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 |
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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 | 36,399 | 0 | 72,798 |
Tags: constructive algorithms, implementation, math
Correct Solution:
```
a = input()
b = input()
possible = True
if len(a) != len(b):
possible = False
elif a == b:
pass
else:
if list(a).count('1') == 0:
possible = False
if list(b).count('1') == 0:
possible = False # 00...00 and 00...00 checked in other branch
# WA
if possible:
print("YES")
else:
print("NO")
``` | output | 1 | 36,399 | 0 | 72,799 |
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 | 36,400 | 0 | 72,800 |
Tags: constructive algorithms, implementation, math
Correct Solution:
```
for i1 in range(1):
s=input()
l=input()
if s==l:
print("YES")
elif len(s)==len(l) and s.count('1')!=0 and l.count('1')!=0:
print("YES")
else:
print("NO")
``` | output | 1 | 36,400 | 0 | 72,801 |
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 | 36,401 | 0 | 72,802 |
Tags: constructive algorithms, implementation, math
Correct Solution:
```
a = input()
b = input()
if len(a) != len(b):
print('NO')
else:
if a == b:
print('YES')
elif '1' in a and '1' in b:
print('YES')
else:
print('NO')
``` | output | 1 | 36,401 | 0 | 72,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
a = input()
b = input()
if len(a) != len(b):
print("NO")
exit()
arr1 = [0, 0]
arr2 = [0, 0]
for i in a:
arr1[int(i)] += 1
for i in b:
arr2[int(i)] += 1
if arr1[1] == 0 and arr2[1] != 0:
print("NO")
elif arr1[1] != 0 and arr2[1] == 0:
print("NO")
else:
print("YES")
``` | instruction | 0 | 36,402 | 0 | 72,804 |
Yes | output | 1 | 36,402 | 0 | 72,805 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
from collections import Counter, defaultdict, OrderedDict, deque
from bisect import bisect_left, bisect_right
from functools import reduce, lru_cache
from typing import List
import itertools
import sys
import math
import heapq
import string
import random
MIN, MAX, MOD = -0x3f3f3f3f, 0x3f3f3f3f, 1000000007
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
a = input()
b = input()
if a==b: print("YES")
elif len(a) != len(b) or '1' not in a or '1' not in b: print("NO")
else: print("YES")
``` | instruction | 0 | 36,403 | 0 | 72,806 |
Yes | output | 1 | 36,403 | 0 | 72,807 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
from sys import stdin
import math
a = stdin.readline().rstrip()
b = stdin.readline().rstrip()
a_all_zero = all(map(lambda x: x == '0', a))
b_all_zero = all(map(lambda x: x == '0', b))
if len(a) != len(b):
print('NO')
exit()
if a == b:
print('YES')
exit()
if b_all_zero != a_all_zero:
print('NO')
exit()
print('YES')
``` | instruction | 0 | 36,404 | 0 | 72,808 |
Yes | output | 1 | 36,404 | 0 | 72,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
from sys import stdin,stdout
from math import gcd
nmbr = lambda: int(stdin.readline())
lst = lambda: list(map(int,stdin.readline().split()))
for i in range(1):#nmbr()):
a=input()
b=input()
na=len(a)
nb=len(b)
if na!=nb:
print("NO")
continue
if a==b:
print('YES')
continue
if '1' in a and '1' in b:
print('YES')
continue
print('NO')
``` | instruction | 0 | 36,405 | 0 | 72,810 |
Yes | output | 1 | 36,405 | 0 | 72,811 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
a=input()
b=input()
flag='0'
if a=='0'*len(a)==b: print('YES')
else:
if len(a)!= len(b):print('NO')
else:
for i in range(len(a)):
if a[i]=='1':
flag=1
if flag=='0':print('NO')
else:
for i in range(len(a)):
if b[i]=='1':
flag=1
if flag =='0':
print('NO')
else:print('YES')
``` | instruction | 0 | 36,406 | 0 | 72,812 |
No | output | 1 | 36,406 | 0 | 72,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
a=input();print("YES"if len(a)>1and"1"in a else"NO")
``` | instruction | 0 | 36,407 | 0 | 72,814 |
No | output | 1 | 36,407 | 0 | 72,815 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
import sys
first = sys.stdin.readline().strip()
second = sys.stdin.readline().strip()
if len(first) != len(second):
print("NO")
sys.exit(0)
count1, count2 = 0, 0
for i in first:
if i == '1':
count1 += 1
for i in second:
