<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)
        s1 = s[:n//2]
        s2 = s[n//2:][::-1]

        prefix_sums1 = [Counter()]
        prefix_sums2 = [Counter()]
        mismatches = [0]

        for c1, c2 in zip(s1, s2):
            prefix_sums1.append(Counter(prefix_sums1[-1]))
            prefix_sums1[-1][c1] += 1
            prefix_sums2.append(Counter(prefix_sums2[-1]))
            prefix_sums2[-1][c2] += 1
            mismatches.append(mismatches[-1] + (1 if c1 != c2 else 0))

        def match(a, b):
            a = max(0, a)
            b = min(n//2-1, b)
            if a > b:
                return True
            return mismatches[b+1] - mismatches[a] == 0

        def swapable(a, b):
            return span(0, a, b) == span(1, a, b)

        @cache
        def span(s, l, r):
            if l > r:
                return Counter()
            if s == 0:
                return prefix_sums1[r+1] - prefix_sums1[l]
            return prefix_sums2[r+1] - prefix_sums2[l]

        def contains(a, b):
            for k, v in b.items():
                if a[k] < v:
                    return False
            return True
            
        def check(a, b, c, d):
            l1, r1 = min(a, c), max(b, d)
            l2, r2 = max(a, c), min(b, d)
            if l2 > r2: # No intersection
                return match(0, min(a, c)-1) and swapable(min(a, c), min(b, d)) \
                   and match(min(b, d)+1, max(a, c)-1) and swapable(max(a, c), max(b, d)) \
                   and match(max(b, d)+1, n // 2 - 1)
            elif a <= c and b >= d: # [a, b] covers [c, d]
                return match(0, a-1) and swapable(a, b) and match(b+1, n//2-1)
            elif c <= a and d >= b: # [c, d] covers [a, b]
                return match(0, c-1) and swapable(c, d) and match(d+1, n//2-1)
            elif a < c:
                if not match(0, a-1) or not match(d+1, n // 2 - 1):
                    return False
                if not contains(span(0, a, b), span(1, a, c-1)) or not contains(span(1, c, d), span(0, b+1, d)):
                    return False
                l = span(0, a, b) - span(1, a, c-1)
                r = span(1, c, d) - span(0, b+1, d)
                return l == r
            else: # c < a
                if not match(0, c-1) or not match(b+1, n // 2 - 1):
                    return False
                if not contains(span(0, a, b), span(1, d+1, b)) or not contains(span(1, c, d), span(0, c, a-1)):
                    return False
                l = span(1, c, d) - span(0, c, a-1)
                r = span(0, a, b) - span(1, d+1, b)
                return l == r
        for a, b, c, d in queries:
            yield check(a, b, n - 1 - d, n - 1 - c)
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