Leetcode #2864: Maximum Odd Binary Number
In this guide, we solve Leetcode #2864 Maximum Odd Binary Number in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Greedy, Math, String
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: s = "010"
Output: "001"
Explanation: Because there is just one '1', it must be in the last position. So the answer is "001".
Python Solution
class Solution:
def maximumOddBinaryNumber(self, s: str) -> str:
cnt = s.count("1")
return "1" * (cnt - 1) + (len(s) - cnt) * "0" + "1"
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.