317. Shortest Distance from All Buildings

You want to build a house on an empty land which reaches all buildings in the shortest amount of distance. You can only move up, down, left and right. You are given a 2D grid of values 0, 1 or 2, where:

• Each 0 marks an empty land which you can pass by freely.
• Each 1 marks a building which you cannot pass through.
• Each 2 marks an obstacle which you cannot pass through.

For example, given three buildings at (0,0), (0,4), (2,2), and an obstacle at (0,2):

1 - 0 - 2 - 0 - 1
|   |   |   |   |
0 - 0 - 0 - 0 - 0
|   |   |   |   |
0 - 0 - 1 - 0 - 0

The point (1,2) is an ideal empty land to build a house, as the total travel distance of 3+3+1=7 is minimal. So return 7.

Note:
There will be at least one building. If it is not possible to build such house according to the above rules, return -1.

Similar : 

• 286. Walls and Gates
• 296. Best Meeting Point

public class Solution {
    public int shortestDistance(int[][] grid) {
        if (grid.length==0 || grid[0].length==0)
            return 0;
        
        int m = grid.length;
        int n = grid[0].length;
        int[][] dist = new int[m][n];
        int[][] reach = new int[m][n];
        final int[] shift = {1, 0, -1, 0, 1};
        int bldNum = 0;
        
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    bldNum++;
                    
                    Queue<int[]> queue = new LinkedList<int[]>();
                    queue.offer(new int[] {i,j});
                    boolean[][] visit = new boolean[m][n];
                    int level = 1;
                    while (!queue.isEmpty()) {
                        int size = queue.size();
                        for (int k=0; k<size; k++) {
                            int[] pt = queue.poll();
                            
                            for (int l = 0; l < 4; l++) {
                                int row = pt[0] + shift[l];
                                int col = pt[1] + shift[l+1];
                                if (row >= 0 && row < m && col >= 0 && col < n && grid[row][col] == 0 && !visit[row][col]) {
                                        dist[row][col] += level;
                                        reach[row][col]++;
                                        visit[row][col] = true;
                                        queue.offer(new int[] {row,col});
                                }
                            }
                            
                        }
                        level++;
                    }
                }
            }

        int shortest = Integer.MAX_VALUE;
        for (int i = 0; i < m; i++)
            for(int j = 0; j < n; j++) {
                if (grid[i][j] == 0 && reach[i][j] == bldNum) {
                    shortest = Math.min(shortest, dist[i][j]);
                }
            }
            
        return shortest == Integer.MAX_VALUE ? -1 : shortest;
    }
}