Introduction

In this blog post, we will explore how to solve the LeetCode problem "1864" 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

Given a binary string s, return the minimum number of character swaps to make it alternating, or -1 if it is impossible. The string is called alternating if no two adjacent characters are equal. For example, the strings "010" and "1010" are alternating, while the string "0100" is not. Any two characters may be swapped, even if they are not adjacent. Example 1: Input: s = "111000" Output: 1 Explanation: Swap positions 1 and 4: "111000" -> "101010" The string is now alternating. Example 2: Input: s = "010" Output: 0 Explanation: The string is already alternating, no swaps are needed. Example 3: Input: s = "1110" Output: -1 Constraints: 1 <= s.length <= 1000 s[i] is either '0' or '1'.

Explanation

Here's a breakdown of the approach and the Python code solution:

• Count Mismatches: Calculate the number of swaps needed to make the string alternating starting with '0' and starting with '1'. This involves counting the positions where the character doesn't match the expected character in the alternating pattern.
• Check Feasibility: If the counts of '0's and '1's in the original string don't allow for an alternating pattern, return -1.

• Return Minimum: Return the minimum number of swaps calculated from the two possible alternating patterns.

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

Code

    def minSwaps(s):
    """
    Calculates the minimum number of swaps to make a binary string alternating.

    Args:
        s: The binary string.

    Returns:
        The minimum number of swaps, or -1 if impossible.
    """

    n = len(s)
    zeros = s.count('0')
    ones = n - zeros

    if abs(zeros - ones) > 1:
        return -1

    swaps0 = 0
    swaps1 = 0

    for i in range(n):
        if i % 2 == 0:  # Even position
            if s[i] == '1':
                swaps0 += 1
            if s[i] == '0':
                swaps1 += 1
        else:  # Odd position
            if s[i] == '0':
                swaps0 += 1
            if s[i] == '1':
                swaps1 += 1

    if zeros > ones:
      return swaps1
    elif ones > zeros:
      return swaps0
    else:
      return min(swaps0, swaps1)