LeetCode in Kotlin

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

class Solution {
    fun maximumOddBinaryNumber(s: String): String {
        var len = s.length
        var count = 0
        val sb = StringBuilder()
        for (i in 0 until len) {
            if (s[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()
    }
}

This site is open source. Improve this page.