message stringlengths 2 23.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it...
Input
The first and only line of input contains a single nonempty string s of length at most 1000 composed of lowercase letters (a-z).
Output
Output YES if the string s contains heidi as a subsequence and NO otherwise.
Examples
Input
abcheaibcdi
Output
YES
Input
hiedi
Output
NO
Note
A string s contains another string p as a subsequence if it is possible to delete some characters from s and obtain p. | instruction | 0 | 35,665 | 0 | 71,330 |
Tags: implementation, strings
Correct Solution:
```
s=input()
p="heidi"
j=0
lst=[]
for i in range(len(p)):
while j<len(s):
if p[i]==s[j]:
lst.append(s[j])
j=j+1
break
else:
j=j+1
ls="".join(lst)
if ls==p:
print("YES")
else:
print("NO")
``` | output | 1 | 35,665 | 0 | 71,331 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it...
Input
The first and only line of input contains a single nonempty string s of length at most 1000 composed of lowercase letters (a-z).
Output
Output YES if the string s contains heidi as a subsequence and NO otherwise.
Examples
Input
abcheaibcdi
Output
YES
Input
hiedi
Output
NO
Note
A string s contains another string p as a subsequence if it is possible to delete some characters from s and obtain p. | instruction | 0 | 35,666 | 0 | 71,332 |
Tags: implementation, strings
Correct Solution:
```
string = input()
now = 0
for i in range(len(string)):
if now == 0 and string[i] == "h":
now += 1
if now == 1 and string[i] == "e":
now += 1
if now == 2 and string[i] == "i":
now += 1
if now == 3 and string[i] == "d":
now += 1
if now == 4 and string[i] == "i":
now += 1
if now == 5:
print("YES")
else:
print("NO")
``` | output | 1 | 35,666 | 0 | 71,333 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it...
Input
The first and only line of input contains a single nonempty string s of length at most 1000 composed of lowercase letters (a-z).
Output
Output YES if the string s contains heidi as a subsequence and NO otherwise.
Examples
Input
abcheaibcdi
Output
YES
Input
hiedi
Output
NO
Note
A string s contains another string p as a subsequence if it is possible to delete some characters from s and obtain p. | instruction | 0 | 35,667 | 0 | 71,334 |
Tags: implementation, strings
Correct Solution:
```
test='heidi'
s=input()
count=0
isFake=False
for i in range(len(s)):
if(s[i]==test[count]):
count=count+1
if(count>=len(test)):
isFake=True
break
if(isFake): print('YES')
else : print('NO')
``` | output | 1 | 35,667 | 0 | 71,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 35,693 | 0 | 71,386 |
Tags: constructive algorithms
Correct Solution:
```
k = int(input())
if k == 1:
print('fall')
exit(0)
if k == 3:
print('september')
exit(0)
# от порядка объединения строк ничего не зависит
# для каждой буквы, итоговая стоимость будет n * (n - 1) / 2
# жадно наберём ответ
if not k:
print('a')
exit(0)
op_count = [0, 0]
for i in range(1, 500):
op_count.append(op_count[-1] + i)
if op_count[-1] > k:
break
ans = []
letter = 'a'
for i in range(len(op_count) - 1, 1, -1):
while op_count[i] <= k:
k -= op_count[i]
ans.append(letter * i)
letter = chr(ord(letter) + 1)
print(''.join(ans))
``` | output | 1 | 35,693 | 0 | 71,387 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 35,694 | 0 | 71,388 |
Tags: constructive algorithms
Correct Solution:
```
n=int(input())
S='abcdefghijklmnopqrstuvwxyz'
S1=''
p=0
while n>0 :
l=1
r=100000
while l<r :
m=(r+l)//2
if ((1+m)/2)*m<=n :
l=m+1
else :
r=m
r=r
S1=S1+(S[p]*(r))
p=p+1
n=n-int((r)/2*(r-1))
if S1=='' :
print('ab')
else :
print(S1)
``` | output | 1 | 35,694 | 0 | 71,389 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 35,695 | 0 | 71,390 |
Tags: constructive algorithms
Correct Solution:
```
k,s=int(input()),""
for c in range(26):
i=0
while i<k:
i+=1
k-=i
s+=chr(c+97)*(i+1)
print(s)
``` | output | 1 | 35,695 | 0 | 71,391 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 35,696 | 0 | 71,392 |
Tags: constructive algorithms
Correct Solution:
```
k=int(input())
#print(k)
last=1
counts=[]
arr=[]
if k==0:
print('ab')
else:
for i in range(1,449):
arr.append(int(i*(i-1)/2))
while k>0:
#print(k)
for i in range(len(arr)):
if arr[i]>k:
k-=arr[i-1]
last=arr[i-1]
counts.append(i)
#print(k)
break
#k=k+last
#counts.append(k)
for i in range(len(counts)):
for j in range(counts[i]):
print(chr(i+97),end='')
``` | output | 1 | 35,696 | 0 | 71,393 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 35,697 | 0 | 71,394 |
Tags: constructive algorithms
Correct Solution:
```
from sys import stdin, stdout
import math
def solution():
n = int(stdin.readline().rstrip())
ch = 'a'
k = 0
if n == 0:
stdout.write(ch)
while n>0:
if k>n:
ch = chr(ord(ch) + 1)
k = 0
stdout.write(ch)
n-=k
k+=1
if __name__ == '__main__':
solution()
``` | output | 1 | 35,697 | 0 | 71,395 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 35,698 | 0 | 71,396 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input());s=n