if i == '1':
count2 += 1
if (count1 == 0 and count2 != 0) or \
(count1 != 0 and count2 == 0):
print("NO")
sys.exit(0)
if count1 == count2 or count1-1 == count2:
print("YES")
else:
print("NO")
``` | instruction | 0 | 36,408 | 0 | 72,816 |
No | output | 1 | 36,408 | 0 | 72,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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
Submitted Solution:
```
x = str(input())
y = str(input())
s = True
n = len(x)
if n == 1:
s = x == y
elif len(x) == len(y):
if not("1" in x):
s = x == y
else:
s = False
if s:
print("YES")
else:
print("NO")
``` | instruction | 0 | 36,409 | 0 | 72,818 |
No | output | 1 | 36,409 | 0 | 72,819 |
Provide a correct Python 3 solution for this coding contest problem.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO | instruction | 0 | 36,728 | 0 | 73,456 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
S = input()
T = input()
q = int(input())
ls = len(S)
lt = len(T)
cs_s = [0] * (ls+1)
cs_t = [0] * (lt+1)
for i in range(ls):
tmp = 1 if S[i] == 'A' else -1
cs_s[i+1] = cs_s[i] + tmp
for i in range(lt):
tmp = 1 if T[i] == 'A' else -1
cs_t[i+1] = cs_t[i] + tmp
for _ in range(q):
a, b, c, d = map(int, input().split())
# print(cs_s[b] - cs_s[a-1], cs_t[d] - cs_t[c-1])
if (cs_s[b] - cs_s[a-1]) % 3 == (cs_t[d] - cs_t[c-1]) % 3:
print('YES')
else:
print('NO')
``` | output | 1 | 36,728 | 0 | 73,457 |
Provide a correct Python 3 solution for this coding contest problem.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO | instruction | 0 | 36,729 | 0 | 73,458 |
"Correct Solution:
```
s = input()
t = input()
ls = len(s)
lt = len(t)
sl = [0]*(ls+1)
tl = [0]*(lt+1)
for i in range(ls):
if s[i] == "A":
sl[i+1] = sl[i] + 1
else:
sl[i+1] = sl[i] + 2
for i in range(lt):
if t[i] == "A":
tl[i+1] = tl[i] + 1
else:
tl[i+1] = tl[i] + 2
for _ in range(int(input())):
a,b,c,d = map(int, input().split())
if (sl[b]-tl[d]-sl[a-1]+tl[c-1])%3 == 0:
print("YES")
else:
print("NO")
``` | output | 1 | 36,729 | 0 | 73,459 |
Provide a correct Python 3 solution for this coding contest problem.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO | instruction | 0 | 36,730 | 0 | 73,460 |
"Correct Solution:
```
S = input()
T = input()
cumsa = [0]
cumsb = [0]
for c in S:
cumsa.append(cumsa[-1] + int(c == 'A'))
cumsb.append(cumsb[-1] + int(c == 'B'))
cumta = [0]
cumtb = [0]
for c in T:
cumta.append(cumta[-1] + int(c == 'A'))
cumtb.append(cumtb[-1] + int(c == 'B'))
Q = int(input())
ans = []
for i in range(Q):
a,b,c,d = map(int,input().split())
sa = cumsa[b] - cumsa[a-1]
sb = cumsb[b] - cumsb[a-1]
ta = cumta[d] - cumta[c-1]
tb = cumtb[d] - cumtb[c-1]
ans.append('YES' if (sa-sb)%3 == (ta-tb)%3 else 'NO')
print('\n'.join(ans))
``` | output | 1 | 36,730 | 0 | 73,461 |
Provide a correct Python 3 solution for this coding contest problem.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO | instruction | 0 | 36,731 | 0 | 73,462 |
"Correct Solution:
```
import sys
from itertools import accumulate
def input():
return sys.stdin.readline().strip()
def acc(li, n):
res = [0] * (n + 1)
for i in range(n):
if li[i] == "A":
res[i + 1] = 1
elif li[i] == "B":
res[i + 1] = 2
return list(accumulate(res))
def restore(x, y, li):
return (li[y] - li[x - 1]) % 3
S = input()
T = input()
N = len(S)
M = len(T)
acc_S = acc(S, N)
acc_T = acc(T, M)
q = int(input())
for _ in range(q):
a, b, c, d = map(int, input().split())
v_S = restore(a, b, acc_S)
v_T = restore(c, d, acc_T)
print("YES") if v_S == v_T else print("NO")
``` | output | 1 | 36,731 | 0 | 73,463 |
Provide a correct Python 3 solution for this coding contest problem.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO | instruction | 0 | 36,732 | 0 | 73,464 |
"Correct Solution:
```
S = input()
T = input()
cumSA = [0] * (len(S) + 1)
cumSB = [0] * (len(S) + 1)
cumTA = [0] * (len(T) + 1)
cumTB = [0] * (len(T) + 1)
for i, s in enumerate(S, start=1):
cumSA[i] = cumSA[i - 1]
cumSB[i] = cumSB[i - 1]
if s == 'A':
cumSA[i] += 1
else:
cumSB[i] += 1
for i, t in enumerate(T, start=1):
cumTA[i] = cumTA[i - 1]
cumTB[i] = cumTB[i - 1]
if t == 'A':
cumTA[i] += 1
else:
cumTB[i] += 1
