Introduction

In this blog post, we will explore how to solve the LeetCode problem "3398" 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 of length n and an integer numOps. You are allowed to perform the following operation on s at most numOps times: Select any index i (where 0 <= i < n) and flip s[i]. If s[i] == '1', change s[i] to '0' and vice versa. You need to minimize the length of the longest substring of s such that all the characters in the substring are identical. Return the minimum length after the operations. Example 1: Input: s = "000001", numOps = 1 Output: 2 Explanation: By changing s[2] to '1', s becomes "001001". The longest substrings with identical characters are s[0..1] and s[3..4]. Example 2: Input: s = "0000", numOps = 2 Output: 1 Explanation: By changing s[0] and s[2] to '1', s becomes "1010". Example 3: Input: s = "0101", numOps = 0 Output: 1 Constraints: 1 <= n == s.length <= 1000 s consists only of '0' and '1'. 0 <= numOps <= n

Explanation

Here's the breakdown of the solution:

• Binary Search: We'll use binary search on the length of the longest identical substring. The goal is to find the minimum possible length for which a valid configuration (within numOps operations) exists.

• Sliding Window: For each potential length, we'll use a sliding window to check if a substring of that length can be transformed into an identical substring (all 0s or all 1s) using at most numOps flips.

• Feasibility Check: The core of the solution lies in efficiently determining if a given substring of a specific length can be made identical with the allowed number of operations. This is done by checking both possibilities: converting it to all 0s and converting it to all 1s, and choosing the one requiring the fewest flips.

• Runtime & Storage Complexity: O(n log n) time complexity due to binary search and O(1) space complexity (excluding the input string).

Code

    def min_longest_substring(s: str, numOps: int) -> int:
    n = len(s)
    low = 1
    high = n
    ans = n

    def check(length: int) -> bool:
        for i in range(n - length + 1):
            zeros = 0
            ones = 0
            for j in range(i, i + length):
                if s[j] == '0':
                    zeros += 1
                else:
                    ones += 1

            if min(zeros, ones) <= numOps:
                return True
        return False

    while low <= high:
        mid = (low + high) // 2
        if check(mid):
            ans = mid
            high = mid - 1
        else:
            low = mid + 1

    return ans