<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 defaultdict
class Solution:
    def canMakePalindromeQueries(self, s: str, queries: List[List[int]]) -> List[bool]:
        n = len(s)

        # wrong_idx keeps track of where there is a mismatch between left and right half of the string
        wrong_idx = [int(s[0] != s[n - 1])]
        for i in range(1, n // 2):
            wrong_idx.append(int(s[i] != s[n - 1 - i]) + wrong_idx[-1])  

        ret_list = []

        for a, b, c, d in queries:
            # c and d "mirrored" over to left half of the string
            c_mirr = n - 1 - c
            d_mirr = n - 1 - d

            # Check for letter mismatch which is impossible to fix given the intervals [a, b] and [c, d] 
            min_idx = min(a, d_mirr)
            max_idx = max(b, c_mirr)
            left_inner = min(b, c_mirr)
            right_inner = max(a, d_mirr)
            if (min_idx > 0 and wrong_idx[min_idx - 1] > 0 
                or wrong_idx[n // 2 - 1] - wrong_idx[max_idx]
                or right_inner > left_inner and wrong_idx[right_inner-1] - wrong_idx[left_inner] > 0):
                ret_list.append(False)
                continue

            # Create dictionaries for left and right half of string which keep track
            # of "letter budget"
            i = min_idx
            left_letters = defaultdict(int)
            right_letters = defaultdict(int)
            while i <= max_idx:
                if a <= i <= b and c <= n - 1 - i <= d:
                    left_letters[s[i]] += 1
                    right_letters[s[n - 1 - i]] += 1
                    if left_letters[s[i]] == 0:
                        del left_letters[s[i]]
                    if right_letters[s[n - 1 - i]] == 0:
                        del right_letters[s[n - 1 - i]]
                elif a <= i <= b:
                    left_letters[s[i]] += 1
                    left_letters[s[n - 1 - i]] -= 1
                    if left_letters[s[i]] == 0:
                        del left_letters[s[i]]
                    if left_letters[s[n - 1 - i]] == 0:
                        del left_letters[s[n - 1 - i]]
                elif c <= n - 1 - i <= d:
                    right_letters[s[n - 1 - i]] += 1
                    right_letters[s[i]] -= 1
                    if right_letters[s[n - 1 - i]] == 0:
                        del right_letters[s[n - 1 - i]]
                    if right_letters[s[i]] == 0:
                        del right_letters[s[i]]

                # Move ahead--skip to right_inner if we're between intervals
                # e.g., between [a, b] and [d_mirr, c_mirr]
                i += 1
                if left_inner < i < right_inner:
                    i = right_inner

            # Make sure we're not short letters on either side
            if any(val < 0 for val in left_letters.values()):
                ret_list.append(False)

            # Check that the "overlapping" parts have the same letters 
            else:
                ret_list.append(left_letters == right_letters)

        return ret_list
1943