Q = int(input())
ans = []
for _ in range(Q):
lS, rS, lT, rT = map(int, input().split())
sA = cumSA[rS] - cumSA[lS - 1]
sB = cumSB[rS] - cumSB[lS - 1]
tA = cumTA[rT] - cumTA[lT - 1]
tB = cumTB[rT] - cumTB[lT - 1]
if (sA - sB) % 3 == (tA - tB) % 3:
ans.append('YES')
else:
ans.append('NO')
print(*ans, sep='\n')
``` | output | 1 | 36,732 | 0 | 73,465 |
Provide a correct Python 3 solution for this coding contest problem.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO | instruction | 0 | 36,733 | 0 | 73,466 |
"Correct Solution:
```
from operator import add
import sys
input=sys.stdin.readline
class SegTree():
def __init__(self, N, e, operator_func=add):
self.e = e # εδ½ε
self.size = N
self.node = [self.e] * (2*N)
self.operator_func = operator_func # ε¦η(add or xor max minγͺγ©)
def set_list(self, l):
for i in range(self.size):
self.node[i+self.size-1] = l[i]
for i in range(self.size-1)[::-1]:
self.node[i] = self.operator_func(self.node[2*i+1], self.node[2*i+2])
def update(self, k, x):
k += self.size-1
self.node[k] = x
while k >= 0:
k = (k - 1) // 2
self.node[k] = self.operator_func(self.node[2*k+1], self.node[2*k+2])
def get(self, l, r):
# [l, r) γ«γ€γγ¦queryγζ±γγ
x = self.e
l += self.size
r += self.size
while l<r:
if l&1:
x = self.operator_func(x, self.node[l-1])
l += 1
if r&1:
r -= 1
x = self.operator_func(x, self.node[r-1])
l >>= 1
r >>= 1
return x
def main():
S = input()
T = input()
treeS = SegTree(len(S), 0)
treeT = SegTree(len(T), 0)
treeS.set_list([1 if s=="A" else 0 for s in S])
treeT.set_list([1 if s=="A" else 0 for s in T])
q = int(input())
for _ in range(q):
a, b, c, d = map(int, input().split())
sa = treeS.get(a-1, b)
sb = b-a+1-sa
ta = treeT.get(c-1, d)
tb = d-c+1-ta
if (2*(sb-sa)+tb-ta)%3 or (sb-sa+2*(tb-ta))%3:
print("NO")
else:
print("YES")
main()
``` | output | 1 | 36,733 | 0 | 73,467 |
Provide a correct Python 3 solution for this coding contest problem.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO | instruction | 0 | 36,734 | 0 | 73,468 |
"Correct Solution:
```
S = input()
T = input()
q = int(input())
abcd = [[int(i) for i in input().split()] for _ in range(q)]
sdp = [(0, 0)]
tdp = [(0, 0)]
for s in S :
a, b = sdp[-1]
if s == 'A' :
sdp.append((a + 1, b))
else :
sdp.append((a, b + 1))
for t in T :
a, b = tdp[-1]
if t == 'A' :
tdp.append((a + 1, b))
else :
tdp.append((a, b + 1))
for a, b, c, d in abcd :
sa = sdp[b][0] - sdp[a-1][0]
sb = sdp[b][1] - sdp[a-1][1]
ta = tdp[d][0] - tdp[c-1][0]
tb = tdp[d][1] - tdp[c-1][1]
tb -= ta - sa
if abs(sb - tb) % 3 == 0 :
print('YES')
else :
print('NO')
``` | output | 1 | 36,734 | 0 | 73,469 |
Provide a correct Python 3 solution for this coding contest problem.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO | instruction | 0 | 36,735 | 0 | 73,470 |
"Correct Solution:
```
s=input()
t=input()
n=len(s)
m=len(t)
sl=[0]
for i in s:
sl.append(sl[-1])
if i=="A":sl[-1]+=1
else:sl[-1]+=2
sl[-1]%=3
tl=[0]
for i in t:
tl.append(tl[-1])
if i=="A":tl[-1]+=1
else:tl[-1]+=2
tl[-1]%=3
q=int(input())
for _ in range(q):
a,b,c,d=map(int,input().split())
a-=1
c-=1
ss=sl[b]-sl[a]
tt=tl[d]-tl[c]
if ss%3==tt%3:print("YES")
else:print("NO")
``` | output | 1 | 36,735 | 0 | 73,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO
Submitted Solution:
```
import math
#import numpy as np
import queue
from collections import deque,defaultdict
import heapq as hpq
from sys import stdin,setrecursionlimit
#from scipy.sparse.csgraph import dijkstra
#from scipy.sparse import csr_matrix
ipt = stdin.readline
setrecursionlimit(10**7)
def main():
s = input()
t = input()
ns = len(s)+1
nt = len(t)+1
ssa = [0]*ns
ssb = [0]*ns
sta = [0]*nt
stb = [0]*nt
for i,si in enumerate(s):
if si == "A":
da = 1
db = 0
else:
da = 0
db = 1
ssa[i+1] = ssa[i]+da
ssb[i+1] = ssb[i]+db
for i,ti in enumerate(t):
if ti == "A":
da = 1
db = 0
else:
da = 0
db = 1
sta[i+1] = sta[i]+da
stb[i+1] = stb[i]+db
q = int(ipt())
for _ in range(q):
a,b,c,d = map(int,ipt().split())
da = ssa[b]-ssa[a-1]-sta[d]+sta[c-1]
db = ssb[b]-ssb[a-1]-stb[d]+stb[c-1]
if (da-db)%3 == 0:
print("YES")
else:
print("NO")
return
if __name__ == '__main__':
