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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 \(s\).

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

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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";
    }
}

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class Solution {
public:
    string maximumOddBinaryNumber(string s) {
        int cnt = count(s.begin(), s.end(), '1');
        return string(cnt - 1, '1') + string(s.size() - cnt, '0') + '1';
    }
};

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

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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';
}

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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"
    }
}

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