<p>You are given a <strong>0-indexed</strong> string <code>s</code> having an <strong>even</strong> length <code>n</code>.</p>

<p>You are also given a <strong>0-indexed</strong> 2D integer array, <code>queries</code>, where <code>queries[i] = [a<sub>i</sub>, b<sub>i</sub>, c<sub>i</sub>, d<sub>i</sub>]</code>.</p>

<p>For each query <code>i</code>, you are allowed to perform the following operations:</p>

<ul>
	<li>Rearrange the characters within the <strong>substring</strong> <code>s[a<sub>i</sub>:b<sub>i</sub>]</code>, where <code>0 &lt;= a<sub>i</sub> &lt;= b<sub>i</sub> &lt; n / 2</code>.</li>
	<li>Rearrange the characters within the <strong>substring</strong> <code>s[c<sub>i</sub>:d<sub>i</sub>]</code>, where <code>n / 2 &lt;= c<sub>i</sub> &lt;= d<sub>i</sub> &lt; n</code>.</li>
</ul>

<p>For each query, your task is to determine whether it is possible to make <code>s</code> a <strong>palindrome</strong> by performing the operations.</p>

<p>Each query is answered <strong>independently</strong> of the others.</p>

<p>Return <em>a <strong>0-indexed</strong> array </em><code>answer</code><em>, where </em><code>answer[i] == true</code><em> if it is possible to make </em><code>s</code><em> a palindrome by performing operations specified by the </em><code>i<sup>th</sup></code><em> query, and </em><code>false</code><em> otherwise.</em></p>

<ul>
	<li>A <strong>substring</strong> is a contiguous sequence of characters within a string.</li>
	<li><code>s[x:y]</code> represents the substring consisting of characters from the index <code>x</code> to index <code>y</code> in <code>s</code>, <strong>both inclusive</strong>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcabc&quot;, queries = [[1,1,3,5],[0,2,5,5]]
<strong>Output:</strong> [true,true]
<strong>Explanation:</strong> In this example, there are two queries:
In the first query:
- a<sub>0</sub> = 1, b<sub>0</sub> = 1, c<sub>0</sub> = 3, d<sub>0</sub> = 5.
- So, you are allowed to rearrange s[1:1] =&gt; a<u>b</u>cabc and s[3:5] =&gt; abc<u>abc</u>.
- To make s a palindrome, s[3:5] can be rearranged to become =&gt; abc<u>cba</u>.
- Now, s is a palindrome. So, answer[0] = true.
In the second query:
- a<sub>1</sub> = 0, b<sub>1</sub> = 2, c<sub>1</sub> = 5, d<sub>1</sub> = 5.
- So, you are allowed to rearrange s[0:2] =&gt; <u>abc</u>abc and s[5:5] =&gt; abcab<u>c</u>.
- To make s a palindrome, s[0:2] can be rearranged to become =&gt; <u>cba</u>abc.
- Now, s is a palindrome. So, answer[1] = true.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abbcdecbba&quot;, queries = [[0,2,7,9]]
<strong>Output:</strong> [false]
<strong>Explanation:</strong> In this example, there is only one query.
a<sub>0</sub> = 0, b<sub>0</sub> = 2, c<sub>0</sub> = 7, d<sub>0</sub> = 9.
So, you are allowed to rearrange s[0:2] =&gt; <u>abb</u>cdecbba and s[7:9] =&gt; abbcdec<u>bba</u>.
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.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;acbcab&quot;, queries = [[1,2,4,5]]
<strong>Output:</strong> [true]
<strong>Explanation: </strong>In this example, there is only one query.
a<sub>0</sub> = 1, b<sub>0</sub> = 2, c<sub>0</sub> = 4, d<sub>0</sub> = 5.
So, you are allowed to rearrange s[1:2] =&gt; a<u>cb</u>cab and s[4:5] =&gt; acbc<u>ab</u>.
To make s a palindrome s[1:2] can be rearranged to become a<u>bc</u>cab.
Then, s[4:5] can be rearranged to become abcc<u>ba</u>.
Now, s is a palindrome. So, answer[0] = true.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n == s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code>queries[i].length == 4</code></li>
	<li><code>a<sub>i</sub> == queries[i][0], b<sub>i</sub> == queries[i][1]</code></li>
	<li><code>c<sub>i</sub> == queries[i][2], d<sub>i</sub> == queries[i][3]</code></li>
	<li><code>0 &lt;= a<sub>i</sub> &lt;= b<sub>i</sub> &lt; n / 2</code></li>
	<li><code>n / 2 &lt;= c<sub>i</sub> &lt;= d<sub>i</sub> &lt; n </code></li>
	<li><code>n</code> is even.</li>
	<li><code>s</code> consists of only lowercase English letters.</li>
</ul>

from collections import Counter
class Solution:
    def canMakePalindromeQueries(self, s: str, queries: List[List[int]]) -> List[bool]:
        
        def get_prefix_count_diff(prefix_count, i, j):
            return [ x-y for x, y in zip(prefix_count[j+1], prefix_count[i])]
        
        def subtract_count(count1, count2):
            res = []
            for x, y in zip(count1, count2):
                if x - y < 0:
                    return []
                res.append(x-y)
            return res
        
        def is_pali(prefix1, prefix2, start1, end1, start2, end2):
            if diff[start1] > 0 or diff[mid]-diff[max(end1,end2)+1] >0:
                return False
            
            if end2 <= end1:
                return get_prefix_count_diff(prefix1, start1,end1) == get_prefix_count_diff(prefix2,start1,end1)

            if start2 > end1:
                return (diff[start2]-diff[end1+1] == 0) and get_prefix_count_diff(prefix1,start1,end1) == get_prefix_count_diff(prefix2,start1,end1) and get_prefix_count_diff(prefix1,start2,end2)==get_prefix_count_diff(prefix2,start2,end2)
            
            c1 = subtract_count(get_prefix_count_diff(prefix1, start1, end1), get_prefix_count_diff(prefix2,start1,start2-1))
            c2 = subtract_count(get_prefix_count_diff(prefix2, start2, end2), get_prefix_count_diff(prefix1, end1+1, end2))
            return c1 and c2 and c1==c2
        
        n = len(s)
        mid = n // 2
        s_rev = s[mid:][::-1]
        s = s[:mid]
        prefix1 = [[0] * 26 for _ in range(mid+1)]
        prefix2 = [[0] * 26 for _ in range(mid+1)]
        diff = [0] * (mid+1)

        for i, (c1,c2) in enumerate(zip(s, s_rev), 1):
            prefix1[i] = prefix1[i-1][:]
            prefix2[i] = prefix2[i-1][:]
            prefix1[i][ord(c1)-ord('a')] += 1
            prefix2[i][ord(c2)-ord('a')] += 1
            diff[i] = diff[i-1] + int((c1!=c2))
        
        res = []
        for a,b,c,d in queries:
            c,d = n-1-d, n-1-c
            is_possible = is_pali(prefix1, prefix2, a,b,c,d) if a <= c else is_pali(prefix2,prefix1,c,d,a,b)
            res.append(is_possible)
        return res

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