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

Count Substrings That Satisfy K-Constraint I (LeetCode #3258) 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: The sliding window technique is powerful for problems involving contiguous subarrays or substrings.

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

The brute force approach for Count Substrings That Satisfy K-Constraint I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible substrings of the binary string and checking each one to see if it satisfies the k-constraint. This is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Substrings That Satisfy K-Constraint I?

Count Substrings That Satisfy K-Constraint I uses the following concepts: String, Sliding Window. The recommended patterns to study are: Sliding Window, Two Pointers.

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

Always consider edge cases, such as strings with all '0's or all '1's. Explain your thought process clearly while coding; it helps interviewers understand your logic. Practice writing both brute force and optimal solutions to solidify your understanding.

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

Not properly managing the counts of '0's and '1's when moving the left pointer. Forgetting to count valid substrings when the window is valid.

#3258

Count Substrings That Satisfy K-Constraint I

Easy

StringSliding WindowSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal approach uses the sliding window technique to efficiently count valid substrings without generating all of them. This method maintains a window of characters and adjusts its size based on the counts of '0's and '1's.

⚙️

Algorithm

5 steps

1Step 1: Initialize two pointers (left and right) to represent the current window.
2Step 2: Expand the right pointer to include more characters until the window violates the k-constraint.
3Step 3: For each valid window, count all possible substrings that can be formed with the current left pointer.
4Step 4: Move the left pointer to shrink the window until it satisfies the k-constraint again.
5Step 5: Repeat until the right pointer reaches the end of the string.

solution.py17 lines

1def count_substrings(s, k):
2    left = 0
3    count = 0
4    zeros = ones = 0
5    for right in range(len(s)):
6        if s[right] == '0':
7            zeros += 1
8        else:
9            ones += 1
10        while zeros > k and ones > k:
11            if s[left] == '0':
12                zeros -= 1
13            else:
14                ones -= 1
15            left += 1
16        count += right - left + 1
17    return count

ℹ

Complexity note: The time complexity is O(n) because each character is processed at most twice (once by the right pointer and once by the left pointer). The space complexity is O(1) since we are using a fixed amount of extra space.

1The sliding window technique is powerful for problems involving contiguous subarrays or substrings.
2Understanding how to maintain counts of elements while adjusting window boundaries is crucial.

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