arr =[ ]
while (True):
for i in range(0,s+3):
if (i*(i-1)*(1/2)<=n):continue
else :
arr.append(i-1);n-=(i-1)*(i-2)*(1/2);break
if (n<=0):break
j=0
for i in 'abcdefghijklmnobqrstuvwxyz':
while (arr[j]!=0):print(i,end='');arr[j]-=1
j+=1
if (j==len(arr)):break
``` | output | 1 | 35,698 | 0 | 71,397 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 35,699 | 0 | 71,398 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
alpha = "abcdefghijklmnopqrstuvwxyz"
s = alpha
ind = 0
cur = 1
while n > 0:
if cur > n:
ind += 1
cur = 1
else:
n -= cur
cur += 1
s += alpha[ind]
print(s)
``` | output | 1 | 35,699 | 0 | 71,399 |
Provide tags and a correct Python 3 solution for this coding contest problem.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | instruction | 0 | 35,700 | 0 | 71,400 |
Tags: constructive algorithms
Correct Solution:
```
k=int(input())
if k==0:
print("a")
exit()
ans=[]
al="a"
while k>0:
m=0
while m*(m+1)//2<k:
m+=1
if m*(m+1)//2>k:
m-=1
k-=m*(m+1)//2
ans.extend([al]*(m+1))
al=chr(ord(al)+1)
print("".join(ans))
``` | output | 1 | 35,700 | 0 | 71,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
def main():
n = int(input())
if n == 0:
print("a")
return
x = [1]
c = 2
while x[-1] < n:
x.append(x[-1] + c)
c += 1
y = [0] * len(x)
for i in range(len(x) - 1, -1, -1):
while x[i] * (y[i] + 1) <= n:
y[i] += 1
n -= x[i] * y[i]
a = []
c = ord('a')
for i in range(len(x)):
if y[i] != 0:
for j in range(y[i]):
t = chr(c)
for z in range(i + 2):
a.append(t)
c += 1
print("".join(a))
main()
``` | instruction | 0 | 35,701 | 0 | 71,402 |
Yes | output | 1 | 35,701 | 0 | 71,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
from collections import defaultdict
from itertools import permutations
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
k=int(input())
copy=k
r=list()
c=97
while k>0:
i=1
s=0
while s+i<=k:
s+=i
i+=1
for j in range (i):
r.append(chr(c))
c+=1
k-=s
if copy==0:
print("a")
else:
print(*r,sep='')
``` | instruction | 0 | 35,702 | 0 | 71,404 |
Yes | output | 1 | 35,702 | 0 | 71,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
from sys import stdin
from math import sqrt
n = int(stdin.readline().rstrip())
if n == 0:
print('a')
N = n * 2
charset = 'abcdefghijklmnopqrstuvwxyz'
sqr = int(sqrt(N)) // 2 + 1
for i in range(1, sqr + 1):
for j in range(1, i + 1):
for k in range(1, 26 + 1):
N_ = N - k * i * (i + 1)
if N_ == 0:
s1 = charset[0:k] * (i + 1)
print(s1)
exit()
if N_ % (j * (j + 1)) == 0 and N_ >= 0:
l = N_ // (j * (j + 1))
if l > 26 - k:
continue
s1 = charset[0:k] * (i + 1)
s2 = charset[k:k + l] * (j + 1)
print(s1 + s2)
exit()
``` | instruction | 0 | 35,703 | 0 | 71,406 |
Yes | output | 1 | 35,703 | 0 | 71,407 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
k = int(input())
ans = ""
cost = 0
for c in [chr(ord('a') + i) for i in range(26)]:
i = 0
while cost + i <= k:
ans += c
cost += i
i += 1
print(ans)
``` | instruction | 0 | 35,704 | 0 | 71,408 |
Yes | output | 1 | 35,704 | 0 | 71,409 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
n = int(input())
lis=[0]*(n+10)
for i in range(1,n+5):
lis[i]=i+lis[i-1]
i=n+1
#print(lis)
cc='abcdefghijklmnopqrstuvwxyz'
c=0
ans=''
while n>0:
while n<lis[i]:
i-=1
z=min(lis[i],n)
n-=z
ans+=(cc[c]*(i+1))
c+=1
print(ans)
``` | instruction | 0 | 35,705 | 0 | 71,410 |
No | output | 1 | 35,705 | 0 | 71,411 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
import string
l = string.ascii_lowercase
n = int(input())
if n==0:
print(l[0])
else:
fact = []
temp = 1
i = 1
while temp<=n:
fact.append(temp)
i+=1
temp = temp*i
fact.reverse()
for i in range(len(fact)):
if n%fact[i]==0:
temp = len(fact)-i
break
if fact[temp-1]==1:
s = ""
for i in range(n):
s+=l[i]*2
print(s)
else:
print(l[:(n//fact[len(fact)-(temp)])]*(temp+1))
``` | instruction | 0 | 35,706 | 0 | 71,412 |
No | output | 1 | 35,706 | 0 | 71,413 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
n = int(input())
cost = [None] * 100001
cost[0] = 0
# def get_cost(k):
# global cost
# if cost[k] is None:
# cost[k] = compute_cost(k)
# return cost[k]
# def compute_cost(k):
# global cost
# less = k // 2
# more = k - less
# return get_cost(less) + get_cost(more) + less * more
# for i in range(50000, len(cost)):
# cost[i] = compute_cost(i)
for i in range(1, len(cost)):
cost[i] = cost[i - 1] + i - 1
def compute_ans(k):
global cost
i = 0
while cost[i] < k:
i += 1
if cost[i] == k:
return [i]
return [i - 1] + compute_ans(k - cost[i - 1])
s = compute_ans(n)
res = ''
for i, cnt in zip('abcdefghijklmnopqrstuvwxyz', s):
res += i * cnt
print(res)
``` | instruction | 0 | 35,707 | 0 | 71,414 |
No | output | 1 | 35,707 | 0 | 71,415 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:
* Remove any two elements s and t from the set, and add their concatenation s + t to the set.
