LeetCode in Kotlin
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
class Solution {
fun secondsToRemoveOccurrences(s: String): Int {
var lastOne = -1
var result = 0
var prevResult: Int
var curResult = 0
var countOne = 0
var countZero = 0
var diff: Int
var pTarget: Int
var pWait: Int
var cTarget: Int
for (i in 0 until s.length) {
if (s[i] == '0') {
++countZero
continue
}
++countOne
diff = i - lastOne - 1
prevResult = curResult
cTarget = countOne - 1
pTarget = cTarget - 1
pWait = prevResult - (lastOne - pTarget)
curResult = if (diff > pWait) {
countZero
} else {
if (countZero == 0) 0 else pWait - diff + 1 + countZero
}
result = curResult
lastOne = i
}
return result
}
}