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:
0's in the string is at most k.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.
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 <= 105s[i] is either '0' or '1'.1 <= k <= s.length1 <= queries.length <= 105queries[i] == [li, ri]0 <= li <= ri < s.lengthIn #3261 Count Substrings That Satisfy K-Constraint II, the goal is to efficiently count substrings that satisfy a constraint related to the number of characters within a limit k. Since brute force checking all substrings would be O(n^2) and too slow, an optimized strategy is required.
A common approach combines a sliding window with prefix sums. The sliding window helps determine the farthest valid boundary for each starting index while maintaining counts of characters that must stay within the k constraint. Once valid ranges are known, prefix sums allow fast aggregation of how many substrings end within specific boundaries.
For query-based ranges, binary search can be used to find the first index where the window exceeds the valid limit, and prefix sums are then used to compute the contribution of valid substrings in constant time. This reduces the overall complexity significantly, making the solution scalable for large inputs.
The optimized solution typically runs in around O(n + q log n) time with linear auxiliary space.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Sliding Window + Prefix Sum + Binary Search | O(n + q log n) | O(n) |
| Brute Force Substring Checking | O(n^2) | O(1) |
codestorywithMIK
Use these hints if you're stuck. Try solving on your own first.
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.
Watch expert explanations and walkthroughs
Jot down your thoughts, approach, and key learnings
Hard string and sliding window problems similar to this frequently appear in FAANG-style interviews. The combination of prefix sums, window optimization, and query handling tests strong algorithmic reasoning and performance optimization skills.
Prefix sums allow fast computation of substring counts across ranges without recalculating values repeatedly. After determining valid boundaries with a sliding window, prefix sums aggregate the number of valid substrings in constant time per query.
The optimal approach combines a sliding window to maintain valid substring ranges with prefix sums to quickly count substrings. Binary search can be used during queries to locate valid boundaries efficiently. This significantly reduces the complexity compared to checking all substrings.
Key techniques include sliding window pointers, prefix sum arrays, and binary search over precomputed boundaries. These structures help efficiently process constraints and handle multiple substring queries.