Introduction

In this blog post, we will explore how to solve the LeetCode problem "3258" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a binary string s and an integer k. 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 denoting the number of substrings of s that satisfy the k-constraint. Example 1: Input: s = "10101", k = 1 Output: 12 Explanation: Every substring of s except the substrings "1010", "10101", and "0101" satisfies the k-constraint. Example 2: Input: s = "1010101", k = 2 Output: 25 Explanation: Every substring of s except the substrings with a length greater than 5 satisfies the k-constraint. Example 3: Input: s = "11111", k = 1 Output: 15 Explanation: All substrings of s satisfy the k-constraint. Constraints: 1 <= s.length <= 50 1 <= k <= s.length s[i] is either '0' or '1'.

Explanation

Here's a solution approach, complexity analysis, and Python code to address the problem:

• High-Level Approach:

  • Iterate through all possible substrings of the given string.
  • For each substring, count the number of '0's and '1's.
  • Check if either the count of '0's or the count of '1's is less than or equal to k. If so, increment the result.

• Complexity:

  • Runtime: O(n^3), where n is the length of the string. This is due to the nested loops for generating substrings and the O(n) operation to count 0s/1s within each substring.
  • Storage: O(1). The solution uses a constant amount of extra space.

Code

    def k_constraint_substrings(s, k):
    """
    Calculates the number of substrings of s that satisfy the k-constraint.

    Args:
        s: The binary string.
        k: The integer constraint.

    Returns:
        The number of substrings that satisfy the k-constraint.
    """

    n = len(s)
    count = 0

    for i in range(n):
        for j in range(i, n):
            substring = s[i:j+1]
            zeros = substring.count('0')
            ones = substring.count('1')

            if zeros <= k or ones <= k:
                count += 1

    return count