README_EN.md

1702. Maximum Binary String After Change

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Description

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

Python3

class Solution:
    def maximumBinaryString(self, binary: str) -> str:
        k = binary.find('0')
        if k == -1:
            return binary
        k += binary[k + 1 :].count('0')
        return '1' * k + '0' + '1' * (len(binary) - k - 1)

Java

class Solution {
    public String maximumBinaryString(String binary) {
        int k = binary.indexOf('0');
        if (k == -1) {
            return binary;
        }
        int n = binary.length();
        for (int i = k + 1; i < n; ++i) {
            if (binary.charAt(i) == '0') {
                ++k;
            }
        }
        char[] ans = binary.toCharArray();
        Arrays.fill(ans, '1');
        ans[k] = '0';
        return String.valueOf(ans);
    }
}

C++

class Solution {
public:
    string maximumBinaryString(string binary) {
        int k = binary.find('0');
        if (k == binary.npos) return binary;
        int n = binary.size();
        for (int i = k + 1; i < n; ++i) {
            if (binary[i] == '0') {
                ++k;
            }
        }
        return string(k, '1') + '0' + string(n - k - 1, '1');
    }
};

Go

func maximumBinaryString(binary string) string {
	k := strings.IndexByte(binary, '0')
	if k == -1 {
		return binary
	}
	for _, c := range binary[k+1:] {
		if c == '0' {
			k++
		}
	}
	ans := []byte(binary)
	for i := range ans {
		ans[i] = '1'
	}
	ans[k] = '0'
	return string(ans)
}

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