Walls And Gates

Walls And Gates first search

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
``````

Walls And Gates Solution

``````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);
}
}``````