<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>

class Solution:
    def canMakePalindromeQueries(self, s: str, queries: List[List[int]]) -> List[bool]:
        n = len(s)
        m = len(queries)
        if ''.join(sorted(s[:n//2])) != ''.join(sorted(s[n//2:])):
            return [False] * m

        s1 = s[:n//2]
        s2 = s[n//2:][::-1]
        prefix_sum_s1 = [[0] * 26 for i in range(n//2 + 1)]
        prefix_sum_s2 = [[0] * 26 for i in range(n//2 + 1)]

        for i in range(n//2):
            char_index = ord(s1[i]) - ord('a')
            for j in range(26):
                prefix_sum_s1[i+1][j] = prefix_sum_s1[i][j]
            prefix_sum_s1[i+1][char_index] += 1
        
        for i in range(n//2):
            char_index = ord(s2[i]) - ord('a')
            for j in range(26):
                prefix_sum_s2[i+1][j] = prefix_sum_s2[i][j]
            prefix_sum_s2[i+1][char_index] += 1

        def get_char_freq(prefix_sum, a, b):
            return [prefix_sum[b+1][i] - prefix_sum[a][i] for i in range(26)]

        tot_mismatch = 0
        prefix_sum_mismatch = [0]
        for i in range(n // 2):
            prefix_sum_mismatch.append(prefix_sum_mismatch[-1])
            if s1[i] != s2[i]:
                prefix_sum_mismatch[-1] += 1
                tot_mismatch += 1

        def get_mismatch(a, b):
            return prefix_sum_mismatch[b+1] - prefix_sum_mismatch[a]

        res = []
        change_flag = False
        for ai, bi, ci, di in queries:
            ci, di = n - 1 - di, n - 1 - ci
            if change_flag is True:
                s1, s2 = s2, s1
                prefix_sum_s1, prefix_sum_s2 = prefix_sum_s2, prefix_sum_s1
                change_flag = False
            if ai > ci or (ai == ci and bi < di):
                ai, bi, ci, di = ci, di, ai, bi
                s1, s2 = s2, s1
                prefix_sum_s1, prefix_sum_s2 = prefix_sum_s2, prefix_sum_s1
                change_flag = True
            # [ [ ] ]
            # a c d b
            if bi >= di:
                res.append(get_mismatch(ai, bi) == tot_mismatch)
                continue
            # [ ] [ ] 
            # a b c d
            elif ci > bi:
                freq1_s1 = get_char_freq(prefix_sum_s1, ai, bi)
                freq1_s2 = get_char_freq(prefix_sum_s2, ai, bi)
                freq2_s1 = get_char_freq(prefix_sum_s1, ci, di)
                freq2_s2 = get_char_freq(prefix_sum_s2, ci, di)
                if get_mismatch(ai, bi) + get_mismatch(ci, di) != tot_mismatch:
                    res.append(False)
                    continue
                if not (all(freq1_s1[i] == freq1_s2[i] for i in range(26)) and all(freq2_s1[i] == freq2_s2[i] for i in range(26))):
                    res.append(False)
                    continue
                res.append(True)
                continue
            # [ [ ] ]
            # a c b d
            else:
                freq1_s1 = get_char_freq(prefix_sum_s1, ai, bi)
                freq_left_s2 = get_char_freq(prefix_sum_s2, ai, ci-1)
                freq2_s2 = get_char_freq(prefix_sum_s2, ci, di)
                freq_right_s1 = get_char_freq(prefix_sum_s1, bi+1, di)
                if get_mismatch(ai, di) != tot_mismatch:
                    res.append(False)
                    continue
                if not (all(freq1_s1[i] >= freq_left_s2[i] for i in range(26)) and all(freq2_s2[i] >= freq_right_s1[i] for i in range(26))):
                    res.append(False)
                    continue
                res.append(True)
                continue
        return res
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