README.md

1702. Maximum Binary String After Change

You are given a binary string binary consisting of only 0's or 1's. You can apply each of the following operations any number of times:

• Operation 1: If the number contains the substring "00", you can replace it with "10".
  • For example, "00010" -> "10010"
• Operation 2: If the number contains the substring "10", you can replace it with "01".
  • For example, "00010" -> "00001"

Return the maximum binary string you can obtain after any number of operations. Binary string x is greater than binary string y if x's decimal representation is greater than y's decimal representation.

Example 1:

Input: binary = "000110"
Output: "111011"
Explanation: A valid transformation sequence can be:
"000110" -> "000101"
"000101" -> "100101"
"100101" -> "110101"
"110101" -> "110011"
"110011" -> "111011"

Example 2:

Input: binary = "01"
Output: "01"
Explanation: "01" cannot be transformed any further.

Constraints:

• 1 <= binary.length <= 105
• binary consist of '0' and '1'.

Solutions (Rust)

1. Solution

impl Solution {
    pub fn maximum_binary_string(binary: String) -> String {
        let count0 = binary.bytes().filter(|&b| b == b'0').count();
        let count1 = binary.bytes().take_while(|&b| b == b'1').count();
        let mut binary = binary.into_bytes();

        if count0 > 1 {
            binary = vec![b'1'; binary.len()];
            binary[count0 + count1 - 1] = b'0';
        }

        String::from_utf8(binary).unwrap()
    }
}