Palindrome Rearrangement Queries
<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>
Hash Table; String; Prefix Sum
function canMakePalindromeQueries(s: string, queries: number[][]): boolean[] {
    const n: number = s.length;
    const m: number = n >> 1;
    const t: string = s.slice(m).split('').reverse().join('');
    s = s.slice(0, m);

    const pre1: number[][] = Array.from({ length: m + 1 }, () => Array(26).fill(0));
    const pre2: number[][] = Array.from({ length: m + 1 }, () => Array(26).fill(0));
    const diff: number[] = Array(m + 1).fill(0);
    for (let i = 1; i <= m; ++i) {
        pre1[i] = [...pre1[i - 1]];
        pre2[i] = [...pre2[i - 1]];
        ++pre1[i][s.charCodeAt(i - 1) - 'a'.charCodeAt(0)];
        ++pre2[i][t.charCodeAt(i - 1) - 'a'.charCodeAt(0)];
        diff[i] = diff[i - 1] + (s[i - 1] === t[i - 1] ? 0 : 1);
    }

    const ans: boolean[] = Array(queries.length).fill(false);
    for (let i = 0; i < queries.length; ++i) {
        const q: number[] = queries[i];
        const [a, b] = [q[0], q[1]];
        const [c, d] = [n - 1 - q[3], n - 1 - q[2]];
        ans[i] = a <= c ? check(pre1, pre2, diff, a, b, c, d) : check(pre2, pre1, diff, c, d, a, b);
    }
    return ans;
}

function check(
    pre1: number[][],
    pre2: number[][],
    diff: number[],
    a: number,
    b: number,
    c: number,
    d: number,
): boolean {
    if (diff[a] > 0 || diff[diff.length - 1] - diff[Math.max(b, d) + 1] > 0) {
        return false;
    }

    if (d <= b) {
        return arraysEqual(count(pre1, a, b), count(pre2, a, b));
    }

    if (b < c) {
        return (
            diff[c] - diff[b + 1] === 0 &&
            arraysEqual(count(pre1, a, b), count(pre2, a, b)) &&
            arraysEqual(count(pre1, c, d), count(pre2, c, d))
        );
    }

    const cnt1: number[] | null = sub(count(pre1, a, b), count(pre2, a, c - 1));
    const cnt2: number[] | null = sub(count(pre2, c, d), count(pre1, b + 1, d));

    return cnt1 !== null && cnt2 !== null && arraysEqual(cnt1, cnt2);
}

function count(pre: number[][], i: number, j: number): number[] {
    const cnt: number[] = new Array(26).fill(0);
    for (let k = 0; k < 26; ++k) {
        cnt[k] = pre[j + 1][k] - pre[i][k];
    }
    return cnt;
}

function sub(cnt1: number[], cnt2: number[]): number[] | null {
    const cnt: number[] = new Array(26).fill(0);
    for (let i = 0; i < 26; ++i) {
        cnt[i] = cnt1[i] - cnt2[i];
        if (cnt[i] < 0) {
            return null;
        }
    }
    return cnt;
}

function arraysEqual(arr1: number[], arr2: number[]): boolean {
    return JSON.stringify(arr1) === JSON.stringify(arr2);
}
