LeetCode #3261 — HARD

Count Substrings That Satisfy K-Constraint II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given a binary string s and an integer k.

You are also given a 2D integer array queries, where queries[i] = [l_i, r_i].

A binary string satisfies the k-constraint if either of the following conditions holds:

The number of 0's in the string is at most k.
The number of 1's in the string is at most k.

Return an integer array answer, where answer[i] is the number of substrings of s[l_i..r_i] that satisfy the k-constraint.

Example 1:

Input: s = "0001111", k = 2, queries = [[0,6]]

Output: [26]

Explanation:

For the query [0, 6], all substrings of s[0..6] = "0001111" satisfy the k-constraint except for the substrings s[0..5] = "000111" and s[0..6] = "0001111".

Example 2:

Input: s = "010101", k = 1, queries = [[0,5],[1,4],[2,3]]

Output: [15,9,3]

Explanation:

The substrings of s with a length greater than 3 do not satisfy the k-constraint.

Constraints:

1 <= s.length <= 10⁵
s[i] is either '0' or '1'.
1 <= k <= s.length
1 <= queries.length <= 10⁵
queries[i] == [l_i, r_i]
0 <= l_i <= r_i < s.length
All queries are distinct.

Step 01

Brute Force Baseline

Problem summary: You are given a binary string s and an integer k. You are also given a 2D integer array queries, where queries[i] = [li, ri]. A binary string satisfies the k-constraint if either of the following conditions holds: The number of 0's in the string is at most k. The number of 1's in the string is at most k. Return an integer array answer, where answer[i] is the number of substrings of s[li..ri] that satisfy the k-constraint.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Sliding Window

Example 1

"0001111"
2
[[0,6]]

Example 2

"010101"
1
[[0,5],[1,4],[2,3]]

Step 02

Core Insight

What unlocks the optimal approach

Answering online queries is tough. Try to answer them offline since the queries are known beforehand.
For each index, how do you calculate the left boundary so that the given condition is satisfied?
Using the precomputed left boundaries and a range data structure, you can now answer the queries optimally.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3261: Count Substrings That Satisfy K-Constraint II
class Solution {
    public long[] countKConstraintSubstrings(String s, int k, int[][] queries) {
        int[] cnt = new int[2];
        int n = s.length();
        int[] d = new int[n];
        Arrays.fill(d, n);
        long[] pre = new long[n + 1];
        for (int i = 0, j = 0; j < n; ++j) {
            cnt[s.charAt(j) - '0']++;
            while (cnt[0] > k && cnt[1] > k) {
                d[i] = j;
                cnt[s.charAt(i++) - '0']--;
            }
            pre[j + 1] = pre[j] + j - i + 1;
        }
        int m = queries.length;
        long[] ans = new long[m];
        for (int i = 0; i < m; ++i) {
            int l = queries[i][0], r = queries[i][1];
            int p = Math.min(r + 1, d[l]);
            long a = (1L + p - l) * (p - l) / 2;
            long b = pre[r + 1] - pre[p];
            ans[i] = a + b;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3261: Count Substrings That Satisfy K-Constraint II
func countKConstraintSubstrings(s string, k int, queries [][]int) (ans []int64) {
	cnt := [2]int{}
	n := len(s)
	d := make([]int, n)
	for i := range d {
		d[i] = n
	}
	pre := make([]int, n+1)
	for i, j := 0, 0; j < n; j++ {
		cnt[s[j]-'0']++
		for cnt[0] > k && cnt[1] > k {
			d[i] = j
			cnt[s[i]-'0']--
			i++
		}
		pre[j+1] = pre[j] + j - i + 1
	}
	for _, q := range queries {
		l, r := q[0], q[1]
		p := min(r+1, d[l])
		a := (1 + p - l) * (p - l) / 2
		b := pre[r+1] - pre[p]
		ans = append(ans, int64(a+b))
	}
	return
}

# Accepted solution for LeetCode #3261: Count Substrings That Satisfy K-Constraint II
class Solution:
    def countKConstraintSubstrings(
        self, s: str, k: int, queries: List[List[int]]
    ) -> List[int]:
        cnt = [0, 0]
        i, n = 0, len(s)
        d = [n] * n
        pre = [0] * (n + 1)
        for j, x in enumerate(map(int, s)):
            cnt[x] += 1
            while cnt[0] > k and cnt[1] > k:
                d[i] = j
                cnt[int(s[i])] -= 1
                i += 1
            pre[j + 1] = pre[j] + j - i + 1
        ans = []
        for l, r in queries:
            p = min(r + 1, d[l])
            a = (1 + p - l) * (p - l) // 2
            b = pre[r + 1] - pre[p]
            ans.append(a + b)
        return ans

// Accepted solution for LeetCode #3261: Count Substrings That Satisfy K-Constraint II
/**
 * [3261] Count Substrings That Satisfy K-Constraint II
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn count_k_constraint_substrings(s: String, k: i32, queries: Vec<Vec<i32>>) -> Vec<i64> {
        let s: Vec<u32> = s.chars().map(|x| x.to_digit(10).unwrap()).collect();
        let n = s.len();

        let mut window = (0, 0);
        let mut right_array = vec![n; n];
        let mut prefix = vec![0; n + 1];

        let mut left = 0;
        for right in 0..n {
            if s[right] == 0 {
                window.0 += 1;
            } else {
                window.1 += 1;
            }

            while window.0 > k && window.1 > k {
                if s[left] == 0 {
                    window.0 -= 1;
                } else {
                    window.1 -= 1;
                }
                right_array[left] = right;
                left += 1;
            }

            prefix[right + 1] = prefix[right] + (right - left + 1) as i64;
        }

        queries
            .into_iter()
            .map(|query| {
                let (l, r) = (query[0] as usize, query[1] as usize);
                let min_r = right_array[l].min(r + 1) as i64;
                let l = l as i64;

                (min_r - l + 1) * (min_r - l) / 2 + prefix[r + 1] - prefix[min_r as usize]
            })
            .collect()
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3261() {
        assert_eq!(
            vec![26],
            Solution::count_k_constraint_substrings("0001111".to_owned(), 2, vec![vec![0, 6]])
        );
        assert_eq!(
            vec![15, 9, 3],
            Solution::count_k_constraint_substrings(
                "010101".to_owned(),
                1,
                vec![vec![0, 5], vec![1, 4], vec![2, 3]]
            )
        );
        assert_eq!(
            vec![1, 3, 1],
            Solution::count_k_constraint_substrings(
                "00".to_owned(),
                1,
                vec![vec![0, 0], vec![0, 1], vec![1, 1]]
            )
        );
    }
}

// Accepted solution for LeetCode #3261: Count Substrings That Satisfy K-Constraint II
function countKConstraintSubstrings(s: string, k: number, queries: number[][]): number[] {
    const cnt: [number, number] = [0, 0];
    const n = s.length;
    const d: number[] = Array(n).fill(n);
    const pre: number[] = Array(n + 1).fill(0);
    for (let i = 0, j = 0; j < n; ++j) {
        cnt[+s[j]]++;
        while (Math.min(cnt[0], cnt[1]) > k) {
            d[i] = j;
            cnt[+s[i++]]--;
        }
        pre[j + 1] = pre[j] + j - i + 1;
    }
    const ans: number[] = [];
    for (const [l, r] of queries) {
        const p = Math.min(r + 1, d[l]);
        const a = ((1 + p - l) * (p - l)) / 2;
        const b = pre[r + 1] - pre[p];
        ans.push(a + b);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n + m)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.