main()
``` | instruction | 0 | 36,736 | 0 | 73,472 |
Yes | output | 1 | 36,736 | 0 | 73,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO
Submitted Solution:
```
import sys
input=sys.stdin.readline
S,T=input().strip(),input().strip()
Acc_S,Acc_T=[0],[0]
for s in S:
if s=='A':
Acc_S.append(Acc_S[-1]+1)
else:
Acc_S.append(Acc_S[-1]+2)
for t in T:
if t=='A':
Acc_T.append(Acc_T[-1]+1)
else:
Acc_T.append(Acc_T[-1]+2)
q=int(input())
for _ in range(q):
a,b,c,d=map(int,input().split())
print('YES' if (Acc_S[b]-Acc_S[a-1])%3==(Acc_T[d]-Acc_T[c-1])%3 else 'NO')
``` | instruction | 0 | 36,737 | 0 | 73,474 |
Yes | output | 1 | 36,737 | 0 | 73,475 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO
Submitted Solution:
```
S = list(input())
T = list(input())
# S=list("BBAAABA")
# T=list("BBBA")
q = int(input())
l = []
for i in range(q):
l.append(list(map(int,input().split(" "))))
sta = [0 for i in range(len(S))]
stb = [0 for i in range(len(S))]
tta = [0 for i in range(len(T))]
ttb = [0 for i in range(len(T))]
if S[0] == "A":
sta[0] = 1
else:
stb[0] = 1
if T[0] == "A":
tta[0] = 1
else:
ttb[0] = 1
for i in range(1,len(S)):
if S[i] == "A":
sta[i] = sta[i-1]+1
stb[i] = stb[i-1]
else:
sta[i] = sta[i-1]
stb[i] = stb[i-1]+1
for i in range(1,len(T)):
if T[i] == "A":
tta[i] = tta[i-1]+1
ttb[i] = ttb[i-1]
else:
tta[i] = tta[i-1]
ttb[i] = ttb[i-1]+1
def get(l,a,b):
if a == 0:
return(l[b])
else:
return(l[b]-l[a-1])
for a,b,c,d in l:
sa = get(sta,a-1,b-1)
sb = get(stb,a-1,b-1)
ta = get(tta,c-1,d-1)
tb = get(ttb,c-1,d-1)
if (sa+sb*2)%3 == (ta+tb*2)%3:
print("YES")
else:
print("NO")
``` | instruction | 0 | 36,738 | 0 | 73,476 |
Yes | output | 1 | 36,738 | 0 | 73,477 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO
Submitted Solution:
```
#from collections import deque,defaultdict
printn = lambda x: print(x,end='')
inn = lambda : int(input())
inl = lambda: list(map(int, input().split()))
inm = lambda: map(int, input().split())
ins = lambda : input().strip()
DBG = True # and False
BIG = 10**18
R = 10**9 + 7
#R = 998244353
def ddprint(x):
if DBG:
print(x)
s = ins()
t = ins()
sacc = [0]
for c in s:
sacc.append(((1 if c=='A' else 2)+sacc[-1])%3)
tacc = [0]
for c in t:
tacc.append(((1 if c=='A' else 2)+tacc[-1])%3)
#ddprint(sacc)
#ddprint(tacc)
q = inn()
for i in range(q):
a,b,c,d = inm()
#ddprint('{} {} {} {}'.format(sacc[b],sacc[a-1],tacc[d],tacc[c-1]))
print('YES' if (sacc[b]-sacc[a-1])%3==(tacc[d]-tacc[c-1])%3 else 'NO')
``` | instruction | 0 | 36,739 | 0 | 73,478 |
Yes | output | 1 | 36,739 | 0 | 73,479 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO
Submitted Solution:
```
S = input()
T = input()
N = len(S)
M = len(T)
Ss = [0]*(N+1)
Ts = [0]*(M+1)
for i in range(N):
if S[i] == "A":
temp = 1
else:
temp = -1
Ss[i+1] = Ss[i] + temp
for i in range(M):
if T[i] == "A":
temp = 1
else:
temp = -1
Ts[i+1] = Ts[i] + temp
Q = int(input())
for i in range(Q):
a, b, c, d = (int(i) for i in input().split())
if a == b:
print("NO")
continue
else:
if (Ss[b] - Ss[a-1] - Ts[d] + Ts[c-1])%3==0:
print("YES")
else:
print("NO")
``` | instruction | 0 | 36,740 | 0 | 73,480 |
No | output | 1 | 36,740 | 0 | 73,481 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO
Submitted Solution:
```
s = input()
t = input()
q = int(input())
for _ in range(q):
a, b, c, d = map(int, input().split())
cnts = cntt = 0
for i in range(a-1, b):
cnts += 1 if s[i] == 'A' else 2
for i in range(c-1, d):
cntt += 1 if t[i] == 'A' else 2
print("YES" if cnts%3 == cntt%3 else "NO")
``` | instruction | 0 | 36,741 | 0 | 73,482 |
No | output | 1 | 36,741 | 0 | 73,483 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO
Submitted Solution:
```
import math
inp=lambda:(int,input().split())
s=input()
t=input()
q=int(input())
for _ in range(q):
a,b,c,d=inp()
if abs((s[a:b].count('A')-s[a:b].count('B')))==abs((t[a:b].count('A')-t[a:b].count('B'))):
print('YES')
else:
print('NO')
``` | instruction | 0 | 36,742 | 0 | 73,484 |
No | output | 1 | 36,742 | 0 | 73,485 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us consider the following operations on a string consisting of `A` and `B`:
