LeetCode-in-Java

2864. Maximum Odd Binary Number

Easy

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'.

Solution

public class Solution {
    public String maximumOddBinaryNumber(String s) {
        int len = s.length();
        int count = 0;
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < len; i++) {
            if (s.charAt(i) == '1') {
                count++;
            }
        }
        if (count == len) {
            return s;
        }
        len -= count;
        while (count > 1) {
            sb.append('1');
            count--;
        }
        while (len > 0) {
            sb.append('0');
            len--;
        }
        sb.append('1');
        return sb.toString();
    }
}

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