<p>You are given a <strong>binary</strong> string <code>s</code> and an integer <code>k</code>.</p>

<p>You are also given a 2D integer array <code>queries</code>, where <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code>.</p>

<p>A <strong>binary string</strong> satisfies the <strong>k-constraint</strong> if <strong>either</strong> of the following conditions holds:</p>

<ul>
	<li>The number of <code>0</code>&#39;s in the string is at most <code>k</code>.</li>
	<li>The number of <code>1</code>&#39;s in the string is at most <code>k</code>.</li>
</ul>

<p>Return an integer array <code>answer</code>, where <code>answer[i]</code> is the number of <span data-keyword="substring-nonempty">substrings</span> of <code>s[l<sub>i</sub>..r<sub>i</sub>]</code> that satisfy the <strong>k-constraint</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;0001111&quot;, k = 2, queries = [[0,6]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[26]</span></p>

<p><strong>Explanation:</strong></p>

<p>For the query <code>[0, 6]</code>, all substrings of <code>s[0..6] = &quot;0001111&quot;</code> satisfy the k-constraint except for the substrings <code>s[0..5] = &quot;000111&quot;</code> and <code>s[0..6] = &quot;0001111&quot;</code>.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;010101&quot;, k = 1, queries = [[0,5],[1,4],[2,3]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[15,9,3]</span></p>

<p><strong>Explanation:</strong></p>

<p>The substrings of <code>s</code> with a length greater than 3 do not satisfy the k-constraint.</p>
</div>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
	<li><code>1 &lt;= k &lt;= s.length</code></li>
	<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code>queries[i] == [l<sub>i</sub>, r<sub>i</sub>]</code></li>
	<li><code>0 &lt;= l<sub>i</sub> &lt;= r<sub>i</sub> &lt; s.length</code></li>
	<li>All queries are distinct.</li>
</ul>

from collections import Counter
from itertools import accumulate

class Solution:
    def countKConstraintSubstrings(self, s: str, k: int, queries: List[List[int]]) -> List[int]:

        # Use sliding window to find maximum windows for right index and left index seperately. Use that to calculate the results. 

        n = len(s)
        left_to_right = [0] * n
        right_to_left = [0] * n

        left = 0
        counts = Counter()
        for right in range(n):
            counts[s[right]] += 1
            while counts['0'] > k and counts['1'] > k:
                counts[s[left]] -= 1
                left += 1
            right_to_left[right] = left

        right = n - 1
        counts = Counter()
        for left in reversed(range(n)):
            counts[s[left]] += 1
            while counts['0'] > k and counts['1'] > k:
                counts[s[right]] -= 1
                right -= 1
            left_to_right[left] = right

        pref_sum = [0] + list(accumulate([(right - left + 1) for right, left in enumerate(right_to_left)]))        
        result = []
        for left_bound, right_bound in queries:
            mid = min(right_bound, left_to_right[left_bound])
            length = mid - left_bound + 1
            curr = length * (length + 1) // 2
            curr += pref_sum[right_bound + 1] - pref_sum[mid + 1]
            result += curr,

        return result


        
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