Introduction

In this blog post, we will explore how to solve the LeetCode problem "2938" 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

There are n balls on a table, each ball has a color black or white. You are given a 0-indexed binary string s of length n, where 1 and 0 represent black and white balls, respectively. In each step, you can choose two adjacent balls and swap them. Return the minimum number of steps to group all the black balls to the right and all the white balls to the left. Example 1: Input: s = "101" Output: 1 Explanation: We can group all the black balls to the right in the following way: - Swap s[0] and s[1], s = "011". Initially, 1s are not grouped together, requiring at least 1 step to group them to the right. Example 2: Input: s = "100" Output: 2 Explanation: We can group all the black balls to the right in the following way: - Swap s[0] and s[1], s = "010". - Swap s[1] and s[2], s = "001". It can be proven that the minimum number of steps needed is 2. Example 3: Input: s = "0111" Output: 0 Explanation: All the black balls are already grouped to the right. Constraints: 1 <= n == s.length <= 105 s[i] is either '0' or '1'.

Explanation

Here's the breakdown of the solution:

• Key Idea: The core idea is to iterate through the string and count the number of swaps needed to move each black ball to its correct position on the right side. The correct position for the i-th black ball is n - count, where count is the number of black balls encountered so far.
• Efficient Counting: We maintain a running count of the number of white balls encountered before each black ball. This directly represents the number of swaps needed to move that particular black ball to its target position on the right.

• Accumulate Swaps: Summing up the number of swaps needed for each black ball gives us the overall minimum number of swaps required to group all black balls to the right.

• Complexity: O(n) time complexity, O(1) space complexity

Code

    def min_moves_to_group(s: str) -> int:
    """
    Calculates the minimum number of swaps to group all black balls to the right.

    Args:
        s: A binary string representing the colors of the balls.

    Returns:
        The minimum number of swaps needed.
    """

    n = len(s)
    white_count = 0
    swaps = 0

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

    return swaps