Medium
You are given an n x n
integer matrix board
where the cells are labeled from 1
to n2
in a Boustrophedon style starting from the bottom left of the board (i.e. board[n - 1][0]
) and alternating direction each row.
You start on square 1
of the board. In each move, starting from square curr
, do the following:
next
with a label in the range [curr + 1, min(curr + 6, n2)]
.
next
has a snake or ladder, you must move to the destination of that snake or ladder. Otherwise, you move to next
.n2
.A board square on row r
and column c
has a snake or ladder if board[r][c] != -1
. The destination of that snake or ladder is board[r][c]
. Squares 1
and n2
do not have a snake or ladder.
Note that you only take a snake or ladder at most once per move. If the destination to a snake or ladder is the start of another snake or ladder, you do not follow the subsequent snake or ladder.
[[-1,4],[-1,3]]
, and on the first move, your destination square is 2
. You follow the ladder to square 3
, but do not follow the subsequent ladder to 4
.Return the least number of moves required to reach the square n2
. If it is not possible to reach the square, return -1
.
Example 1:
Input: board = [[-1,-1,-1,-1,-1,-1],[-1,-1,-1,-1,-1,-1],[-1,-1,-1,-1,-1,-1],[-1,35,-1,-1,13,-1],[-1,-1,-1,-1,-1,-1],[-1,15,-1,-1,-1,-1]]
Output: 4
Explanation:
In the beginning, you start at square 1 (at row 5, column 0).
You decide to move to square 2 and must take the ladder to square 15.
You then decide to move to square 17 and must take the snake to square 13.
You then decide to move to square 14 and must take the ladder to square 35.
You then decide to move to square 36, ending the game.
This is the lowest possible number of moves to reach the last square, so return 4.
Example 2:
Input: board = [[-1,-1],[-1,3]]
Output: 1
Constraints:
n == board.length == board[i].length
2 <= n <= 20
grid[i][j]
is either -1
or in the range [1, n2]
.1
and n2
do not have any ladders or snakes.import java.util.LinkedList
import java.util.Queue
class Solution {
private var size = 0
fun snakesAndLadders(board: Array<IntArray>): Int {
val queue: Queue<Int> = LinkedList()
size = board.size
val target = size * size
val visited = BooleanArray(target)
queue.add(1)
visited[0] = true
var step = 0
while (queue.isNotEmpty()) {
val queueSize = queue.size
for (i in 0 until queueSize) {
val previousLabel = queue.poll()
if (previousLabel == target) {
return step
}
for (currentLabel in previousLabel + 1..Math.min(target, previousLabel + 6)) {
if (visited[currentLabel - 1]) {
continue
}
visited[currentLabel - 1] = true
val position = indexToPosition(currentLabel)
if (board[position[0]][position[1]] == -1) {
queue.add(currentLabel)
} else {
queue.add(board[position[0]][position[1]])
}
}
}
step++
}
return -1
}
private fun indexToPosition(index: Int): IntArray {
val vertical = size - 1 - (index - 1) / size
val horizontal: Int = if ((size - vertical) % 2 == 1) {
(index - 1) % size
} else {
size - 1 - (index - 1) % size
}
return intArrayOf(vertical, horizontal)
}
}