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# 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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