Question

2864. Maximum Odd Binary Number

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

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";
  }
}

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';
  }
};

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 {
  const cnt = s.length - s.replace(/1/g, '').length;
  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"
  }
}