1. Select a character in a string. If it is `A`, replace it with `BB`. If it is `B`, replace with `AA`.
2. Select a substring that is equal to either `AAA` or `BBB`, and delete it from the string.
For example, if the first operation is performed on `ABA` and the first character is selected, the string becomes `BBBA`. If the second operation is performed on `BBBAAAA` and the fourth through sixth characters are selected, the string becomes `BBBA`.
These operations can be performed any number of times, in any order.
You are given two string S and T, and q queries a_i, b_i, c_i, d_i. For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.
Constraints
* 1 \leq |S|, |T| \leq 10^5
* S and T consist of letters `A` and `B`.
* 1 \leq q \leq 10^5
* 1 \leq a_i \leq b_i \leq |S|
* 1 \leq c_i \leq d_i \leq |T|
Input
Input is given from Standard Input in the following format:
S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q
Output
Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print `YES`. Otherwise, print `NO`.
Examples
Input
BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4
Output
YES
NO
YES
NO
Input
AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14
Output
YES
YES
YES
YES
YES
YES
NO
NO
NO
NO
Submitted Solution:
```
s = input()
t = input()
q = int(input())
abcd = [tuple(map(int,input().split())) for i in range(q)]
def to_one(word):
if len(word)==0:
return 0
if len(word)==1:
return word
if word[:2] == 'AB' or word[:2] == 'BA':
return to_one(word[2:])
if word[:3] == 'AAA':
return to_one(word[3:])
if word[:3] == 'BBB':
return to_one(word[3:])
if word[:2] == 'AA':
return to_one('B'+word[2:])
if word[:2] == 'BB':
return to_one('A'+word[2:])
for a,b,c,d in abcd:
s_i = s[a-1:b]
t_i = t[c-1:d]
if to_one(s_i) == to_one(t_i):
print('YES')
else:
print('NO')
``` | instruction | 0 | 36,743 | 0 | 73,486 |
No | output | 1 | 36,743 | 0 | 73,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted. | instruction | 0 | 36,846 | 0 | 73,692 |
Tags: brute force, implementation, strings
Correct Solution:
```
a = input()[::-1]
b = input()[::-1]
k = 0
for i in range(min(len(a), len(b))):
if a[i] == b[i]:
k += 1
else:
break
print(len(a) + len(b) - 2 * k)
``` | output | 1 | 36,846 | 0 | 73,693 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted. | instruction | 0 | 36,847 | 0 | 73,694 |
Tags: brute force, implementation, strings
Correct Solution:
```
a = input()
b = input()
la = len(a)
lb = len(b)
ans = 0
if la==0 or lb==0 :
print (la + lb)
elif(a[la-1]!=b[lb-1]):
print(la + lb)
else:
m = min(la,lb)
for i in range(m):
if a[la-i-1]==b[lb-i-1]:
ans = ans + 1
else:
break
print(la + lb - ans - ans)
``` | output | 1 | 36,847 | 0 | 73,695 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted. | instruction | 0 | 36,848 | 0 | 73,696 |
Tags: brute force, implementation, strings
Correct Solution:
```
a = input()
b = input()
count = -0
if a == b:
print(0)
else:
while count > -min(len(b), len(a)):
if a[count-1] == b[count-1]:
count -= 1
# print(count, a[-count - 1], b[-count - 1])
else:
# count += 1
break
pass
print(len(a)+len(b)+count*2)
``` | output | 1 | 36,848 | 0 | 73,697 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted. | instruction | 0 | 36,849 | 0 | 73,698 |
Tags: brute force, implementation, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
s = input()
t = input()
ks = len(s) - 1
kt = len(t) - 1
cur = 0
while s[ks] == t[kt]:
cur += 1
ks -= 1
kt -= 1
if ks < 0 or kt < 0:
break