The cost of such operation is defined to be <image>, where f(s, c) denotes the number of times character c appears in string s.
Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.
Input
The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.
Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Examples
Input
12
Output
abababab
Input
3
Output
codeforces
Note
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
* {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
* {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "a", "b", "a", "b"}, with a cost of 1;
* {"abab", "ab", "a", "b"}, with a cost of 0;
* {"abab", "aba", "b"}, with a cost of 1;
* {"abab", "abab"}, with a cost of 1;
* {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
Submitted Solution:
```
n = int(input())
let = ord('a')
res = ""
while n != 0:
i = 1
while i * (i - 1) // 2 <= n:
i += 1
i -= 1
res += chr(let) * i
n = n - i * (i - 1) // 2
let += 1
print(res)
``` | instruction | 0 | 35,708 | 0 | 71,416 |
No | output | 1 | 35,708 | 0 | 71,417 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 36,190 | 0 | 72,380 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
import string
def solve():
s = input()
n = int(input())
alphabets = len(string.ascii_lowercase)
count = [[0] * alphabets for _ in range(len(s) + 1)]
for i, c in enumerate(s, 1):
count[i][ord(c) - ord('a')] += 1
for c in range(alphabets):
count[i][c] += count[i - 1][c]
def query(l, r):
if l == r:
return 'Yes'
if s[l - 1] != s[r - 1]:
return 'Yes'
unique = 0
for x, y in zip(count[l - 1], count[r]):
if y - x > 0:
unique += 1
if unique > 2:
return 'Yes'
return 'No'
for _ in range(n):
print(query(*map(int, input().split())))
solve()
``` | output | 1 | 36,190 | 0 | 72,381 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 36,191 | 0 | 72,382 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
s=input()
m=[]
for i in range(len(s)+1):
m.append([])
for j in range(26):
m[-1].append(0)
for i in range(len(s)):
for j in range(26):
if j==ord(s[i])-97:
m[i+1][j]=m[i][j]+1
else:
m[i+1][j]=m[i][j]
q=int(input())
for _ in range(q):
l,r=map(int,input().split())
if l==r:
print('Yes')
continue
count=0
for i in range(26):
if m[l-1][i]!=m[r][i]:
count+=1
if count==1:
print('No')
else:
if count>=3:
print('Yes')
else:
if s[l-1]==s[r-1]:
print('No')
else:
print('Yes')
``` | output | 1 | 36,191 | 0 | 72,383 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 36,192 | 0 | 72,384 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
S = input()
N = len(S)
X = [[0] * 26 for _ in range(N+1)]
for i in range(N):
for j in range(26):
X[i+1][j] = X[i][j]
a = ord(S[i]) - 97
X[i+1][a] += 1
Q = int(input())
for _ in range(Q):
l, r = map(int, input().split())
if l == r:
print("Yes")
continue
l -= 1
A = [X[r][i] - X[l][i] for i in range(26)]
if A.count(0) < 24:
print("Yes")
elif A.count(0) > 24:
print("No")
elif S[l] == S[r-1]:
print("No")
else:
print("Yes")
``` | output | 1 | 36,192 | 0 | 72,385 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 36,193 | 0 | 72,386 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
import sys
import math
input = sys.stdin.readline
from functools import cmp_to_key;
def pi():
return(int(input()))
def pl():
return(int(input(), 16))
def ti():
return(list(map(int,input().split())))
def ts():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
mod = 998244353;
f = [];
def fact(n,m):
global f;
f = [1 for i in range(n+1)];
f[0] = 1;
for i in range(1,n+1):
f[i] = (f[i-1]*i)%m;
def fast_mod_exp(a,b,m):
res = 1;
while b > 0:
if b & 1:
res = (res*a)%m;
a = (a*a)%m;
b = b >> 1;
return res;
def inverseMod(n,m):
return fast_mod_exp(n,m-2,m);
def ncr(n,r,m):
if r == 0: return 1;
return ((f[n]*inverseMod(f[n-r],m))%m*inverseMod(f[r],m))%m;
def main():
B();
def B():
s = ts();
t = pi();
c = [[0 for j in range(len(s))] for i in range(26)];
for i in range(26):
for j in range(len(s)):
if ord(s[j])-ord("a") == i:
c[i][j] = c[i][j-1] + 1 if j > 0 else 1;
else:
c[i][j] = c[i][j-1] if j > 0 else 0;
for i in range(t):
[l,r] = ti();
count = 0;
for j in range(26):
count += (1 if c[j][r-1]-c[j][l-1] > 0 else 0);
if s[l-1] != s[r-1] or l == r:
print("Yes");
elif count >= 3:
print("Yes");
else:
print("No");
main();
``` | output | 1 | 36,193 | 0 | 72,387 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 36,194 | 0 | 72,388 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
import sys
import math
from collections import defaultdict,Counter
input=sys.stdin.readline
def print(x):
sys.stdout.write(str(x)+"\n")
# sys.stdout=open("CP3/output.txt",'w')
# sys.stdin=open("CP3/input.txt",'r')
# m=pow(10,9)+7
s=input()
pref=defaultdict(dict)
pref[0]={s[0]:1}
