Time Needed to Rearrange a Binary String LeetCode Solution

Problem – Time Needed to Rearrange a Binary String

You are given a binary string s. In one second, all occurrences of "01" are simultaneously replaced with "10". This process repeats until no occurrences of "01" exist.

Return the number of seconds needed to complete this process.

Example 1:

Input: s = "0110101"
Output: 4
Explanation: 
After one second, s becomes "1011010".
After another second, s becomes "1101100".
After the third second, s becomes "1110100".
After the fourth second, s becomes "1111000".
No occurrence of "01" exists any longer, and the process needed 4 seconds to complete,
so we return 4.

Example 2:

Input: s = "11100"
Output: 0
Explanation:
No occurrence of "01" exists in s, and the processes needed 0 seconds to complete,
so we return 0.

Constraints:

• 1 <= s.length <= 1000
• s[i] is either '0' or '1'.

Time Needed to Rearrange a Binary String LeetCode Solution in Java

public int secondsToRemoveOccurrences(String s) {
    int zeros = 0, seconds = 0;
    for (int i = 0; i < s.length(); ++i) {
        zeros += s.charAt(i) == '0' ? 1 : 0;
        if (s.charAt(i) == '1' && zeros > 0)
            seconds = Math.max(seconds + 1, zeros);
    }
    return seconds;        
}

Time Needed to Rearrange a Binary String LeetCode Solution in C++

int secondsToRemoveOccurrences(string s) {
    int zeros = 0, seconds = 0;
    for (int i = 0; i < s.size(); ++i) {
        zeros += s[i] == '0';
        if (s[i] == '1' && zeros)
            seconds = max(seconds + 1, zeros);
    }
    return seconds;
}

Time Needed to Rearrange a Binary String LeetCode Solution in Python

class Solution:
    def secondsToRemoveOccurrences(self, s: str) -> int:
        cnt = 0
        while ('01' in s):
            s = s.replace('01', '10')
            cnt += 1
        return cnt

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