LeetCode in Kotlin
957. Prison Cells After N Days
Medium
There are 8 prison cells in a row and each cell is either occupied or vacant.
Each day, whether the cell is occupied or vacant changes according to the following rules:
- If a cell has two adjacent neighbors that are both occupied or both vacant, then the cell becomes occupied.
- Otherwise, it becomes vacant.
Note that because the prison is a row, the first and the last cells in the row can’t have two adjacent neighbors.
You are given an integer array cells where cells[i] == 1 if the ith cell is occupied and cells[i] == 0 if the ith cell is vacant, and you are given an integer n.
Return the state of the prison after n days (i.e., n such changes described above).
Example 1:
Input: cells = [0,1,0,1,1,0,0,1], n = 7
Output: [0,0,1,1,0,0,0,0]
Explanation: The following table summarizes the state of the prison on each day:
Day 0: [0, 1, 0, 1, 1, 0, 0, 1]
Day 1: [0, 1, 1, 0, 0, 0, 0, 0]
Day 2: [0, 0, 0, 0, 1, 1, 1, 0]
Day 3: [0, 1, 1, 0, 0, 1, 0, 0]
Day 4: [0, 0, 0, 0, 0, 1, 0, 0]
Day 5: [0, 1, 1, 1, 0, 1, 0, 0]
Day 6: [0, 0, 1, 0, 1, 1, 0, 0]
Day 7: [0, 0, 1, 1, 0, 0, 0, 0]
Example 2:
Input: cells = [1,0,0,1,0,0,1,0], n = 1000000000
Output: [0,0,1,1,1,1,1,0]
Constraints:
cells.length == 8cells[i]is either0or1.1 <= n <= 109
Solution
@Suppress("NAME_SHADOWING")
class Solution {
fun prisonAfterNDays(cells: IntArray, n: Int): IntArray {
var n = n
if (n == 0) {
return cells
}
var first: IntArray? = null
var prev = cells
var period: Int
var day = 0
while (n > 0) {
day++
n--
val next = getNextDay(prev)
if (next.contentEquals(first)) {
period = day - 1
n %= period
}
if (day == 1) {
first = next
}
prev = next
}
return prev
}
private fun getNextDay(arr: IntArray): IntArray {
val ret = IntArray(8)
for (i in 1..6) {
if (arr[i - 1] == arr[i + 1]) {
ret[i] = 1
}
}
return ret
}
}