for j in range(1,len(s)):
pref[j]=pref[j-1].copy()
pref[j][s[j]]=pref[j].get(s[j],0)+1
# print(pref)
q=int(input())
for i in range(q):
l,r=map(int,input().split())
if l==r or s[l-1]!=s[r-1]:
print("Yes")
continue
c=0
if l==1:
c=len(pref[r-1])
else:
for j in pref[r-1]:
if pref[l-2].get(j,0)<pref[r-1][j]:
c+=1
if c>2:
print("Yes")
else:
print("No")
``` | output | 1 | 36,194 | 0 | 72,389 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 36,195 | 0 | 72,390 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 2/3/20
"""
import collections
import time
import os
import sys
import bisect
import heapq
from typing import List
def solve(s, queries):
N = len(s)
charcount = [[0 for _ in range(N+1)] for _ in range(26)]
for i, v in enumerate(s):
for ch in range(26):
charcount[ch][i+1] = charcount[ch][i]
ch = ord(v) - ord('a')
charcount[ch][i+1] += 1
# for i in range(26):
# print(charcount[i])
ans = []
for l, r in queries:
if l == r:
ans.append(True)
elif s[l] != s[r]:
ans.append(True)
else:
c = 0
for ch in range(26):
c += 1 if charcount[ch][r+1] - charcount[ch][l] > 0 else 0
ans.append(True if c > 2 else False)
print('\n'.join(['Yes' if v else 'No' for v in ans]))
s = input()
nq = int(input())
queries = []
for i in range(nq):
l, r = map(int, input().split())
queries.append((l-1, r-1))
solve(s, queries)
``` | output | 1 | 36,195 | 0 | 72,391 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 36,196 | 0 | 72,392 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
s = input()
n = len(s)
cnt = [[0]*26 for i in range(n+1)]
for i in range(1, n+1):
for c in range(26):
cnt[i][c] = cnt[i-1][c] + (ord(s[i-1])-ord('a') == c)
q = int(input())
for i in range(q):
l, r = map(int, input().split())
l -= 1
r -= 1
flag = 0
for c in range(26):
flag += (cnt[r+1][c]-cnt[l][c] > 0)
if l == r or s[l] != s[r] or flag >= 3:
print("Yes")
else:
print("No")
``` | output | 1 | 36,196 | 0 | 72,393 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 36,197 | 0 | 72,394 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
s = input()
n = len(s)
num = [[0]*26 for i in range(n)]
for i in range(n):
p = ord(s[i]) - ord('a')
if i != 0:
for j in range(26):
num[i][j] = num[i-1][j]
num[i][p] += 1
def solve(l: int, r: int) -> bool:
if l == r:
return True
elif s[l] != s[r]:
return True
else:
cnt = 0
for i in range(26):
if l > 0:
if num[r][i]-num[l-1][i] > 0:
cnt += 1
else:
if num[r][i] > 0:
cnt += 1
return True if cnt > 2 else False
q = int(input())
while q > 0:
l, r = map(int, input().split())
print('Yes') if solve(l - 1, r - 1) else print('No')
q -= 1
``` | output | 1 | 36,197 | 0 | 72,395 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 36,198 | 0 | 72,396 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
for i in arr:
stdout.write(str(i)+' ')
stdout.write('\n')
range = xrange # not for python 3.0+
s=' '+raw_input().strip()
n=len(s)-1
dp=[[0 for i in range(26)] for j in range(n+1)]
for i in range(1,n+1):
dp[i][ord(s[i])-97]+=1
for j in range(26):
dp[i][j]+=dp[i-1][j]
for q in range(input()):
l,r=in_arr()
if l==r or s[l]!=s[r]:
pr('Yes\n')
continue
c=0
for i in range(26):
c+=int(dp[r][i]-dp[l-1][i]!=0)
if c>=3:
pr('Yes\n')
else:
pr('No\n')
``` | output | 1 | 36,198 | 0 | 72,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
import sys
readline = sys.stdin.buffer.readline
readlines = sys.stdin.readlines
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
prn = lambda x: print(*x, sep='\n')
def solve():
s = list(map(lambda x: x - 97, list(ns())))
n = len(s)
q = ni()
f = [[0]*26 for _ in range(n+1)]
for i in range(n):
f[i+1] = f[i][:]
f[i+1][s[i]] += 1
for _ in range(q):
l, r = nm()
m = 0
c = 0
for i in range(26):
x = f[r][i] - f[l-1][i]
if x: c += 1
m = max(m, x)
if r == l or c > 2 or c == 2 and s[l-1] != s[r-1]:
print('Yes')
else:
print('No')
return
solve()
# T = ni()
# for _ in range(T):
# solve()
``` | instruction | 0 | 36,199 | 0 | 72,398 |
Yes | output | 1 | 36,199 | 0 | 72,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
from collections import defaultdict
s = input().strip()
d = defaultdict(int)
vs = [[0]*26 for _ in range(len(s)+1)]
S = "abcdefghijklmnopqrstuvwxyz"
dd = dict()
for i in range(26):
dd[S[i]] = i
for i, c in enumerate(s):
vs[i+1] = vs[i][:]
vs[i+1][dd[c]] += 1
N = int(input())
for _ in range(N):
L, R = map(int, input().split())
if R == L:
print("Yes")
else:
if s[L-1] != s[R-1]:
print("Yes")
else:
r = 0
for i in range(26):
c = vs[R][i] - vs[L-1][i]
# print(i, c)
if c > 0:
r += 1
# print(r)
if r < 3:
print("No")
else:
print("Yes")
``` | instruction | 0 | 36,200 | 0 | 72,400 |
Yes | output | 1 | 36,200 | 0 | 72,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
import sys
input=sys.stdin.readline
