LeetCode-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.

Analysis:

Simply BFS on every building point.

Solution:

public class Solution {
    public void addPointToQueue(int[][] grid, int[][] sum, int[][] dis, boolean[][] visited, int curDis,
            int nextX, int nextY, Queue<Point> q) {
        int row = sum.length;
        int col = sum[0].length;
        if (nextX < 0 || nextX >= row || nextY < 0 || nextY >= col || visited[nextX][nextY] || grid[nextX][nextY] > 0) {
            return;
        }

        visited[nextX][nextY] = true;
        dis[nextX][nextY] = curDis + 1;
        q.add(new Point(nextX, nextY));

        sum[nextX][nextY] += curDis + 1;
    }

    public void BuildingUpdate(int[][] grid, int[][] sum, boolean[][] allReachable, Point start) {
        int row = grid.length;
        int col = grid[0].length;
        int[][] dis = new int[row][col];
        boolean[][] visited = new boolean[row][col];

        Queue<Point> q = new LinkedList<Point>();
        q.add(start);
        while (!q.isEmpty()) {
            Point cur = q.poll();
            int d = dis[cur.x][cur.y];

            addPointToQueue(grid, sum, dis, visited, d, cur.x - 1, cur.y, q);
            addPointToQueue(grid, sum, dis, visited, d, cur.x + 1, cur.y, q);
            addPointToQueue(grid, sum, dis, visited, d, cur.x, cur.y - 1, q);
            addPointToQueue(grid, sum, dis, visited, d, cur.x, cur.y + 1, q);
        }

        // Check reachability
        for (int i = 0; i < row; i++)
            for (int j = 0; j < col; j++)
                if (!visited[i][j]) {
                    allReachable[i][j] = false;
                }
    }

    public int shortestDistance(int[][] grid) {
        if (grid.length == 0 || grid[0].length == 0)
            return -1;

        int row = grid.length;
        int col = grid[0].length;
        int[][] sum = new int[row][col];
        boolean[][] allReachable = new boolean[row][col];
        for (int i = 0; i < row; i++) {
            Arrays.fill(allReachable[i], true);
        }

        for (int i = 0; i < row; i++)
            for (int j = 0; j < col; j++)
                if (grid[i][j] == 1) {
                    BuildingUpdate(grid, sum, allReachable, new Point(i, j));
                }

        int res = Integer.MAX_VALUE;
        for (int i = 0; i < row; i++)
            for (int j = 0; j < col; j++)
                if (allReachable[i][j]) {
                    res = Math.min(res, sum[i][j]);
                }

        if (res == Integer.MAX_VALUE)
            return -1;
        return res;
    }
}

Note: the code can be shorter, just I like this which makes the code clear and organized.