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.

Using BFS. For every building, get distance from it to all empty spaces. Then sum up and find minimum. In order to avoid creating to many tracking matrices, every time we finishing updating distance for one building, decrement the empty space by 1, thus we know this space is visited and is valid for next building to traverse.

    public int shortestDistance(int[][] grid) {
        if (grid.length == 0 || grid[0].length == 0) {
            return -1;
        }
        int m = grid.length, n = grid[0].length;
        int[][] sum = new int[m][n];
        int[][] neighbors = {
            {-1, 0},
            {1, 0},
            {0, -1},
            {0, 1}
        };

        int val = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    findMin(grid, sum, neighbors, i, j, val);
                    val--;
                }
            }
        }
        int rst = Integer.MAX_VALUE;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] < 0) {
                    rst = Math.min(rst, sum[i][j]);
                }
            }
        }
        return rst;
    }

    private void findMin(int[][] grid, int[][]sum, int[][] neighbors, int i, int j, int val) {
        int m = grid.length, n = grid[0].length;
        int[][] dist = new int[grid.length][grid[0].length];
        Queue queue = new LinkedList<>();
        queue.add(i * n + j);
        while (!queue.isEmpty()) {
            int curr = queue.poll();
            int x = curr / n;
            int y = curr % n;
            for (int[] nei : neighbors) {
                int x1 = x + nei[0];
                int y1 = y + nei[1];
                if (x1 >= 0 && x1 < m && y1 >= 0 && y1 < n && grid[x1][y1] == val) {
                    dist[x1][y1] = dist[x][y] + 1;
                    grid[x1][y1]--;
                    sum[x1][y1] += dist[x1][y1];
                    queue.add(x1 * n + y1);
                }
            }
        }
    }