<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]:
        
        # building the cumulative sum array
        # right to left
        cumSumR2L = [0] * len(s)

        rep_2_loc = {0:defaultdict(lambda: len(s)), 1:defaultdict(lambda: len(s))}
        count1 = 0
        count0 = 0

        pointer = -1

        for i in range(len(s)):
            if s[i] == "1":
                count1 += 1
                rep_2_loc[1][count1] = i 
            else:
                count0 += 1
                rep_2_loc[0][count0] = i

            if count1 > k and count0 > k:

                pointer = min(rep_2_loc[1][count1 - k], rep_2_loc[0][count0 - k])

            cumSumR2L[i] = cumSumR2L[i-1] + i - pointer #  first iteration goes to -1
        
        #print(cumSumR2L)

        # finding longest window
        # left to right
        maxBoundL2R = [0] * len(s)

        rep_2_loc = {0:defaultdict(lambda: len(s)), 1:defaultdict(lambda: len(s))}
        count1 = 0
        count0 = 0
        pointer = len(s)

        for i in range(len(s)-1,-1,-1):
            if s[i] == "1":
                count1 += 1
                rep_2_loc[1][count1] = i 
            else:
                count0 += 1
                rep_2_loc[0][count0] = i

            if count1 > k and count0 > k:

                pointer = max(rep_2_loc[1][count1 - k], rep_2_loc[0][count0 - k])

            maxBoundL2R[i] = pointer - 1

        #print(maxBoundL2R)

        ans = []

        for l_bound, r_bound in queries:
            
            split = min(maxBoundL2R[l_bound], r_bound)

            maxSegmentL2R = (split - l_bound + 1) * (split - l_bound + 2) // 2 # n*(n-1)/2

            substringsR2L = cumSumR2L[r_bound] - cumSumR2L[split]

            ans.append(maxSegmentL2R + substringsR2L)


        return ans
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