<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 RabinKarp:
    def __init__(self, A):
        self.P = P = 37
        self.MOD = MOD = 344555666677777
        Pinv = 46561576578078
        self.pre = pre = [0]
        self.pinvs = pinvs = [1]

        ha = 0
        pwr = pwrinv = 1
        for i, x in enumerate(A):
            ha = (ha + x * pwr) % MOD
            pwr = pwr * P % MOD
            pwrinv = pwrinv * Pinv % MOD
            pre.append(ha)
            pinvs.append(pwrinv)

    def query(self, i, j):
        # return hash of s[i..j]
        return (self.pre[j + 1] - self.pre[i]) * self.pinvs[i] % self.MOD


class Solution:
    def canMakePalindromeQueries(self, S: str, queries: List[List[int]]) -> List[bool]:
        N = len(S) // 2
        A = [ord(c) - ord("a") for c in S[:N]]
        B = [ord(c) - ord("a") for c in S[: N - 1 : -1]]
        rkA = RabinKarp(A)
        rkB = RabinKarp(B)

        PA = [[0] * 26]
        PB = [[0] * 26]

        for i in range(N):
            PA.append(PA[-1][:])
            PA[-1][A[i]] += 1
            PB.append(PB[-1][:])
            PB[-1][B[i]] += 1

        def cover(A, PA, i1, j1, B, PB, i2, j2):
            return all(
                PA[j1 + 1][k] - PA[i1][k] >= PB[j2 + 1][k] - PB[i2][k]
                for k in range(26)
            )

        def check(A, PA, i1, j1, B, PB, i2, j2):
            if i1 > i2:
                return check(B, PB, i2, j2, A, PA, i1, j1)

            lo = min(i1, i2) - 1
            hi = max(j1, j2) + 1
            if rkA.query(0, lo) != rkB.query(0, lo) or rkA.query(
                hi, N - 1
            ) != rkB.query(hi, N - 1):
                return False
            if j2 <= j1:  # A overlaps B
                return cover(A, PA, i1, j1, B, PB, i1, j1)
            if i2 <= j1:  # AAAAxxxxBBBB
                return (
                    cover(A, PA, i1, j1, B, PB, i1, i2 - 1)
                    and cover(B, PB, i2, j2, A, PA, j1 + 1, j2)
                    and cover(A, PA, i1, j2, B, PB, i1, j2)
                )
            if j1 < i2:  # AAAA  BBBB
                return (
                    rkA.query(j1 + 1, i2 - 1) == rkB.query(j1 + 1, i2 - 1)
                    and cover(A, PA, i1, j1, B, PB, i1, j1)
                    and cover(A, PA, i2, j2, B, PB, i2, j2)
                )

        ans = []
        for i1, j1, c, d in queries:
            i2, j2 = 2 * N - 1 - d, 2 * N - 1 - c
            ans.append(check(A, PA, i1, j1, B, PB, i2, j2))
        return ans
1834