LeetCode #2864 — EASY

Maximum Odd Binary Number

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

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'.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Greedy

Example 1

"010"

Example 2

"0101"

Step 02

Core Insight

What unlocks the optimal approach

The binary representation of an odd number contains <code>'1'</code> in the least significant place.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2864: Maximum Odd Binary Number
class Solution {
    public String maximumOddBinaryNumber(String s) {
        int cnt = s.length() - s.replace("1", "").length();
        return "1".repeat(cnt - 1) + "0".repeat(s.length() - cnt) + "1";
    }
}

// Accepted solution for LeetCode #2864: Maximum Odd Binary Number
func maximumOddBinaryNumber(s string) string {
	cnt := strings.Count(s, "1")
	return strings.Repeat("1", cnt-1) + strings.Repeat("0", len(s)-cnt) + "1"
}

# Accepted solution for LeetCode #2864: Maximum Odd Binary Number
class Solution:
    def maximumOddBinaryNumber(self, s: str) -> str:
        cnt = s.count("1")
        return "1" * (cnt - 1) + (len(s) - cnt) * "0" + "1"

// Accepted solution for LeetCode #2864: Maximum Odd Binary Number
impl Solution {
    pub fn maximum_odd_binary_number(s: String) -> String {
        let cnt = s.chars().filter(|&c| c == '1').count();
        "1".repeat(cnt - 1) + &"0".repeat(s.len() - cnt) + "1"
    }
}

// Accepted solution for LeetCode #2864: Maximum Odd Binary Number
function maximumOddBinaryNumber(s: string): string {
    const cnt = s.length - s.replace(/1/g, '').length;
    return '1'.repeat(cnt - 1) + '0'.repeat(s.length - cnt) + '1';
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.