s=[i for i in input() if i!='\n']
freq=[[0 for i in range(26)] for _ in range(len(s))]
for i in range(len(s)):
freq[i][ord(s[i])-97]+=1
if i>0:
for j in range(26):
freq[i][j]+=freq[i-1][j]
#print(freq)
q=int(input())
for i in range(q):
l,r=map(int,input().split())
count=0
for j in range(26):
temp=freq[r-1][j]
if l>1:
temp-=freq[l-2][j]
if temp:
count+=1
#print(sum(freq[r-1])-minus)
if r==l or count>=3 or s[l-1]!=s[r-1]:
print("Yes")
else:
print("No")
``` | instruction | 0 | 36,201 | 0 | 72,402 |
Yes | output | 1 | 36,201 | 0 | 72,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
def charToInt(c): #'a'->0
return ord(c)-ord('a')
def main():
s=input()
n=len(s)
arr=[charToInt(c) for c in s]
p=makeArr(0,n,26)
for i in range(n):
p[i][arr[i]]+=1
for i in range(1,n):
for j in range(26):
p[i][j]+=p[i-1][j]
def getCnts(char,l,r):
if l==0: return p[r][char]
else: return p[r][char]-p[l-1][char]
allans=[]
q=int(input())
for _ in range(q):
l,r=readIntArr()
l-=1
r-=1
charTypeCnt=0
for j in range(26):
if getCnts(j,l,r)>0:
charTypeCnt+=1
if l==r:
allans.append('Yes')
elif charTypeCnt==1:
allans.append('No')
elif charTypeCnt==2:
if arr[l]==arr[r]:
allans.append('No')
else:
allans.append('Yes') # swap both sides
else: # more than 2 types
allans.append('Yes')
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(*args):
"""
*args : (default value, dimension 1, dimension 2,...)
Returns : arr[dim1][dim2]... filled with default value
"""
assert len(args) >= 2, "makeArr args should be (default value, dimension 1, dimension 2,..."
if len(args) == 2:
return [args[0] for _ in range(args[1])]
else:
return [makeArr(args[0],*args[2:]) for _ in range(args[1])]
def queryInteractive(x,y):
print('? {} {}'.format(x,y))
sys.stdout.flush()
return int(input())
def answerInteractive(ans):
print('! {}'.format(ans))
sys.stdout.flush()
inf=float('inf')
MOD=10**9+7
for _abc in range(1):
main()
``` | instruction | 0 | 36,202 | 0 | 72,404 |
Yes | output | 1 | 36,202 | 0 | 72,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
def diff(a,b):
ans = []
for i in range(len(a)):
ans.append(a[i]-b[i])
return ans
a = input()
n = len(a)
q = int(input())
count = [[0 for i in range(26)] for j in range(n)]
count[0][ord(a[0])-97] += 1
for i in range(1,n):
for j in range(26):
count[i][j] = count[i][j-1]
count[i][ord(a[i])-97] += 1
# for i in count:
# print (*i)
for i in range(q):
l,r = map(int,input().split())
if l==r:
print ("Yes")
else:
if a[l-1]!=a[r-1]:
print ("No")
else:
d = diff(count[r-1],count[l-1])
if d.count(0)<=23:
print("Yes")
else:
print ("No")
``` | instruction | 0 | 36,203 | 0 | 72,406 |
No | output | 1 | 36,203 | 0 | 72,407 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
from bisect import bisect_left, bisect_right
def solve(l, r):
if (l == r):
return "Yes"
if (s[l] != s[r]):
return "Yes"
j = ord(s[l]) - ord('a')
res = []
for i in range(26):
if (i == j):
continue
a = bisect_left(d[i], r)
if (a >= len(d[i])):
a = len(d[i]) - 1
b = bisect_right(d[i], l)
if (b < len(d[i]) and a != b):
temp = [d[i][a], d[i][b]]
for x, y in res:
if (x < temp[1] or y > temp[0]):
return "Yes"
res.append(temp)
return "No"
s = input()
d = {i: [] for i in range(26)}
for i, j in enumerate(s):
d[ord(j) - ord('a')].append(i)
for q in range(int(input())):
l, r = [int(i) - 1 for i in input().split()]
print(solve(l, r))
``` | instruction | 0 | 36,204 | 0 | 72,408 |
No | output | 1 | 36,204 | 0 | 72,409 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
s = input()
# print("s: {}".format(s))
n = len(s)
letters_set = set(s)
# print("letters_set: {}".format(letters_set))
prefix_letter_occurence = {}
for l in letters_set:
prefix_letter_occurence[l] = [0 for i in range(n)]
for i in range(n):
cur_letter = s[i]
if i == 0:
prefix_letter_occurence[cur_letter][0] = 1
else:
for l in letters_set:
prefix_letter_occurence[l][i] = prefix_letter_occurence[l][i-1] + (l == cur_letter)
# for j in range(i, n):
# prefix_letter_occurence[cur_letter][j] += 1
# print("prefix_letter_occurence: {}".format(prefix_letter_occurence))
def has_inreducible(l, r):
if l == r:
return "Yes"
# letters_counts = {}
non_zeroes = 0
for letter in letters_set:
if l == 0:
let_count = prefix_letter_occurence[letter][r]
else:
let_count = prefix_letter_occurence[letter][r] - prefix_letter_occurence[letter][l-1]
if let_count > 0:
non_zeroes += 1
if non_zeroes == 1:
return "No"
else:
if s[l] != s[r]:
return "Yes"
else:
return "No"
q = int(input())
for _ in range(q):
l, r = map(int, input().split())
l -= 1
r -= 1
print(has_inreducible(l, r))
``` | instruction | 0 | 36,205 | 0 | 72,410 |
No | output | 1 | 36,205 | 0 | 72,411 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
