Solution to prison after n days problem. Time Complexity O(M^N)

PrisonAfterNDays.kt

fun prisonAfterNDays(cells: IntArray, N: Int): IntArray {
    var currentIndex = 0
    val prisonMap = HashMap<List<Int>, Int>()

    while (currentIndex < N) {
        var tempArr = cells.clone()

        if (prisonMap[tempArr.toList()]?.hashCode() != null) {
            val cycleLength = currentIndex - prisonMap[tempArr.toList()]!!
            val remainingDays = N - currentIndex
            val numOfRepeats = remainingDays / cycleLength
            val newIndex = currentIndex + (cycleLength * numOfRepeats)
            tempArr = prisonMap.filterValues { it == (N - newIndex).plus(prisonMap[tempArr.toList()]!!) }.keys.first().toIntArray()
            cells.forEachIndexed { index, _ -> cells[index] = tempArr[index] }
            break
        } else {
            prisonMap[tempArr.toList()] = currentIndex
        }

        for (prisonIndex in 0 until cells.size) {
            if (prisonIndex == 0 || prisonIndex == cells.size - 1) cells[prisonIndex] = 0
            else if (tempArr[prisonIndex - 1] == tempArr[prisonIndex + 1]) cells[prisonIndex] = 1
            else cells[prisonIndex] = 0
        }

        currentIndex++
    }

    return cells
}