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286. Walls and Gates (Solution 2)

package LeetCode_286

import java.util.*

/**
 * 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 1: DFS, Time complexity:O(m^2*n^2), Space complexity:O(4^n) * Solution 2: BFS, Time complexity:O(mn), Space complexity:O(mn) * */ 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 bfs from where is a gate if (grid[i][j] == 0) { bfs(Pair(i, j), grid) } } } } private fun bfs(pair: Pair<Int, Int>, grid: Array<IntArray>) { val directions = intArrayOf(0, 1, 0, -1, 0) val queue
= LinkedList<Pair<Int, Int>>() queue.offer(pair) while (queue.isNotEmpty()) { val cur = queue.pop() //expand 4 directions for (i in 0 until 4) { val x = cur.first + directions[i] val y = cur.second + directions[i + 1] //return when reach gate,wall,or confirm min distance if (x < 0 || x >= grid.size || y < 0 || y >= grid[0].size || grid[x][y] < grid[cur.first][cur.second] + 1) { continue } //update new count grid[x][y] = grid[cur.first][cur.second] + 1 queue.offer(Pair(x, y)) } } } }