import copy
s = input()
count = [[0] * 26]
for index, e in enumerate(s):
new_char_acc = copy.deepcopy(count[index])
new_char_acc[ord(e) - ord('a')] += 1
count.append(new_char_acc)
n = int(input())
for _ in range(n):
l, r = map(int, input().split())
element_size = 0
for i in range(26):
if count[r][i] - count[l - 1][i] > 0:
element_size += 1
if l == r or element_size >= 3 or s[l] == s[r]:
print("Yes")
else:
print("No")
``` | instruction | 0 | 36,206 | 0 | 72,412 |
No | output | 1 | 36,206 | 0 | 72,413 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6 | instruction | 0 | 36,243 | 0 | 72,486 |
Tags: math
Correct Solution:
```
for _ in range(int(input())):
a=input()
m,n=0,0
l=len(a)
for i in range(l):
if a[i]=="-":
m+=1
else:
n+=1
if m>n:
l+=(i+1)
m-=1
print(l)
``` | output | 1 | 36,243 | 0 | 72,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6 | instruction | 0 | 36,245 | 0 | 72,490 |
Tags: math
Correct Solution:
```
import time,math,bisect,sys
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
return int(stdin.readline())
def IO(): # to take string input
return stdin.readline()
def IP(): # to take tuple as input
return map(int,stdin.readline().split())
def L(): # to take list as input
return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
return stdout.write(str(x))
def PI(x,y): # to print tuple separatedly
return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
if a==0:
return b
elif b==0:
return a
if a>b:
return gcd(a%b,b)
else:
return gcd(a,b%a)
def readTree(): # to read tree
v=int(input())
adj=[set() for i in range(v+1)]
for i in range(v-1):
u1,u2=In()
adj[u1].add(u2)
adj[u2].add(u1)
return adj,v
def bfs(adj,v): # a schema of bfs
visited=[False]*(v+1)
q=deque()
while q:
pass
def sieve():
li=[True]*1000001
li[0],li[1]=False,False
for i in range(2,len(li),1):
if li[i]==True:
for j in range(i*i,len(li),i):
li[j]=False
prime=[]
for i in range(1000001):
if li[i]==True:
prime.append(i)
return prime
def setBit(n):
count=0
while n!=0:
n=n&(n-1)
count+=1
return count
#####################################################################################
mx=10**9+7
def solve():
s=input()
n=len(s)
pref=[0]*n
cur=0
for i in range(n):
if s[i]=='-':
cur-=1
else:
cur+=1
pref[i]=cur
inf=max(pref)
last=0
sm=0
for i in range(n):
if pref[i]<0:
x=pref[i]*(-1)
if last<x:
sm+=(i+1)*(x-last)
last=x
print(sm+(i+1))
t=II()
for i in range(t):
solve()
#######
#
#
####### # # # #### # # #
# # # # # # # # # # #
# #### # # #### #### # #
###### # # #### # # # # #
``` | output | 1 | 36,245 | 0 | 72,491 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6 | instruction | 0 | 36,246 | 0 | 72,492 |
Tags: math
Correct Solution:
```
for i in range(int(input())):
s = input()
a = 0
b = [0]*len(s)
for i in range(len(s)):
if s[i] == '+':
a = a+1
b[i] = a
else:
a = a-1
b[i] = a
res = 0
min = 0
minp = 0
for i in range(len(s)):
if (b[i]<0) & (b[i]<min):
minp = min
min = b[i]
res = res+(minp - min)* (i+1)
print(res+len(s))
``` | output | 1 | 36,246 | 0 | 72,493 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6 | instruction | 0 | 36,247 | 0 | 72,494 |
Tags: math
Correct Solution:
```
t = int(input())
for s in range(t) :
pm = input()
cur = 0
res = 0
for i in range(len(pm)) :
if pm[i] == '+' :
cur += 1
else :
cur -= 1
if cur < 0 :
res += i + 1
cur = 0
else :
print(res + len(pm))
``` | output | 1 | 36,247 | 0 | 72,495 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6 | instruction | 0 | 36,248 | 0 | 72,496 |
Tags: math
Correct Solution:
```
def countFreq(arr, n,mp,s):
for i in range(n):
if arr[i] not in mp.keys():mp[arr[i]] = i+1
for x in mp:
if x < 0:s += mp[x]
return s
for _ in range(int(input())):
a = list(str(input()));br,cur = [0]*len(a),0
for i in range(len(a)):
cur = (cur + 1 if a[i] == '+' else cur - 1)
br[i] = cur
print(countFreq(br,len(br),dict(),0) + len(a))
``` | output | 1 | 36,248 | 0 | 72,497 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6 | instruction | 0 | 36,249 | 0 | 72,498 |
Tags: math
Correct Solution:
```
t = int(input())
for _ in range(t):
s = list(input())
cur = 0
num = 0
ans = 0
for i in range(len(s)):
if s[i] == "+":
cur += 1
else:
cur -= 1
if cur < 0:
ans += i+1
cur = 0
ans += len(s)
print(ans)
``` | output | 1 | 36,249 | 0 | 72,499 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6 | instruction | 0 | 36,250 | 0 | 72,500 |
Tags: math
Correct Solution:
```
for test in range(int(input())):
s = input()
neg = 0
res = 0
cur = 0
for i in range(len(s)):
res += 1
if s[i] == '-':
neg += 1
if neg > cur:
cur += 1
res += i + 1
else:
cur += 1
print(res)
``` | output | 1 | 36,250 | 0 | 72,501 |
Provide tags and a correct Python 2 solution for this coding contest problem.
