LeetCode 3261: Count Substrings That Satisfy K-Constraint II Solution

Master LeetCode problem 3261 (Count Substrings That Satisfy K-Constraint II), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3261. Count Substrings That Satisfy K-Constraint II

Hard

Array String Binary Search Sliding Window Prefix Sum

Problem Explanation

Explanation:

We can solve this problem using a sliding window approach.
We will iterate over the queries and for each query, we will maintain two pointers left and right to represent the current substring.
We will keep track of the counts of 0's and 1's in the substring.
As we move the right pointer, we will update the counts and check if the current substring satisfies the k-constraint.
If it does, we increment the count of valid substrings.
Finally, we return the count of valid substrings for each query.

Time Complexity: O(n * q) where n is the length of the string and q is the number of queries.
Space Complexity: O(n) for storing the counts of 0's and 1's.

Solution Code

class Solution {
    public int[] countSubstrings(String s, int k, int[][] queries) {
        int n = s.length();
        int[] ans = new int[queries.length];
        
        for (int i = 0; i < queries.length; i++) {
            int[] query = queries[i];
            int left = query[0];
            int right = query[1];
            int[] counts = new int[2];
            int count = 0;
            
            for (int j = left; j <= right; j++) {
                counts[s.charAt(j) - '0']++;
                if (Math.min(counts[0], counts[1]) <= k) {
                    count += j - left + 1;
                } else {
                    while (Math.min(counts[0], counts[1]) > k) {
                        counts[s.charAt(left) - '0']--;
                        left++;
                    }
                    count += j - left + 1;
                }
            }
            
            ans[i] = count;
        }
        
        return ans;
    }
}

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