<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)
    # mirroredDiffs[i] := the number of different letters between the first i
    # letters of s[0..n / 2) and the first i letters of s[n / 2..n)[::-1]
    mirroredDiffs = self._getMirroredDiffs(s)
    # counts[i] := the count of s[0..i)
    counts = self._getCounts(s)
    ans = []

    def subtractArrays(a: List[int], b: List[int]):
      return [x - y for x, y in zip(a, b)]

    for a, b, c, d in queries:
      # Use left-closed, right-open intervals to facilitate the calculation.
      #   ...... [a, b) ...|... [rb, ra) ......
      #   .... [rd, rc) .....|..... [c, d) ....
      b += 1
      d += 1
      ra = n - a  # the reflected index of a in s[n / 2..n)
      rb = n - b  # the reflected index of b in s[n / 2..n)
      rc = n - c  # the reflected index of c in s[n / 2..n)
      rd = n - d  # the reflected index of d in s[n / 2..n)
      # No difference is allowed outside the query ranges.
      if (min(a, rd) > 0 and mirroredDiffs[min(a, rd)] > 0) or \
         (n // 2 > max(b, rc) and mirroredDiffs[n // 2] - mirroredDiffs[max(b, rc)] > 0) or \
         (rd > b and mirroredDiffs[rd] - mirroredDiffs[b] > 0) or \
         (a > rc and mirroredDiffs[a] - mirroredDiffs[rc] > 0):
        ans.append(False)
      else:
        # The `count` map of the intersection of [a, b) and [rd, rc) in
        # s[0..n / 2) must equate to the `count` map of the intersection of
        # [c, d) and [rb, ra) in s[n / 2..n).
        leftRangeCount = subtractArrays(counts[b], counts[a])
        rightRangeCount = subtractArrays(counts[d], counts[c])
        if a > rd:
          rightRangeCount = subtractArrays(
              rightRangeCount, subtractArrays(counts[min(a, rc)], counts[rd]))
        if rc > b:
          rightRangeCount = subtractArrays(
              rightRangeCount, subtractArrays(counts[rc], counts[max(b, rd)]))
        if c > rb:
          leftRangeCount = subtractArrays(
              leftRangeCount, subtractArrays(counts[min(c, ra)], counts[rb]))
        if ra > d:
          leftRangeCount = subtractArrays(
              leftRangeCount, subtractArrays(counts[ra], counts[max(d, rb)]))
        ans.append(min(leftRangeCount) >= 0
                   and min(rightRangeCount) >= 0
                   and leftRangeCount == rightRangeCount)

    return ans

  def _getMirroredDiffs(self, s: str) -> List[int]:
    diffs = [0]
    for i, j in zip(range(len(s)), reversed(range(len(s)))):
      if i >= j:
        break
      diffs.append(diffs[-1] + (s[i] != s[j]))
    return diffs

  def _getCounts(self, s: str) -> List[List[int]]:
    count = [0] * 26
    counts = [count.copy()]
    for c in s:
      count[ord(c) - ord('a')] += 1
      counts.append(count.copy())
    return counts
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