You are given a string s consisting only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6 | instruction | 0 | 36,251 | 0 | 72,502 |
Tags: math
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return map(int,raw_input().split())
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
for t in range(ni()):
s=raw_input().strip()
c=0
n=len(s)
ans=10**6+1
for i in range(n):
if s[i]=='+':
c+=1
else:
c-=1
ans=min(ans,c)
ans=min(0,ans)
ans=int(abs(ans))
d=Counter()
cnt=0
ans1=0
for i in range(n):
if s[i]=='+':
cnt+=1
else:
cnt-=1
ans1+=ans
if cnt<0 and not d[cnt]:
d[cnt]=1
ans-=1
ans1+=n
pn(ans1)
``` | output | 1 | 36,251 | 0 | 72,503 |
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 only of characters + and -. You perform some process with this string. This process can be described by the following pseudocode:
res = 0
for init = 0 to inf
cur = init
ok = true
for i = 1 to |s|
res = res + 1
if s[i] == '+'
cur = cur + 1
else
cur = cur - 1
if cur < 0
ok = false
break
if ok
break
Note that the inf denotes infinity, and the characters of the string are numbered from 1 to |s|.
You have to calculate the value of the res after the process ends.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The only lines of each test case contains string s (1 ≤ |s| ≤ 10^6) consisting only of characters + and -.
It's guaranteed that sum of |s| over all test cases doesn't exceed 10^6.
Output
For each test case print one integer — the value of the res after the process ends.
Example
Input
3
--+-
---
++--+-
Output
7
9
6
Submitted Solution:
```
t=int(input())
for _ in range(t):
mp={'+':1,'-':-1}
a=[mp[c] for c in input()]
#LOOP with value init takes stops at index i where a[0]+...+a[i]=(-init-1) to find this for each init just go over once on prefix array. store only when you reach smaller values.
#let j= smallest index where min_over_i{a[0]+...+a[i]} =minv happens. Thus outer loop breaks when init=minv which runs for length of s.
cumsum,currmincumsum,ans=0,0,0
for i in range(len(a)):
cumsum+=a[i]
if cumsum<currmincumsum: ans+=i+1; currmincumsum=cumsum
ans+=len(a)
print(ans)
``` | instruction | 0 | 36,252 | 0 | 72,504 |
Yes | output | 1 | 36,252 | 0 | 72,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bitlandians are quite weird people. They do everything differently. They have a different alphabet so they have a different definition for a string.
A Bitlandish string is a string made only of characters "0" and "1".
BitHaval (the mayor of Bitland) loves to play with Bitlandish strings. He takes some Bitlandish string a, and applies several (possibly zero) operations to it. In one operation the mayor may take any two adjacent characters of a string, define one of them as x and the other one as y. Then he calculates two values p and q: p = x xor y, q = x or y. Then he replaces one of the two taken characters by p and the other one by q.
The xor operation means the bitwise excluding OR operation. The or operation is the bitwise OR operation.
So for example one operation can transform string 11 to string 10 or to string 01. String 1 cannot be transformed into any other string.
You've got two Bitlandish strings a and b. Your task is to check if it is possible for BitHaval to transform string a to string b in several (possibly zero) described operations.
Input
The first line contains Bitlandish string a, the second line contains Bitlandish string b. The strings can have different lengths.
It is guaranteed that the given strings only consist of characters "0" and "1". The strings are not empty, their length doesn't exceed 106.
Output
Print "YES" if a can be transformed into b, otherwise print "NO". Please do not print the quotes.
Examples
Input
11
10
Output
YES
Input
1
01
Output
NO
Input
000
101
Output
NO | instruction | 0 | 36,394 | 0 | 72,788 |
Tags: constructive algorithms, implementation, math
Correct Solution:
```
from sys import *
s=input()
r=input()
if len(s)==len(r):
g1,g2=0,0
for i in s:
if i=="1":
g1=1
break
for i in r:
if i=="1":
g2=1
break
if not g1^g2:
print("YES")
exit(0)
print("NO")
``` | output | 1 | 36,394 | 0 | 72,789 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bitlandians are quite weird people. They do everything differently. They have a different alphabet so they have a different definition for a string.
A Bitlandish string is a string made only of characters "0" and "1".
BitHaval (the mayor of Bitland) loves to play with Bitlandish strings. He takes some Bitlandish string a, and applies several (possibly zero) operations to it. In one operation the mayor may take any two adjacent characters of a string, define one of them as x and the other one as y. Then he calculates two values p and q: p = x xor y, q = x or y. Then he replaces one of the two taken characters by p and the other one by q.
The xor operation means the bitwise excluding OR operation. The or operation is the bitwise OR operation.
So for example one operation can transform string 11 to string 10 or to string 01. String 1 cannot be transformed into any other string.
You've got two Bitlandish strings a and b. Your task is to check if it is possible for BitHaval to transform string a to string b in several (possibly zero) described operations.
Input
The first line contains Bitlandish string a, the second line contains Bitlandish string b. The strings can have different lengths.
It is guaranteed that the given strings only consist of characters "0" and "1". The strings are not empty, their length doesn't exceed 106.
Output
Print "YES" if a can be transformed into b, otherwise print "NO". Please do not print the quotes.
