问题

### Instructions: You are an expert Python programmer. You will be given a question (problem specification) and the first lines of Python solution to this problem, and will write in Python the remaining lines of the program to produce a correct Python program that matches the specification and passes all tests. You will NOT return anything except for the second part of the program that you wrote.
### Question:
You are given a 0-indexed string s having an even length n.
You are also given a 0-indexed 2D integer array, queries, where queries[i] = [a_i, b_i, c_i, d_i].
For each query i, you are allowed to perform the following operations:

Rearrange the characters within the substring s[a_i:b_i], where 0 <= a_i <= b_i < n / 2.
Rearrange the characters within the substring s[c_i:d_i], where n / 2 <= c_i <= d_i < n.

For each query, your task is to determine whether it is possible to make s a palindrome by performing the operations.
Each query is answered independently of the others.
Return a 0-indexed array answer, where answer[i] == true if it is possible to make s a palindrome by performing operations specified by the i^th query, and false otherwise.

A substring is a contiguous sequence of characters within a string.
s[x:y] represents the substring consisting of characters from the index x to index y in s, both inclusive.


Example 1:

Input: s = "abcabc", queries = [[1,1,3,5],[0,2,5,5]]
Output: [true,true]
Explanation: In this example, there are two queries:
In the first query:
- a_0 = 1, b_0 = 1, c_0 = 3, d_0 = 5.
- So, you are allowed to rearrange s[1:1] => abcabc and s[3:5] => abcabc.
- To make s a palindrome, s[3:5] can be rearranged to become => abccba.
- Now, s is a palindrome. So, answer[0] = true.
In the second query:
- a_1 = 0, b_1 = 2, c_1 = 5, d_1 = 5.
- So, you are allowed to rearrange s[0:2] => abcabc and s[5:5] => abcabc.
- To make s a palindrome, s[0:2] can be rearranged to become => cbaabc.
- Now, s is a palindrome. So, answer[1] = true.

Example 2:

Input: s = "abbcdecbba", queries = [[0,2,7,9]]
Output: [false]
Explanation: In this example, there is only one query.
a_0 = 0, b_0 = 2, c_0 = 7, d_0 = 9.
So, you are allowed to rearrange s[0:2] => abbcdecbba and s[7:9] => abbcdecbba.
It is not possible to make s a palindrome by rearranging these substrings because s[3:6] is not a palindrome.
So, answer[0] = false.
Example 3:

Input: s = "acbcab", queries = [[1,2,4,5]]
Output: [true]
Explanation: In this example, there is only one query.
a_0 = 1, b_0 = 2, c_0 = 4, d_0 = 5.
So, you are allowed to rearrange s[1:2] => acbcab and s[4:5] => acbcab.
To make s a palindrome s[1:2] can be rearranged to become abccab.
Then, s[4:5] can be rearranged to become abccba.
Now, s is a palindrome. So, answer[0] = true.

Constraints:

2 <= n == s.length <= 10^5
1 <= queries.length <= 10^5
queries[i].length == 4
a_i == queries[i][0], b_i == queries[i][1]
c_i == queries[i][2], d_i == queries[i][3]
0 <= a_i <= b_i < n / 2
n / 2 <= c_i <= d_i < n
n is even.
s consists of only lowercase English letters.

### Format: You will use the following starter code to write the solution to the problem and enclose your code within delimiters.
```python
class Solution(object):
def canMakePalindromeQueries(self, s, queries):
"""
:type s: str
:type queries: List[List[int]]
:rtype: List[bool]
"""
def check(left1, right1, left2, right2):
def same(left, right):
return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d))

min_left, max_left = min(left1, left2), max(left1, left2)
min_right, max_right = min(right1, right2), max(right1, right2)
if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0):
return False
if min_right < max_left: # non-overlapped
return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right)
# overlapped
if (left1 == min_left) == (right1 == max_right): # inside another
return same(min_left, max_right)
# not inside another
p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1)
diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)]
diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)]
return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]]

lookup = [-1]*26
d = 0
for x in s:
if lookup[ord(x)-ord('a')] != -1:
continue
lookup[ord(x)-ord('a')] = d
d += 1
prefix = [0]*(len(s)//2+1)
prefixs1 = [[0]*d for _ in range(len(s)//2+1)]
prefixs2 = [[0]*d for _ in range(len(s)//2+1)]
```

### Answer: (enclose your partial completion in backticks. Only write the missing portion of the code, not the entire code. Be very careful to match the appropriate indentation. Directly appending your code after the partial code should produce a correct completion solution.)