result = len(s) - 2*cur + len(t)
print(result)
``` | output | 1 | 36,849 | 0 | 73,699 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted. | instruction | 0 | 36,850 | 0 | 73,700 |
Tags: brute force, implementation, strings
Correct Solution:
```
s,t=input(),input()
x,flag=min(len(s),len(t)),0
for i in range(1,x+1):
if s[-i]==t[-i]:
continue
else:
flag=1
break
if flag==0:
print(len(s)-i+len(t)-i)
else:
print((len(s)-(i-1))+(len(t)-(i-1)))
``` | output | 1 | 36,850 | 0 | 73,701 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted. | instruction | 0 | 36,851 | 0 | 73,702 |
Tags: brute force, implementation, strings
Correct Solution:
```
a=list(input())
b=list(input())
a=a[::-1]
b=b[::-1]
i=0
while i<min(len(a),len(b)):
if a[i]==b[i]:
i+=1
else:
break
print(len(a)+len(b)-2*i)
``` | output | 1 | 36,851 | 0 | 73,703 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted. | instruction | 0 | 36,852 | 0 | 73,704 |
Tags: brute force, implementation, strings
Correct Solution:
```
a = input()
b = input()
a_len,b_len = len(a), len(b)
len_Sum = a_len+b_len
counter = 0
for ind in range(-1,(min(a_len,b_len)*-1 -1),-1):
if a[ind] != b[ind]:
break
else:
counter+=2
print(len_Sum - counter)
``` | output | 1 | 36,852 | 0 | 73,705 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted. | instruction | 0 | 36,853 | 0 | 73,706 |
Tags: brute force, implementation, strings
Correct Solution:
```
a=input()
b=input()
x=len(a)
y=len(b)
while x and y and a[x-1]==b[y-1]:
x-=1
y-=1
print(x+y)
``` | output | 1 | 36,853 | 0 | 73,707 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted.
Submitted Solution:
```
def main():
s1 = input()
s2 = input()
i = len(s1) - 1
j = len(s2) - 1
while i >= 0 and j >= 0:
if s1[i] == s2[j]:
i -= 1
j -= 1
else:
break
print((i + 1) + (j + 1))
if __name__ == "__main__":
main()
``` | instruction | 0 | 36,854 | 0 | 73,708 |
Yes | output | 1 | 36,854 | 0 | 73,709 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted.
Submitted Solution:
```
A = input()
B = input()
A = A[::-1]
B = B[::-1]
lena = len(A)
lenb = len(B)
i = 0
ans = 0
while True:
if lena!=0 and lenb!=0 and A[i] == B[i] :
lena -= 1
lenb -= 1
i += 1
else:
break
ans = lena + lenb
print(ans)
``` | instruction | 0 | 36,855 | 0 | 73,710 |
Yes | output | 1 | 36,855 | 0 | 73,711 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted.
Submitted Solution:
```
w1, w2 = input()[::-1], input()[::-1]
neq = 0
while neq < min(len(w1),len(w2)) and w1[neq] == w2[neq]: neq += 1
print(len(w1) + len(w2) - 2*neq)
``` | instruction | 0 | 36,856 | 0 | 73,712 |
Yes | output | 1 | 36,856 | 0 | 73,713 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted.
Submitted Solution:
```
s = list(reversed(input()))
t = list(reversed(input()))
i = 0
while i < len(s) and i < len(t) and s[i] == t[i]:
i += 1
print(len(s) - i + len(t) - i)
``` | instruction | 0 | 36,857 | 0 | 73,714 |
Yes | output | 1 | 36,857 | 0 | 73,715 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted.
Submitted Solution:
```
a=input()
b=input()
def f(a,b):
t=0
z1=len(a)
zz=len(b)
z=min(z1,zz)
p=abs(z1-zz)
#print(z,zz,p)
if len(a)==z:
for i in range(z-1,-1,-1):
if a[i]==b[i+p]:
t+=1
elif len(b)==z:
for i in range(z-1,-1,-1):
if b[i]==a[i+p]:
t+=1
return t
#print(a,b+'w',f(a,b))
z=len(a)
b2=len(b)
#print(f(a,b))
print(z+b2-2*f(a,b))
``` | instruction | 0 | 36,858 | 0 | 73,716 |
No | output | 1 | 36,858 | 0 | 73,717 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted.
