# Solving Leetcode Interviews in Seconds with AI: Maximum Odd Binary Number


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```