答案

模型 模型答案 正确答案
金声玉亮2.0 ```python
for i in range(len(s)//2):
prefix[i+1] = prefix[i] + (1 if s[i] != s[len(s)-1-i] else 0)
for j in range(d):
prefixs1[i+1][j] = prefixs1[i][j] + (1 if lookup[ord(s[i])-ord('a')] == j else 0)
prefixs2[i+1][j] = prefixs2[i][j] + (1 if lookup[ord(s[len(s)-1-i])-ord('a')] == j else 0)

result = []
for a, b, c, d in queries:
result.append(check(a, b, len(s)-1-d, len(s)-1-c))

return result
```		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]
deepseek-r1

```python
half = len(s) // 2
s1 = s[:half]
s2_rev = s[half:][::-1]
for i in range(half):
prefix[i+1] = prefix[i] + (s1[i] != s2_rev[i])
c = lookup[ord(s1[i]) - ord('a')]
prefixs1[i+1] = list(prefixs1[i])
prefixs1[i+1][c] += 1
c = lookup[ord(s2_rev[i]) - ord('a')]
prefixs2[i+1] = list(prefixs2[i])
prefixs2[i+1][c] += 1

answer = []
for a, b, c, d in queries:
left2 = (len(s)-1) - d
right2 = (len(s)-1) - c
answer.append(check(a, b, left2, right2))
return answer
```		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]
gpto1 ```python
n = len(s)
N = n // 2
for i in range(N):
c1 = s[i]
c2 = s[n - 1 - i]
idx1 = lookup[ord(c1) - ord('a')]
idx2 = lookup[ord(c2) - ord('a')]
prefixs1[i + 1] = prefixs1[i][:]
prefixs1[i + 1][idx1] += 1
prefixs2[i + 1] = prefixs2[i][:]
prefixs2[i + 1][idx2] += 1
if c1 != c2:
prefix[i + 1] = prefix[i] + 1
else:
prefix[i + 1] = prefix[i]
answer = []
for a_i, b_i, c_i, d_i in queries:
left1, right1 = a_i, b_i
left2 = n - 1 - d_i
right2 = n - 1 - c_i
if left2 > right2:
left2, right2 = right2, left2
if check(left1, right1, left2, right2):
answer.append(True)
else:
answer.append(False)
return answer
```		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]
gpt-4o-2024-08-06 ```python
for i in range(len(s)//2):
prefix[i+1] = prefix[i] + (1 if s[i] != s[len(s)-1-i] else 0)
for j in range(d):
prefixs1[i+1][j] = prefixs1[i][j]
prefixs2[i+1][j] = prefixs2[i][j]
if lookup[ord(s[i])-ord('a')] != -1:
prefixs1[i+1][lookup[ord(s[i])-ord('a')]] += 1
if lookup[ord(s[len(s)-1-i])-ord('a')] != -1:
prefixs2[i+1][lookup[ord(s[len(s)-1-i])-ord('a')]] += 1

result = []
for a, b, c, d in queries:
if check(a, b, c-len(s)//2, d-len(s)//2):
result.append(True)
else:
result.append(False)
return result
```		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]
Meta-Llama-3.1-405B-Instruct-Turbo ```python
for i in range(len(s)//2):
prefix[i+1] = prefix[i]
prefixs1[i+1] = prefixs1[i][:]
prefixs2[i+1] = prefixs2[i][:]
if s[i] != s[-i-1]:
prefix[i+1] += 1
prefixs1[i+1][lookup[ord(s[i])-ord('a')]] += 1
prefixs2[i+1][lookup[ord(s[-i-1])-ord('a')]] += 1
return [check(a, b, c-len(s)//2, d-len(s)//2) for a, b, c, d in queries]
```		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]
Meta-Llama-3.1-70B-Instruct-Turbo ```python
for i in range(len(s)//2):
prefix[i+1] = prefix[i] + (lookup[ord(s[i])-ord('a')] != -1)
prefixs1[i+1] = prefixs1[i][:]
prefixs2[i+1] = prefixs2[i][:]
if lookup[ord(s[i])-ord('a')] != -1:
prefixs1[i+1][lookup[ord(s[i])-ord('a')]] += 1
if lookup[ord(s[-i-1])-ord('a')] != -1:
prefixs2[i+1][lookup[ord(s[-i-1])-ord('a')]] += 1
return [check(a, b, c-len(s)//2, d-len(s)//2) for a, b, c, d in queries]
```		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]
DeepSeek-Coder-V2-Lite-Instruct ```python
for i in range(len(s)//2):
prefix[i+1] = prefix[i] + (1 if s[i] != s[~i] else 0)
for j in range(d):
prefixs1[i+1][j] = prefixs1[i][j] + (1 if lookup[ord(s[i])-ord('a')] == j else 0)
prefixs2[i+1][j] = prefixs2[i][j] + (1 if lookup[ord(s[~i])-ord('a')] == j else 0)

answer = []
for a, b, c, d in queries:
if prefix[b+1] - prefix[a] + prefix[d+1] - prefix[c] == 0:
answer.append(True)
else:
answer.append(False)
return answer
```		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]
Qwen2-72B-Instruct ```python
for i, x in enumerate(s[:len(s)//2]):
prefix[i+1] = prefix[i] + (lookup[ord(x)-ord('a')]+1)
for j in range(d):
prefixs1[i+1][j] = prefixs1[i][j] + (j == lookup[ord(x)-ord('a')])
for i, x in enumerate(s[len(s)//2:]):
prefix[i+len(s)//2+1] = prefix[i+len(s)//2] + (lookup[ord(x)-ord('a')]+1)
for j in range(d):
prefixs2[i+1][j] = prefixs2[i][j] + (j == lookup[ord(x)-ord('a')])

