Introduction

In this blog post, we will explore how to solve the LeetCode problem "1529" 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 0-indexed binary string target of length n. You have another binary string s of length n that is initially set to all zeros. You want to make s equal to target. In one operation, you can pick an index i where 0 <= i < n and flip all bits in the inclusive range [i, n - 1]. Flip means changing '0' to '1' and '1' to '0'. Return the minimum number of operations needed to make s equal to target. Example 1: Input: target = "10111" Output: 3 Explanation: Initially, s = "00000". Choose index i = 2: "00000" -> "00111" Choose index i = 0: "00111" -> "11000" Choose index i = 1: "11000" -> "10111" We need at least 3 flip operations to form target. Example 2: Input: target = "101" Output: 3 Explanation: Initially, s = "000". Choose index i = 0: "000" -> "111" Choose index i = 1: "111" -> "100" Choose index i = 2: "100" -> "101" We need at least 3 flip operations to form target. Example 3: Input: target = "00000" Output: 0 Explanation: We do not need any operations since the initial s already equals target. Constraints: n == target.length 1 <= n <= 105 target[i] is either '0' or '1'.

Explanation

Here's the solution:

• Approach: Iterate through the target string. Keep track of the number of flips performed so far. If the current bit in the s string (which is implicitly maintained based on the flips) doesn't match the target bit, increment the flip count. Each flip affects all subsequent bits, so we only need to track the total number of flips.

• Complexity:

  • Runtime: O(n)
  • Storage: O(1)

Code

    def minFlips(target):
    """
    Calculates the minimum number of flips to transform an all-zeros string to the target string.

    Args:
        target (str): The target binary string.

    Returns:
        int: The minimum number of flips needed.
    """

    flips = 0
    current_state = 0  # 0 represents the initial state (all zeros)

    for bit in target:
        if (current_state % 2) != int(bit):
            flips += 1
            current_state += 1

    return flips