// You are given a m x n 2D grid initialized with these three possible values.

// -1 - A wall or an obstacle.

// 0 - A gate.

// 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.

// For 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

public class Solution {

public void wallsAndGates(int[][] rooms) {

//iterate through the matrix calling dfs on all indices that contain a zero

for(int i = 0; i < rooms.length; i++) {

for(int j = 0; j < rooms[0].length; j++) {

if(rooms[i][j] == 0) {

dfs(rooms, i, j, 0);

}

}

}

}

void dfs(int[][] rooms, int i, int j, int distance) {

//if you have gone out of the bounds of the array or you have run into a wall/obstacle, return

// room[i][j] < distance also ensure that we do not overwrite any previously determined distance if it is shorter than our current distance

if(i < 0 || i >= rooms.length || j < 0 || j >= rooms[0].length || rooms[i][j] < distance) {

return;

}

//set current index's distance to distance

rooms[i][j] = distance;

//recurse on all adjacent neighbors of rooms[i][j]

dfs(rooms, i + 1, j, distance + 1);

dfs(rooms, i - 1, j, distance + 1);

dfs(rooms, i, j + 1, distance + 1);

dfs(rooms, i, j - 1, distance + 1);

}

}