)
```
|
output
| 1
| 11,930
| 0
| 23,861
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
|
instruction
| 0
| 11,931
| 0
| 23,862
|
Tags: implementation
Correct Solution:
```
s = input()
y = s.count("y")
x = s.count("x")
t = abs(y-x)
if y > x:
print("y"*t)
else:
print("x"*t)
```
|
output
| 1
| 11,931
| 0
| 23,863
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
|
instruction
| 0
| 11,932
| 0
| 23,864
|
Tags: implementation
Correct Solution:
```
s = input()
x = 0
y = 0
for i in range(len(s)):
if s[i] == 'x':
x += 1
else:
y += 1
if x > y :
print('x' * (x - y))
elif x < y :
print('y' * (y - x))
else:
print('')
```
|
output
| 1
| 11,932
| 0
| 23,865
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
|
instruction
| 0
| 11,933
| 0
| 23,866
|
Tags: implementation
Correct Solution:
```
s = input()
x, y = s.count("x"), s.count("y")
if x > y:
print("x"*(x-y))
else:
print("y"*(y-x))
```
|
output
| 1
| 11,933
| 0
| 23,867
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
|
instruction
| 0
| 11,934
| 0
| 23,868
|
Tags: implementation
Correct Solution:
```
def main():
stack = []
for c in input():
if stack and stack[-1] != c:
del stack[-1]
else:
stack.append(c)
print(''.join(reversed(stack)))
if __name__ == '__main__':
main()
```
|
output
| 1
| 11,934
| 0
| 23,869
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
|
instruction
| 0
| 11,935
| 0
| 23,870
|
Tags: implementation
Correct Solution:
```
s = input()
xs = s.count("x")
ys = s.count("y")
if xs > ys:
print("x" * (xs - ys))
else:
print("y" * (ys - xs))
```
|
output
| 1
| 11,935
| 0
| 23,871
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
Submitted Solution:
```
a = list(input())
x = [x for x in a if x != 'y']
y = [x for x in a if x != 'x']
lenx = len(x)
leny = len(y)
m = abs(leny-lenx)
if lenx > leny :
jawab = 'x'
else :
jawab = 'y'
print(jawab*m)
```
|
instruction
| 0
| 11,936
| 0
| 23,872
|
Yes
|
output
| 1
| 11,936
| 0
| 23,873
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
Submitted Solution:
```
from sys import stdin
# ~ from collections import defaultdict
input = stdin.readline
# ~ T = int(input())
T = 1
for t in range(1,T + 1):
s = input()
x = 0; y = 0
for i in s:
x += (i == 'x')
y += (i == 'y')
if x > y:
for i in range((x - y)):
print('x',end = "")
else:
for i in range((y - x)):
print('y',end = "")
```
|
instruction
| 0
| 11,937
| 0
| 23,874
|
Yes
|
output
| 1
| 11,937
| 0
| 23,875
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
Submitted Solution:
```
s = input()
x = s.count('x')
y = s.count('y')
if x > y:
print('x' * (x - y))
else:
print('y' * (y - x))
```
|
instruction
| 0
| 11,938
| 0
| 23,876
|
Yes
|
output
| 1
| 11,938
| 0
| 23,877
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
Submitted Solution:
```
string = input()
y = 0
for s in string:
if s == 'y':
y += 1
x = len(string) - y
if x > y:
print('x'*(x-y))
elif y > x:
print('y'*(y-x))
```
|
instruction
| 0
| 11,939
| 0
| 23,878
|
Yes
|
output
| 1
| 11,939
| 0
| 23,879
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
Submitted Solution:
```
s = input()
s = list(s)
c = 0
size = len(s)
index = 0
while c==0:
if index==len(s)-1:
c=1
elif s[index]=='y' and s[index+1]=='x':
s[index]='x'
s[index+1]='y'
index += 1
index = 0
while c==1:
if index+1 == len(s):
c=2
elif s[index]=='x' and s[index+1]=='y':
del(s[index])
del(s[index])
index = 0
else:
index += 1
s = ''.join(s)
print(s)
```
|
instruction
| 0
| 11,940
| 0
| 23,880
|
No
|
output
| 1
| 11,940
| 0
| 23,881
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
Submitted Solution:
```
s=input()
while "yx" in s:
s=s.replace("yx","xy").replace("xy","")
print(s.replace("xy",""))
```
|
instruction
| 0
| 11,941
| 0
| 23,882
|
No
|
output
| 1
| 11,941
| 0
| 23,883
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
Submitted Solution:
```
d=input()
print(d.replace("yx", "").replace("xy", ""))
```
|
instruction
| 0
| 11,942
| 0
| 23,884
|
No
|
output
| 1
| 11,942
| 0
| 23,885
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Vitaly loves different algorithms. Today he has invented a new algorithm just for you. Vitaly's algorithm works with string s, consisting of characters "x" and "y", and uses two following operations at runtime:
1. Find two consecutive characters in the string, such that the first of them equals "y", and the second one equals "x" and swap them. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
2. Find in the string two consecutive characters, such that the first of them equals "x" and the second one equals "y". Remove these characters from the string. If there are several suitable pairs of characters, we choose the pair of characters that is located closer to the beginning of the string.
The input for the new algorithm is string s, and the algorithm works as follows:
1. If you can apply at least one of the described operations to the string, go to step 2 of the algorithm. Otherwise, stop executing the algorithm and print the current string.
2. If you can apply operation 1, then apply it. Otherwise, apply operation 2. After you apply the operation, go to step 1 of the algorithm.
Now Vitaly wonders, what is going to be printed as the result of the algorithm's work, if the input receives string s.
Input
The first line contains a non-empty string s.
It is guaranteed that the string only consists of characters "x" and "y". It is guaranteed that the string consists of at most 106 characters. It is guaranteed that as the result of the algorithm's execution won't be an empty string.
Output
In the only line print the string that is printed as the result of the algorithm's work, if the input of the algorithm input receives string s.
Examples
Input
x
Output
x
Input
yxyxy
Output
y
Input
xxxxxy
Output
xxxx
Note
In the first test the algorithm will end after the first step of the algorithm, as it is impossible to apply any operation. Thus, the string won't change.
In the second test the transformation will be like this:
1. string "yxyxy" transforms into string "xyyxy";
2. string "xyyxy" transforms into string "xyxyy";
3. string "xyxyy" transforms into string "xxyyy";
4. string "xxyyy" transforms into string "xyy";
5. string "xyy" transforms into string "y".
As a result, we've got string "y".
In the third test case only one transformation will take place: string "xxxxxy" transforms into string "xxxx". Thus, the answer will be string "xxxx".
Submitted Solution:
```
# link: https://codeforces.com/problemset/problem/255/B
def solve(s):
n = len(s)
if n == 1:
return (s)
# else:
answer = ""
i = 0
while i<n-1:
if s[i] + s[i+1] in ["xy", "yx"]:
i += 2
if i==n-1:
answer += s[i]
else:
answer += s[i]
i += 1
return (answer)
if __name__ == "__main__":
s = input()
result = solve(s)
print(result)
```
|
instruction
| 0
| 11,943
| 0
| 23,886
|
No
|
output
| 1
| 11,943
| 0
| 23,887
|
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