317 Shortest Distance from All Buildings

317. Shortest Distance from All Buildings

1. Question

You want to build a house on anemptyland 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.

2. Implementation

(1) BFS

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

        int m = grid.length;
        int n = grid[0].length;

        int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

        int[][] distance = new int[m][n];
        int[][] reach = new int[m][n];

        int buildings = 0;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    ++buildings;
                    getDistanceByBFS(i, j, distance, reach, directions, grid);
                }
            }
        }

        int res = 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] == buildings) {
                    res = Math.min(res, distance[i][j]);
                }
            }
        }
        return res == Integer.MAX_VALUE ? -1 : res;
    }

    public void getDistanceByBFS(int row, int col, int[][] distance, int[][] reach, int[][] directions, int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;

        boolean[][] visited = new boolean[m][n];

        Queue<int[]> queue = new LinkedList<>();
        queue.add(new int[] {row, col});

        int dist = 0;

        while (!queue.isEmpty()) {
            ++dist;
            int size = queue.size();

            for (int i = 0; i < size; i++) {
                int[] curCell = queue.remove();

                for (int[] direction : directions) {
                    int nextRow = curCell[0] + direction[0];
                    int nextCol = curCell[1] + direction[1];

                    if (isValid(nextRow, nextCol, visited, grid)) {
                        distance[nextRow][nextCol] += dist;
                        reach[nextRow][nextCol] += 1;
                        queue.add(new int[] {nextRow, nextCol});
                        visited[nextRow][nextCol] = true;
                    }
                }
            }
        }
    }

    public boolean isValid(int row, int col, boolean[][] visited, int[][] grid) {
        return row >= 0 && row < grid.length && col >= 0 && col < grid[0].length && !visited[row][col] && grid[row][col] == 0;
    }
}

3. Time & Space Complexity

BFS: 时间复杂度O(M^2N^2),BFS的时间复杂度是O(mn), 总的时间复杂度是 O(# of buildings ) * O(mn), 最坏的情况是每个cell都是building， 总的时间复杂度为O(mn) * O(mn) => O(m^2n^2), 空间复杂度O(mn)