ans = []
for a, b, c, d in queries:
ans.append(check(a, b, c-len(s)//2, d-len(s)//2))
return ans
```		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]
baichuan4 ```python
for i in range(len(s)//2):
prefix[i+1] = prefix[i] + lookup[ord(s[i])-ord('a')]
prefixs1[i+1][lookup[ord(s[i])-ord('a')]] += 1
for i in range(len(s)//2, len(s)):
prefixs2[i-len(s)//2+1][lookup[ord(s[i])-ord('a')]] += 1

return [check(a, b, c-len(s)//2, d-len(s)//2) for a, b, c, d in queries]
```		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]
doubao ```python
result = []
for query in queries:
left1, right1, left2, right2 = query
result.append(check(left1, right1, left2, right2))
return result
```		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]
gpto1 ```python
n = len(s)
N = n // 2
for i in range(N):
c1 = s[i]
c2 = s[n - 1 - i]
idx1 = lookup[ord(c1) - ord('a')]
idx2 = lookup[ord(c2) - ord('a')]
prefixs1[i + 1] = prefixs1[i][:]
prefixs1[i + 1][idx1] += 1
prefixs2[i + 1] = prefixs2[i][:]
prefixs2[i + 1][idx2] += 1
if c1 != c2:
prefix[i + 1] = prefix[i] + 1
else:
prefix[i + 1] = prefix[i]
answer = []
for a_i, b_i, c_i, d_i in queries:
left1, right1 = a_i, b_i
left2 = n - 1 - d_i
right2 = n - 1 - c_i
if left2 > right2:
left2, right2 = right2, left2
if check(left1, right1, left2, right2):
answer.append(True)
else:
answer.append(False)
return answer
```		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]
kimi ```python
for i in range(len(s)//2):
prefix[i+1] = prefix[i] + 1 if s[2*i] == s[2*i+1] else prefix[i]
for j in range(d):
prefixs1[i+1][j] = prefixs1[i][j] + 1 if s[2*i] == chr(ord('a')+j) else prefixs1[i][j]
prefixs2[i+1][j] = prefixs2[i][j] + 1 if s[2*i+1] == chr(ord('a')+j) else prefixs2[i][j]
result = []
for a, b, c, d in queries:
result.append(check(a, b, c, d))
return result
```		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]
minimax ```python
for i in range(1, len(s)//2+1):
prefix[i] = prefix[i-1] + lookup[ord(s[i-1])-ord('a')]
for j in range(d):
prefixs1[i][j] = prefixs1[i-1][j] + (1 if j == lookup[ord(s[i-1])-ord('a')] else 0)
prefixs2[i][j] = prefixs2[i-1][j] + (1 if j == lookup[ord(s[-i])-ord('a')] else 0)

answer = []
for a, b, c, d in queries:
answer.append(check(a, b, c, d))
return answer
```		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]
wenxinyiyan ```python
for i, x in enumerate(s):
if i < len(s)//2:
prefix[i+1] = prefix[i] + (1<<(lookup[ord(x)-ord('a')]))
for j in range(d):
prefixs1[i+1][j] = prefixs1[i][j]
prefixs1[i+1][lookup[ord(x)-ord('a')]] += 1
else:
for j in range(d):
prefixs2[i-len(s)//2+1][j] = prefixs2[i-len(s)//2][j]
prefixs2[i-len(s)//2+1][lookup[ord(x)-ord('a')]] += 1

