286. Walls and Gates (Solution 1)

package LeetCode_286

/**
 * 286. Walls and Gates
 * (Prime)
 *
 * You are given a m x n 2D grid initialized with these three possible values.
1. -1 - A wall or an obstacle.
2. 0 - A gate.
3. INF - Infinity means an empty room.
We use the value 231 - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647.
Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF.

Example:
Given the 2D grid:
INF  -1  0  INF
INF INF INF  -1
INF  -1 INF  -1
0  -1 INF INF

After running your function, the 2D grid should be:
3  -1   0   1
2   2   1  -1
1  -1   2  -1
0  -1   3   4
 * */
class Solution {
    /*
    * Solution: DFS, Time complexity:O(m^2 * n^2), Space complexity:O(4^n)
    * */
    fun fillEmptyRoom(grid: Array<IntArray>) {
        val m = grid.size
        val n = grid[0].size
        for (i in 0 until m) {
            for (j in 0 until n) {
                //start dfs from where is a gate
                if (grid[i][j] == 0) {
                    dfs(i, j, 0, grid)
                }
            }
        }
    }

    private fun dfs(x: Int, y: Int, count: Int, grid: Array<IntArray>) {
        //when reach gate,wall,or confirm min distance return
        if (x < 0 || x >= grid.size || y < 0 || y >= grid[0].size || grid[x][y] < count) {
            return
        }
        grid[x][y] = count
        //expand 4 directions
        dfs(x + 1, y, count + 1, grid)
        dfs(x - 1, y, count + 1, grid)
        dfs(x, y + 1, count + 1, grid)
        dfs(x, y - 1, count + 1, grid)
    }
}