Introduction

In this blog post, we will explore how to solve the LeetCode problem "1702" 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 binary string binary consisting of only 0's or 1's. You can apply each of the following operations any number of times: Operation 1: If the number contains the substring "00", you can replace it with "10". For example, "00010" -> "10010" Operation 2: If the number contains the substring "10", you can replace it with "01". For example, "00010" -> "00001" Return the maximum binary string you can obtain after any number of operations. Binary string x is greater than binary string y if x's decimal representation is greater than y's decimal representation. Example 1: Input: binary = "000110" Output: "111011" Explanation: A valid transformation sequence can be: "000110" -> "000101" "000101" -> "100101" "100101" -> "110101" "110101" -> "110011" "110011" -> "111011" Example 2: Input: binary = "01" Output: "01" Explanation: "01" cannot be transformed any further. Constraints: 1 <= binary.length <= 105 binary consist of '0' and '1'.

Explanation

Here's the breakdown of the solution:

• Identify the Zero Count and First Zero Position: The core idea is to count the number of zeros in the binary string and locate the index of the first zero. All zeros except one can be converted to ones, and that single zero will be placed optimally.
• Construct the Maximum String: Build the largest possible binary string by setting all bits to '1' except for a single '0'. This '0' will be placed in such a way that the resulting string is maximized. This is achieved by placing the single '0' at the index of the first '0' found in the original string, after converting all preceding zeros to ones.

• Handle Edge Cases: If there are no zeros in the input, the input string is already the maximum.

• Runtime & Storage Complexity: O(n) time complexity, O(n) space complexity.

Code

    def maximumBinaryString(binary: str) -> str:
    """
    Transforms a binary string to its maximum possible value using the given operations.

    Args:
        binary: The input binary string.

    Returns:
        The maximum possible binary string after applying the operations.
    """

    n = len(binary)
    zeros = 0
    first_zero = -1

    for i in range(n):
        if binary[i] == '0':
            zeros += 1
            if first_zero == -1:
                first_zero = i

    if zeros <= 1:
        return binary

    result = ['1'] * n
    result[first_zero + zeros - 1] = '0'

    return "".join(result)