<p>You are given a string <code>s</code> and array <code>queries</code> where <code>queries[i] = [left<sub>i</sub>, right<sub>i</sub>, k<sub>i</sub>]</code>. We may rearrange the substring <code>s[left<sub>i</sub>...right<sub>i</sub>]</code> for each query and then choose up to <code>k<sub>i</sub></code> of them to replace with any lowercase English letter.</p>

<p>If the substring is possible to be a palindrome string after the operations above, the result of the query is <code>true</code>. Otherwise, the result is <code>false</code>.</p>

<p>Return a boolean array <code>answer</code> where <code>answer[i]</code> is the result of the <code>i<sup>th</sup></code> query <code>queries[i]</code>.</p>

<p>Note that each letter is counted individually for replacement, so if, for example <code>s[left<sub>i</sub>...right<sub>i</sub>] = &quot;aaa&quot;</code>, and <code>k<sub>i</sub> = 2</code>, we can only replace two of the letters. Also, note that no query modifies the initial string <code>s</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example :</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcda&quot;, queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]
<strong>Output:</strong> [true,false,false,true,true]
<strong>Explanation:</strong>
queries[0]: substring = &quot;d&quot;, is palidrome.
queries[1]: substring = &quot;bc&quot;, is not palidrome.
queries[2]: substring = &quot;abcd&quot;, is not palidrome after replacing only 1 character.
queries[3]: substring = &quot;abcd&quot;, could be changed to &quot;abba&quot; which is palidrome. Also this can be changed to &quot;baab&quot; first rearrange it &quot;bacd&quot; then replace &quot;cd&quot; with &quot;ab&quot;.
queries[4]: substring = &quot;abcda&quot;, could be changed to &quot;abcba&quot; which is palidrome.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;lyb&quot;, queries = [[0,1,0],[2,2,1]]
<strong>Output:</strong> [false,true]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length, queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= left<sub>i</sub> &lt;= right<sub>i</sub> &lt; s.length</code></li>
	<li><code>0 &lt;= k<sub>i</sub> &lt;= s.length</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def canMakePaliQueries(self, s: str, queries: List[List[int]]) -> List[bool]:
        n = len(s)
        prefix = [[0] * 26 for _ in range(n + 1)]
        
        for i in range(n):
            for j in range(26):
                prefix[i + 1][j] = prefix[i][j]
            prefix[i + 1][ord(s[i]) - ord('a')] += 1
        
        def replaceable(start, end, k):
            odd_count = 0
            for j in range(26):
                if (prefix[end + 1][j] - prefix[start][j]) % 2 != 0:
                    odd_count += 1
            return odd_count // 2 <= k

        result = []
        for start, end, k in queries:
            result.append(replaceable(start, end, k))
        
        return result
1743