LeetCode 2864. Maximum Odd Binary Number Solution in Java, C++, Python & More | Explanation + Code

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.

PythonJavaC++GoTypeScriptRust

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

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"; } }(code-box)

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'; } };(code-box)

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

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'; }(code-box)

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" } }(code-box)