Introduction

In this blog post, we will explore how to solve the LeetCode problem "1758" 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 string s consisting only of the characters '0' and '1'. In one operation, you can change any '0' to '1' or vice versa. The string is called alternating if no two adjacent characters are equal. For example, the string "010" is alternating, while the string "0100" is not. Return the minimum number of operations needed to make s alternating. Example 1: Input: s = "0100" Output: 1 Explanation: If you change the last character to '1', s will be "0101", which is alternating. Example 2: Input: s = "10" Output: 0 Explanation: s is already alternating. Example 3: Input: s = "1111" Output: 2 Explanation: You need two operations to reach "0101" or "1010". Constraints: 1 <= s.length <= 104 s[i] is either '0' or '1'.

Explanation

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

• High-Level Approach:

  • Consider two possible alternating strings: one starting with '0' and the other starting with '1'.
  • Calculate the number of operations needed to transform the input string s into each of these alternating strings.
  • Return the minimum of these two operation counts.

• Complexity:

  • Runtime: O(n), where n is the length of the string.
  • Storage: O(1) - uses a constant amount of extra space.

Code

    def minOperations(s: str) -> int:
    """
    Calculates the minimum number of operations to make a string alternating.

    Args:
        s: The input string consisting of '0' and '1'.

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

    n = len(s)
    count0 = 0
    count1 = 0

    for i in range(n):
        if i % 2 == 0:  # Even index
            if s[i] == '1':
                count0 += 1
            else:
                count1 += 1
        else:  # Odd index
            if s[i] == '1':
                count1 += 1
            else:
                count0 += 1

    return min(count0, count1)