Maximum binary string after change

We will use greedy approach to solve the problem. We will find the first 0 and change it to 1. Then we will find the next 0 and change it to 1. We will repeat this process until we find the last 0. Finally we will return the binary string.

class Solution:
    def maximumBinaryString(self, binary: str) -> str:
        if '0' not in binary:
            return binary
        n = len(binary)
        k = binary.count('1', binary.find('0'))
        return '1' * (n-k-1) + '0' + '1' * k

Time complexity: O(n)
Space complexity: O(1)