Introduction

In this blog post, we will explore how to solve the LeetCode problem "2864" 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 s that contains at least one '1'. You have to rearrange the bits in such a way that the resulting binary number is the maximum odd binary number that can be created from this combination. Return a string representing the maximum odd binary number that can be created from the given combination. Note that the resulting string can have leading zeros. Example 1: Input: s = "010" Output: "001" Explanation: Because there is just one '1', it must be in the last position. So the answer is "001". Example 2: Input: s = "0101" Output: "1001" Explanation: One of the '1's must be in the last position. The maximum number that can be made with the remaining digits is "100". So the answer is "1001". Constraints: 1 <= s.length <= 100 s consists only of '0' and '1'. s contains at least one '1'.

Explanation

Here's the solution:

• Count 1s: Count the number of '1's in the input string.

• Construct the result: Place one '1' at the end to ensure the number is odd. Then, add the remaining '1's at the beginning, followed by '0's to fill the remaining spaces.

• Time Complexity: O(n), where n is the length of the string. Space Complexity: O(n)

Code

    def maximum_odd_binary_number(s: str) -> str:
    """
    Given a binary string s that contains at least one '1'. You have to rearrange the bits in such a way that the resulting binary number is the maximum odd binary number that can be created from this combination.
    Return a string representing the maximum odd binary number that can be created from the given combination. Note that the resulting string can have leading zeros.

    For example:
    maximum_odd_binary_number("010") == "001"
    maximum_odd_binary_number("0101") == "1001"
    """

    ones = s.count('1')
    zeros = len(s) - ones

    result = '1' * (ones - 1) + '0' * zeros + '1'

    return result