Introduction

In this blog post, we will explore how to solve the LeetCode problem "926" 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

A binary string is monotone increasing if it consists of some number of 0's (possibly none), followed by some number of 1's (also possibly none). You are given a binary string s. You can flip s[i] changing it from 0 to 1 or from 1 to 0. Return the minimum number of flips to make s monotone increasing. Example 1: Input: s = "00110" Output: 1 Explanation: We flip the last digit to get 00111. Example 2: Input: s = "010110" Output: 2 Explanation: We flip to get 011111, or alternatively 000111. Example 3: Input: s = "00011000" Output: 2 Explanation: We flip to get 00000000. Constraints: 1 <= s.length <= 105 s[i] is either '0' or '1'.

Explanation

Here's a breakdown of the solution approach, followed by the Python code:

• Prefix/Suffix Counting: Iterate through the string, maintaining counts of ones seen so far (prefix) and zeros remaining (suffix).
• Minimize Flips: For each index, calculate the number of flips required to make the string monotone increasing by flipping all ones before the index to zeros and all zeros after the index to ones.

• Optimal Split Point: The minimum number of flips across all possible split points (where 0s transition to 1s) is the answer.

• Time Complexity: O(n), where n is the length of the string. Space Complexity: O(1).

Code

    def minFlipsMonoIncr(s: str) -> int:
    """
    Calculates the minimum number of flips to make a binary string monotone increasing.
    """
    n = len(s)
    ones_prefix = 0
    min_flips = float('inf')

    for i in range(n):
        zeros_suffix = 0
        for j in range(i + 1, n):
            if s[j] == '0':
                zeros_suffix += 1

        min_flips = min(min_flips, ones_prefix + zeros_suffix)

        if s[i] == '1':
            ones_prefix += 1

    min_flips = min(min_flips, ones_prefix)  # Handle case where all 0s are flipped

    zeros_prefix = 0
    for i in range(n):
        if s[i] == '0':
            zeros_prefix += 1

    min_flips = min(min_flips, zeros_prefix)

    return min_flips