# 2864. Maximum Odd Binary Number

## Description

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'.

## Solutions

Solution 1: Greedy

 First, we count the number of ‘1’s in the string $s$, denoted as $cnt$. Then, we place $cnt - 1$ ‘1’s at the highest position, followed by the remaining $s - cnt$ ‘0’s, and finally add one ‘1’.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.

• class Solution {
public String maximumOddBinaryNumber(String s) {
int cnt = 0;
for (char c : s.toCharArray()) {
if (c == '1') {
++cnt;
}
}
return "1".repeat(cnt - 1) + "0".repeat(s.length() - cnt) + "1";
}
}

• class Solution {
public:
string maximumOddBinaryNumber(string s) {
int cnt = count_if(s.begin(), s.end(), [](char c) { return c == '1'; });
string ans;
for (int i = 1; i < cnt; ++i) {
ans.push_back('1');
}
for (int i = 0; i < s.size() - cnt; ++i) {
ans.push_back('0');
}
ans.push_back('1');
return ans;
}
};

• class Solution:
def maximumOddBinaryNumber(self, s: str) -> str:
cnt = s.count("1")
return "1" * (cnt - 1) + (len(s) - cnt) * "0" + "1"

• func maximumOddBinaryNumber(s string) string {
cnt := strings.Count(s, "1")
return strings.Repeat("1", cnt-1) + strings.Repeat("0", len(s)-cnt) + "1"
}

• function maximumOddBinaryNumber(s: string): string {
let cnt = 0;
for (const c of s) {
cnt += c === '1' ? 1 : 0;
}
return '1'.repeat(cnt - 1) + '0'.repeat(s.length - cnt) + '1';
}

• 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"
}
}