<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, Q: List[List[int]]) -> List[bool]:
        CHRS = set(s)
        n = len(s)
        m = len(Q)
        A = s[:n//2]
        B = s[n//2:][::-1]
        
        # if the character counts do not match -> all false
        if sorted(A) != sorted(B):
            return [False] * m
        
        D = [0]
        chCountA = {}
        chCountB = {}
        
        for i in range(n//2):
            if A[i] != B[i]:
                D.append(D[-1] + 1)
            else:
                D.append(D[-1])
            for x in CHRS:
                if x not in chCountA:
                    chCountA[x] = [0]
                if x not in chCountB:
                    chCountB[x] = [0]
                if A[i] == x:
                    chCountA[x].append(chCountA[x][-1]+1)
                else:
                    chCountA[x].append(chCountA[x][-1])
                if B[i] == x:
                    chCountB[x].append(chCountB[x][-1]+1)
                else:
                    chCountB[x].append(chCountB[x][-1])
        
        def check(x,y):
            return D[y+1] - D[x] == 0
        
        if D[-1] == 0:
            return [True] * m
        
        # non-intersecting
        def check1(a,b,c,d,flag = 0): # flag == 0 check in B else check in A
            # check if all chrs in A[a:b] are available in B[c:d]
            ans = True
            for x in CHRS:
                if flag == 0:
                    count1 = chCountA[x][b+1] - chCountA[x][a]
                    count2 = chCountB[x][d+1] - chCountB[x][c]
                else:
                    count1 = chCountB[x][b+1] - chCountB[x][a]
                    count2 = chCountA[x][d+1] - chCountA[x][c]
                ans &= count2 >= count1
            return ans
        
        
        out = []
        
        for a,b,c,d in Q:
            c, d = n - d - 1, n - c - 1
            ans = True
            if min(a,c) > 0:
                ans &= check(0,min(a,c)-1)
            if b < c:
                ans &= check(b+1,c-1)
            if max(b,d) < (n//2)-1:
                ans &= check(max(b,d)+1,(n//2)-1)
            
            if a < c:
                ans &= check1(a,min(b,c-1),a,b,1)
            if c < a:
                ans &= check1(c,min(d,a-1),c,d,0)
            
            if b > d:
                ans &= check1(max(a,d+1),b,a,b,1)
            if d > b:
                ans &= check1(max(c,b+1),d,c,d,0)
            out.append(ans)
        return out
1701