Examples
Input
11
10
Output
YES
Input
1
01
Output
NO
Input
000
101
Output
NO | instruction | 0 | 36,395 | 0 | 72,790 |
Tags: constructive algorithms, implementation, math
Correct Solution:
```
i=input;a,b=i(),i();print(["YES","NO"][(len(a)!=len(b) or set(a)=={'0'} or set(b) == {'0'} or len(a)<2) and a!=b])
``` | output | 1 | 36,395 | 0 | 72,791 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bitlandians are quite weird people. They do everything differently. They have a different alphabet so they have a different definition for a string.
A Bitlandish string is a string made only of characters "0" and "1".
BitHaval (the mayor of Bitland) loves to play with Bitlandish strings. He takes some Bitlandish string a, and applies several (possibly zero) operations to it. In one operation the mayor may take any two adjacent characters of a string, define one of them as x and the other one as y. Then he calculates two values p and q: p = x xor y, q = x or y. Then he replaces one of the two taken characters by p and the other one by q.
The xor operation means the bitwise excluding OR operation. The or operation is the bitwise OR operation.
So for example one operation can transform string 11 to string 10 or to string 01. String 1 cannot be transformed into any other string.
You've got two Bitlandish strings a and b. Your task is to check if it is possible for BitHaval to transform string a to string b in several (possibly zero) described operations.
Input
The first line contains Bitlandish string a, the second line contains Bitlandish string b. The strings can have different lengths.
It is guaranteed that the given strings only consist of characters "0" and "1". The strings are not empty, their length doesn't exceed 106.
Output
Print "YES" if a can be transformed into b, otherwise print "NO". Please do not print the quotes.
Examples
Input
11
10
Output
YES
Input
1
01
Output
NO
Input
000
101
Output
NO | instruction | 0 | 36,396 | 0 | 72,792 |
Tags: constructive algorithms, implementation, math
Correct Solution:
```
s = [int(i) for i in list(input())]
t = [int(i) for i in list(input())]
l=len(s)
ll=len(t)
a=s.count(0)
b=t.count(0)
if l!=ll:
print("NO")
elif a==l and b==ll:
print("YES")
elif (a==l and b!=ll) or(a!=l and b==ll):
print("NO")
else:
print("YES")
``` | output | 1 | 36,396 | 0 | 72,793 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bitlandians are quite weird people. They do everything differently. They have a different alphabet so they have a different definition for a string.
A Bitlandish string is a string made only of characters "0" and "1".
BitHaval (the mayor of Bitland) loves to play with Bitlandish strings. He takes some Bitlandish string a, and applies several (possibly zero) operations to it. In one operation the mayor may take any two adjacent characters of a string, define one of them as x and the other one as y. Then he calculates two values p and q: p = x xor y, q = x or y. Then he replaces one of the two taken characters by p and the other one by q.
The xor operation means the bitwise excluding OR operation. The or operation is the bitwise OR operation.
So for example one operation can transform string 11 to string 10 or to string 01. String 1 cannot be transformed into any other string.
You've got two Bitlandish strings a and b. Your task is to check if it is possible for BitHaval to transform string a to string b in several (possibly zero) described operations.
Input
The first line contains Bitlandish string a, the second line contains Bitlandish string b. The strings can have different lengths.
It is guaranteed that the given strings only consist of characters "0" and "1". The strings are not empty, their length doesn't exceed 106.
Output
Print "YES" if a can be transformed into b, otherwise print "NO". Please do not print the quotes.
Examples
Input
11
10
Output
YES
Input
1
01
Output
NO
Input
000
101
Output
NO | instruction | 0 | 36,397 | 0 | 72,794 |
Tags: constructive algorithms, implementation, math
Correct Solution:
```
def helper(a,b):
n=len(a)
m=len(b)
if n!=m:
return 'NO'
if a==b:
return 'YES'
if a=='0'*n or b=='0'*m:
return'NO'
return 'YES'
a=input()
b=input()
print(helper(a,b))
``` | output | 1 | 36,397 | 0 | 72,795 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bitlandians are quite weird people. They do everything differently. They have a different alphabet so they have a different definition for a string.
A Bitlandish string is a string made only of characters "0" and "1".
BitHaval (the mayor of Bitland) loves to play with Bitlandish strings. He takes some Bitlandish string a, and applies several (possibly zero) operations to it. In one operation the mayor may take any two adjacent characters of a string, define one of them as x and the other one as y. Then he calculates two values p and q: p = x xor y, q = x or y. Then he replaces one of the two taken characters by p and the other one by q.
The xor operation means the bitwise excluding OR operation. The or operation is the bitwise OR operation.
So for example one operation can transform string 11 to string 10 or to string 01. String 1 cannot be transformed into any other string.
You've got two Bitlandish strings a and b. Your task is to check if it is possible for BitHaval to transform string a to string b in several (possibly zero) described operations.
Input
The first line contains Bitlandish string a, the second line contains Bitlandish string b. The strings can have different lengths.
It is guaranteed that the given strings only consist of characters "0" and "1". The strings are not empty, their length doesn't exceed 106.
Output
Print "YES" if a can be transformed into b, otherwise print "NO". Please do not print the quotes.
Examples
Input
11
10
Output
YES
Input
1
01
Output
NO
Input
000
101
Output
NO | instruction | 0 | 36,398 | 0 | 72,796 |
Tags: constructive algorithms, implementation, math
Correct Solution:
```
s = input()
ans = input()
if len(s)!=len(ans) or ('1' in ans and '1' not in s) or ('1' not in ans and '1' in s):
print("NO")
else:
print("YES")
``` | output | 1 | 36,398 | 0 | 72,797 |
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