Submitted Solution:
```
s = input()
t = input()
if len(t) > len(s):
g = s
s = t
t = g
moves = 0
if (len(s) == len(t)):
if s == t:
print (moves)
else:
while (s != t):
if len(s) <= 0 or len(t) <= 0:
break
s = s[1:]
t = t[1:]
moves += 2
print (moves)
elif len(s) > len(t):
while (len(s) != len(t)):
s = s[1:]
moves += 1
if s == t:
print (moves)
else:
while (s != t):
if len(s) <= 0 or len(t) <= 0:
break
s = s[1:]
t = t[1:]
moves += 2
print (moves)
``` | instruction | 0 | 36,859 | 0 | 73,718 |
No | output | 1 | 36,859 | 0 | 73,719 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted.
Submitted Solution:
```
s=input()
t=input()
z=0
d=0
while z>=0:
if (len(s)-1-z and len(t)-1-z)>=0:
if s[len(s)-1-z]==t[len(t)-1-z]:
z=z+1
d=d+2
else:
break
else:
break
print(len(s)+len(t)-d)
``` | instruction | 0 | 36,860 | 0 | 73,720 |
No | output | 1 | 36,860 | 0 | 73,721 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by 1. You can't choose a string if it is empty.
For example:
* by applying a move to the string "where", the result is the string "here",
* by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings s and t equal.
Input
The first line of the input contains s. In the second line of the input contains t. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2β
10^5, inclusive.
Output
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Examples
Input
test
west
Output
2
Input
codeforces
yes
Output
9
Input
test
yes
Output
7
Input
b
ab
Output
1
Note
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" 8 times. As a result, the string becomes "codeforces" β "es". The move should be applied to the string "yes" once. The result is the same string "yes" β "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted.
Submitted Solution:
```
T_ON = 0
DEBUG_ON = 0
MOD = 998244353
def solve():
s = input()
t = input()
if s == t:
print(0)
return
for i in range(min(len(s), len(t))):
j = -1 - i
if s[j] != t[j]:
break
print(len(s) + len(t) - 2 * i)
def main():
T = read_int() if T_ON else 1
for i in range(T):
solve()
def debug(*xargs):
if DEBUG_ON:
print(*xargs)
from collections import *
import math
#---------------------------------FAST_IO---------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#----------------------------------IO_WRAP--------------------------------------
def read_int():
return int(input())
def read_ints():
return list(map(int, input().split()))
def print_nums(nums):
print(" ".join(map(str, nums)))
def YES():
print("YES")
def Yes():
print("Yes")
def NO():
print("NO")
def No():
print("No")
def First():
print("First")
def Second():
print("Second")
#----------------------------------FIB--------------------------------------
def fib(n):
"""
the nth fib, start from zero
"""
a, b = 0, 1
for _ in range(n):
a, b = b, a + b
return a
def fib_ns(n):
"""
the first n fibs, start from zero
"""
assert n >= 1
f = [0 for _ in range(n + 1)]
f[0] = 0
f[1] = 1
for i in range(2, n + 1):
f[i] = f[i - 1] + f[i - 2]
return f
def fib_to_n(n):
"""
return fibs <= n, start from zero
n=8
f=[0,1,1,2,3,5,8]
"""
f = []
a, b = 0, 1
while a <= n:
f.append(a)
a, b = b, a + b
return f
#----------------------------------MOD--------------------------------------
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def xgcd(a, b):
"""return (g, x, y) such that a*x + b*y = g = gcd(a, b)"""
x0, x1, y0, y1 = 0, 1, 1, 0
while a != 0:
(q, a), b = divmod(b, a), a
y0, y1 = y1, y0 - q * y1
x0, x1 = x1, x0 - q * x1
return b, x0, y0
def lcm(a, b):
d = gcd(a, b)
return a * b // d
def is_even(x):
return x % 2 == 0
def is_odd(x):
return x % 2 == 1
def modinv(a, m):
"""return x such that (a * x) % m == 1"""
g, x, _ = xgcd(a, m)
if g != 1:
raise Exception('gcd(a, m) != 1')
return x % m
def mod_add(x, y):
x += y
while x >= MOD:
x -= MOD
while x < 0:
x += MOD
return x
def mod_mul(x, y):
return (x * y) % MOD
def mod_pow(x, y):
if y == 0:
return 1
if y % 2:
return mod_mul(x, mod_pow(x, y - 1))
p = mod_pow(x, y // 2)
return mod_mul(p, p)
def mod_inv(y):
return mod_pow(y, MOD - 2)
def mod_div(x, y):
# y^(-1): Fermat little theorem, MOD is a prime
return mod_mul(x, mod_inv(y))
#---------------------------------PRIME---------------------------------------
def is_prime(n):
if n == 1:
return False
for i in range(2, int(n ** 0.5) + 1):
debug(n, i)
if n % i == 0:
return False
return True
def gen_primes(n):
"""
generate primes of [1..n] using sieve's method
"""
P = [True for _ in range(n + 1)]
P[0] = P[1] = False
for i in range(int(n ** 0.5) + 1):
if P[i]:
for j in range(2 * i, n + 1, i):
P[j] = False
return P
#---------------------------------MISC---------------------------------------
def is_lucky(n):
return set(list(str(n))).issubset({'4', '7'})
#---------------------------------MAIN---------------------------------------
