<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(queries)
        M = len(s)
        ans = [False for _ in range(N)]

        freq = [[0 for _1 in range(M+1)] for _2 in range(26)]
        for i in range(1,M+1):
            c = ord(s[i-1:i]) - ord('a')
            freq[c][i] += 1
            for j in range(26):
                freq[j][i] += freq[j][i-1]

        for j in range(N):
            q = queries[j]
            num_odd = 0
            for i in range(26):
                if (freq[i][q[1]+1] - freq[i][q[0]])%2 > 0: num_odd += 1
            num_k = 2 * q[2]
            if (q[1]-q[0]+1) % 2 > 0: num_k += 1
            if num_k>=num_odd: ans[j] = True
        return ans

"""
class Solution:
    def canMakePaliQueries(self, s: str, queries: List[List[int]]) -> List[bool]:
        N = len(queries)
        ans = [False for _ in range(N)]

        def canMake(s, k):
            freq = dict()
            for c in s:
                freq[c] = freq.get(c, 0) + 1
            num_odd = 0
            for key in freq:
                if freq[key]%2 > 0: num_odd += 1
            if len(s) % 2 == 0 and num_odd <= 2*k:
                return True
            if len(s) % 2 > 0 and num_odd <= 2*k+1:
                return True
            return False

        for i in range(N):
            q = queries[i]
            if canMake( s[ q[0]:q[1]+1 ], q[2] ): ans[i] = True
        return ans
"""
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