How do you solve Count Substrings That Satisfy K-Constraint II on LeetCode?

Count Substrings That Satisfy K-Constraint II (LeetCode #3261) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Sliding window technique allows us to efficiently count valid substrings without generating all possible substrings.

What is the brute force solution for Count Substrings That Satisfy K-Constraint II?

The brute force approach for Count Substrings That Satisfy K-Constraint II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible substrings and counting those that satisfy the k-constraint. This is straightforward but inefficient for large strings due to the quadratic number of substrings. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Count Substrings That Satisfy K-Constraint II?

Always explain your thought process clearly, especially when transitioning from brute force to optimized solutions. Practice identifying patterns like sliding windows or two pointers in string problems. Be prepared to discuss time and space complexity for both brute force and optimized solutions.

What are common mistakes when solving Count Substrings That Satisfy K-Constraint II?

Students often forget to reset counts for each new starting index in the brute force approach. In the optimal approach, students may incorrectly handle the left pointer when adjusting the window.

#3261

Count Substrings That Satisfy K-Constraint II

Hard

ArrayStringBinary SearchSliding WindowPrefix SumSliding WindowTwo PointersPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a sliding window approach to efficiently count valid substrings by maintaining a window of characters that satisfy the k-constraint. This reduces the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Create a function to count valid substrings within a given range using a sliding window.
2Step 2: For each query, determine the left and right bounds of the substring.
3Step 3: Use two pointers to expand the window and count the number of valid substrings.
4Step 4: Store results for each query and return them.

solution.py33 lines

1# Full working Python code
2
3def count_valid_substrings(s, k, l, r):
4    count = 0
5    left = 0
6    zeros = 0
7    ones = 0
8    for right in range(l, r + 1):
9        if s[right] == '0':
10            zeros += 1
11        else:
12            ones += 1
13        while zeros > k and ones > k:
14            if s[left] == '0':
15                zeros -= 1
16            else:
17                ones -= 1
18            left += 1
19        count += right - left + 1
20    return count
21
22
23def count_substrings(s, k, queries):
24    results = []
25    for l, r in queries:
26        results.append(count_valid_substrings(s, k, l, r))
27    return results
28
29# Example usage
30s = '0001111'
31k = 2
32queries = [[0, 6]]
33print(count_substrings(s, k, queries))

ℹ

Complexity note: This complexity is achieved because we only traverse the string once for each query, maintaining a sliding window, leading to linear time complexity.

1Sliding window technique allows us to efficiently count valid substrings without generating all possible substrings.
2Understanding the constraints helps in optimizing the solution by focusing on the counts of '0's and '1's.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.