main()
``` | instruction | 0 | 36,861 | 0 | 73,722 |
No | output | 1 | 36,861 | 0 | 73,723 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string. | instruction | 0 | 36,879 | 0 | 73,758 |
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
def isPalindrome(word):
if word == word[::-1]:
return 1
else:
return 0
t=int(input())
for j in range(t):
a=list(input())
n=len(a)
half=n//2
if half!=1:
half+=1
for i in range(n-1):
#print(a)
flag=isPalindrome(a)
#print(flag,a)
if flag==0:
print(''.join(a))
break
else:
b=a[i+1]
a[i+1] = a[i]
a[i]=b
else:
print('-1')
"""
3
abcab
aaacaaa
bbbbbbbb
"""
``` | output | 1 | 36,879 | 0 | 73,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string. | instruction | 0 | 36,880 | 0 | 73,760 |
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
for i in range(int(input())):
a=input()
if len(set(a))==1:
print(-1)
else:
p=0
for j in range(0,len(a)//2):
if a[j]==a[len(a)-j-1]:
p+=1
if len(a)%2==0:
if p==len(a)//2:
print(a[1:]+a[0])
else:
print(a)
elif len(a)%2!=0:
if p == (len(a)//2):
print(a[(len(a)//2)]+a[0:(len(a)//2)]+a[(len(a)//2)+1:])
else:
print(a)
``` | output | 1 | 36,880 | 0 | 73,761 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string. | instruction | 0 | 36,881 | 0 | 73,762 |
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
import random
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def main():
n = I()
aa = [S() for _ in range(n)]
rr = []
for s in aa:
c = collections.Counter(s)
if len(c) == 1:
rr.append(-1)
else:
rr.append(''.join(sorted([c for c in s])))
return '\n'.join(map(str, rr))
print(main())
``` | output | 1 | 36,881 | 0 | 73,763 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string. | instruction | 0 | 36,882 | 0 | 73,764 |
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
for i in range(int(input())):
s = sorted(input())
print(-1 if s == s[::-1] else ''.join(s))
``` | output | 1 | 36,882 | 0 | 73,765 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string. | instruction | 0 | 36,883 | 0 | 73,766 |
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
t=int(input())
for i in range(t):
n=input()
c=[]
count=0
for i in range(len(n)):
c.insert(i,n[i])
m=list(reversed(n))
if c!=m:
print(n)
else:
for i in range(len(n)-1):
if n[i]==n[i+1]:
count+=1
if count==len(n)-1:
print("-1")
else:
x=c[len(n)-1]
c.remove(c[len(n)-1])
c.append(x)
for i in range(len(n)):
print(c[i],end="")
print()
``` | output | 1 | 36,883 | 0 | 73,767 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string. | instruction | 0 | 36,884 | 0 | 73,768 |
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
def is_pal(s):
for i in range(len(s)):
if s[i] != s[-(i+1)]:
return False
return True
t = int(input())
for tt in range(t):
s = input()
if is_pal(s):
s0 = s[0]
sw = 0
for i in range(len(s)):
if s[i] != s0:
sw = i
break
if sw == 0:
print(-1)
else:
sl = list(s)
sl[0], sl[sw] = sl[sw], sl[0]
print(''.join(sl))
else:
print(s)
``` | output | 1 | 36,884 | 0 | 73,769 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s consisting only of lowercase Latin letters.
You can rearrange all letters of this string as you wish. Your task is to obtain a good string by rearranging the letters of the given string or report that it is impossible to do it.
Let's call a string good if it is not a palindrome. Palindrome is a string which is read from left to right the same as from right to left. For example, strings "abacaba", "aa" and "z" are palindromes and strings "bba", "xd" are not.
You have to answer t independent queries.
Input
The first line of the input contains one integer t (1 β€ t β€ 100) β number of queries.
Each of the next t lines contains one string. The i-th line contains a string s_i consisting only of lowercase Latin letter. It is guaranteed that the length of s_i is from 1 to 1000 (inclusive).
Output
Print t lines. In the i-th line print the answer to the i-th query: -1 if it is impossible to obtain a good string by rearranging the letters of s_i and any good string which can be obtained from the given one (by rearranging the letters) otherwise.
Example
Input
3
aa
abacaba
xdd
Output
-1
abaacba
xdd
Note
In the first query we cannot rearrange letters to obtain a good string.
Other examples (not all) of correct answers to the second query: "ababaca", "abcabaa", "baacaba".
In the third query we can do nothing to obtain a good string. | instruction | 0 | 36,885 | 0 | 73,770 |
Tags: constructive algorithms, greedy, sortings, strings
Correct Solution:
```
#In the name of GOD!
t = int(input())
for i in range(t):
s = list(input())
s.sort()
if s[0] == s[-1]: print(-1)
else: print(''.join(s))
``` | output | 1 | 36,885 | 0 | 73,771 |
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