result = []
for query in queries:
left1, right1, left2, right2 = query
left2 -= len(s)//2
right2 -= len(s)//2
result.append(check(left1, right1, left2, right2))
return result
```

### Explanation:

The provided solution uses a prefix sum array and bitmasks to efficiently check whether a given query can make the string `s` a palindrome. Here's a step-by-step explanation of the missing portion:

1. **Initialization of Lookup Table and Prefix Arrays**:


* The `lookup` table is used to map each unique character in `s` to a unique index (0 to 25 for lowercase English letters). This helps in creating bitmasks and prefix sums efficiently.
* `d` represents the number of unique characters encountered so far.
* `prefix` is a prefix sum array that stores bitmasks representing the presence of characters in the left half of `s`.
* `prefixs1` and `prefixs2` are 2D prefix sum arrays that store the count of each unique character in the left and right halves of `s`, respectively.
2. **Populating the Prefix Arrays**:


* The code iterates over each character `x` in `s` and its corresponding index `i`.
* If `i` is in the left half of `s`, the bitmask in `prefix` is updated, and the count of the character in `prefixs1` is incremented.
* If `i` is in the right half of `s`, the index is adjusted to match the corresponding position in the left half (`i-len(s)//2`), and the count of the character in `prefixs2` is incremented.
3. **Processing Queries**:


* For each query in `queries`, the left and right indices (`left1`, `right1`, `left2`, `right2`) are extracted.
* The `left2` and `right2` indices are adjusted to match the corresponding positions in the left half of `s` (`left2 -= len(s)//2` and `right2 -= len(s)//2`).
* The `check` function is called with these indices, and its result is appended to the `result` list.
4. **The `check` Function**:


* This function checks whether a given query can make `s` a palindrome.
* It compares the bitmasks and character counts in the prefix arrays to determine if the necessary rearrangements are possible.
* The logic inside `check` is complex and involves multiple cases (non-overlapped, inside another, not inside another) to handle different scenarios of character rearrangements.
5. **Returning the Result**:


* Finally, the `result` list, which contains the answers (`true` or `false`) for each query, is returned.

Note: The provided solution assumes that the input constraints are satisfied and does not include explicit error handling for invalid inputs.		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]
zhipu ```python
for i in range(len(s)//2):
x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[-i-1])-ord('a')]
prefix[i+1] = prefix[i] + (x != y)
for j in range(d):
prefixs1[i+1][j] = prefixs1[i][j] + (j == x)
prefixs2[i+1][j] = prefixs2[i][j] + (j == y)
return [check(a, b, c, d) for a, b, c, d in queries]
```		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]
deepseek-v3 ```python
for i in range(len(s)//2):
prefix[i+1] = prefix[i] + (1 if s[i] != s[len(s)-1-i] else 0)
for j in range(d):
prefixs1[i+1][j] = prefixs1[i][j]
prefixs2[i+1][j] = prefixs2[i][j]
prefixs1[i+1][lookup[ord(s[i])-ord('a')]] += 1
prefixs2[i+1][lookup[ord(s[len(s)-1-i])-ord('a')]] += 1
return [check(a, b, c - len(s)//2, d - len(s)//2) for a, b, c, d in queries]
```		 class Solution(object): def canMakePalindromeQueries(self, s, queries): """ :type s: str :type queries: List[List[int]] :rtype: List[bool] """ def check(left1, right1, left2, right2): def same(left, right): return all(prefixs1[right+1][i]-prefixs1[left][i] == prefixs2[right+1][i]-prefixs2[left][i] for i in range(d)) min_left, max_left = min(left1, left2), max(left1, left2) min_right, max_right = min(right1, right2), max(right1, right2) if not (prefix[min_left]-prefix[0] == prefix[-1]-prefix[max_right+1] == 0): return False if min_right < max_left: # non-overlapped return prefix[max_left]-prefix[min_right+1] == 0 and same(min_left, min_right) and same(max_left, max_right) # overlapped if (left1 == min_left) == (right1 == max_right): # inside another return same(min_left, max_right) # not inside another p1, p2 = (prefixs1, prefixs2) if min_left == left1 else (prefixs2, prefixs1) diff1 = [(p1[min_right+1][i]-p1[min_left][i])-(p2[max_left][i]-p2[min_left][i]) for i in range(d)] diff2 = [(p2[max_right+1][i]-p2[max_left][i])-(p1[max_right+1][i]-p1[min_right+1][i]) for i in range(d)] return diff1 == diff2 and all(x >= 0 for x in diff1) # test case: s = "aabbba", queries = [[0,1,3,4]] lookup = [-1]*26 d = 0 for x in s: if lookup[ord(x)-ord('a')] != -1: continue lookup[ord(x)-ord('a')] = d d += 1 prefix = [0]*(len(s)//2+1) prefixs1 = [[0]*d for _ in range(len(s)//2+1)] prefixs2 = [[0]*d for _ in range(len(s)//2+1)] for i in range(len(s)//2): x, y = lookup[ord(s[i])-ord('a')], lookup[ord(s[~i])-ord('a')] prefix[i+1] = prefix[i]+int(x != y) for j in range(d): prefixs1[i+1][j] = prefixs1[i][j]+int(j == x) prefixs2[i+1][j] = prefixs2[i][j]+int(j == y) return [check(q[0], q[1], (len(s)-1)-q[3], (len(s)-1)-q[2]) for q in queries]