<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>

class Solution:
    def countKConstraintSubstrings(self, s: str, k: int, queries: List[List[int]]) -> List[int]:
        # Sliding window to calculate leftmost valid starting indices
        left = []   # stores the leftmost index till which it has a valid substring
        n = len(s)
        j = 0  # starting index of the window
        count1 = 0  # count of '1's in the window
        count0 = 0  # count of '0's in the window
        
        for i in range(n):
            if s[i] == "1":
                count1 += 1
            else:
                count0 += 1

            while count1 > k and count0 > k:  # Ensure both constraints hold
                if s[j] == "1":
                    count1 -= 1
                else:
                    count0 -= 1
                j += 1

            left.append(j)  # left[i] is the leftmost valid index for substring ending at i

        # Calculate the number of valid substrings ending at each index i
        temp = []
        for i in range(n):
            temp.append(i - left[i] + 1)

        # Cumulative sum of valid substrings up to each index
        cum = [0] * n
        cum[0] = temp[0]
        for i in range(1, n):
            cum[i] = cum[i - 1] + temp[i]

        # Sliding window to calculate rightmost valid ending indices
        right = [0] * n
        j = n - 1  # starting from the right end
        count1 = 0
        count0 = 0
        for i in range(n - 1, -1, -1):
            if s[i] == "1":
                count1 += 1
            else:
                count0 += 1

            while count1 > k and count0 > k:  # Ensure both constraints hold
                if s[j] == "1":
                    count1 -= 1
                else:
                    count0 -= 1
                j -= 1

            right[i] = j  # right[i] is the rightmost valid index for substring starting at i

        # Answer queries
        result = []
        for low, high in queries:
            rightmost = min(high, right[low])
            l = rightmost - low + 1
            total_substrings = (l * (l + 1)) // 2  # number of substrings within the valid range
            if rightmost < high:
                total_substrings += cum[high] - cum[rightmost]  # Add substrings starting after rightmost
            result.append(total_substrings)

        return result

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