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
class Solution {
    public boolean[] canMakePalindromeQueries(String s, int[][] queries) {
        int n = s.length();
        int m = n / 2;
        String t = new StringBuilder(s.substring(m)).reverse().toString();
        s = s.substring(0, m);
        int[][] pre1 = new int[m + 1][0];
        int[][] pre2 = new int[m + 1][0];
        int[] diff = new int[m + 1];
        pre1[0] = new int[26];
        pre2[0] = new int[26];
        for (int i = 1; i <= m; ++i) {
            pre1[i] = pre1[i - 1].clone();
            pre2[i] = pre2[i - 1].clone();
            ++pre1[i][s.charAt(i - 1) - 'a'];
            ++pre2[i][t.charAt(i - 1) - 'a'];
            diff[i] = diff[i - 1] + (s.charAt(i - 1) == t.charAt(i - 1) ? 0 : 1);
        }
        boolean[] ans = new boolean[queries.length];
        for (int i = 0; i < queries.length; ++i) {
            int[] q = queries[i];
            int a = q[0], b = q[1];
            int c = n - 1 - q[3], d = 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;
    }

    private boolean check(int[][] pre1, int[][] pre2, int[] diff, int a, int b, int c, int d) {
        if (diff[a] > 0 || diff[diff.length - 1] - diff[Math.max(b, d) + 1] > 0) {
            return false;
        }
        if (d <= b) {
            return Arrays.equals(count(pre1, a, b), count(pre2, a, b));
        }
        if (b < c) {
            return diff[c] - diff[b + 1] == 0 && Arrays.equals(count(pre1, a, b), count(pre2, a, b))
                && Arrays.equals(count(pre1, c, d), count(pre2, c, d));
        }
        int[] cnt1 = sub(count(pre1, a, b), count(pre2, a, c - 1));
        int[] cnt2 = sub(count(pre2, c, d), count(pre1, b + 1, d));
        return cnt1 != null && cnt2 != null && Arrays.equals(cnt1, cnt2);
    }

    private int[] count(int[][] pre, int i, int j) {
        int[] cnt = new int[26];
        for (int k = 0; k < 26; ++k) {
            cnt[k] = pre[j + 1][k] - pre[i][k];
        }
        return cnt;
    }

    private int[] sub(int[] cnt1, int[] cnt2) {
        int[] cnt = new int[26];
        for (int i = 0; i < 26; ++i) {
            cnt[i] = cnt1[i] - cnt2[i];
            if (cnt[i] < 0) {
                return null;
            }
        }
        return cnt;
    }
}