LeetCode-in-Java
2380. Time Needed to Rearrange a Binary String
Medium
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 <= 1000s[i]is either'0'or'1'.
Follow up:
Can you solve this problem in O(n) time complexity?
Solution
public class Solution {
public int secondsToRemoveOccurrences(String s) {
int lastOne = -1;
int result = 0;
int prevResult = 0;
int curResult = 0;
int countOne = 0;
int countZero = 0;
int diff;
int pTarget;
int pWait;
int cTarget;
for (int i = 0; i < s.length(); ++i) {
if (s.charAt(i) == '0') {
++countZero;
continue;
}
++countOne;
diff = i - lastOne - 1;
prevResult = curResult;
cTarget = countOne - 1;
pTarget = cTarget - 1;
pWait = prevResult - (lastOne - pTarget);
if (diff > pWait) {
curResult = countZero;
} else {
curResult = countZero == 0 ? 0 : pWait - diff + 1 + countZero;
}
result = curResult;
lastOne = i;
}
return result;
}
}