Problem Description
You are given an m x n grid rooms 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.
Solution
This can be solved using multi-source BFS in which the source will be the gates.
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class Solution {
private static final int INF = 2147483647;
public void wallsAndGates(int[][] rooms) {
int n = rooms.length;
int m = rooms[0].length;
Queue<Pair> q = new LinkedList<>();
//add all cells with value 0 (which represent the gates) to the queue
//we will use multi-source BFS to check the distance of other reachable cells
for(int i=0; i<n; i++){
for(int j=0; j<m; j++){
if(rooms[i][j] == 0) q.add(new Pair(i, j));
}
}
while(!q.isEmpty()){
Pair p = q.poll();
int[] x = {0, 0, 1, -1};
int[] y = {1, -1, 0, 0};
for(int k=0; k<4; k++){
int x2 = p.x + x[k];
int y2 = p.y + y[k];
//if the next cell is within the range and its value is INF, which means it's an empty room which has not been visited yet
if(x2 >=0 && y2 >=0 && x2<n && y2<m && rooms[x2][y2] == INF){
rooms[x2][y2] = 1 + rooms[p.x][p.y];
q.add(new Pair(x2, y2));
}
}
}
}
}
class Pair{
int x;
int y;
Pair(int x, int y){
this.x = x;
this.y = y;
}
}