Build Post Office II

Description

Given a 2D grid, each cell is either a wall  2, an house  1 or empty  0 (the number zero, one, two), find a place to build a post office so that the sum of the distance from the post office to all the houses is smallest.

Return the smallest sum of distance. Return -1 if it is not possible.

• You cannot pass through wall and house, but can pass through empty.
• You only build post office on an empty.

 

Example

Example 1:

Input：[[0,1,0,0,0],[1,0,0,2,1],[0,1,0,0,0]]
Output：8
Explanation： Placing a post office at (1,1), the distance that post office to all the house sum is smallest.

Example 2:

Input：[[0,1,0],[1,0,1],[0,1,0]]
Output：4
Explanation： Placing a post office at (1,1), the distance that post office to all the house sum is smallest.

Challenge

Solve this problem within O(n^3) time.

Ideas:

• The title uses bfs, traversing the grid for the first time, to be located in the empty bfs, the search is completed, if present, houses are not vis mark the change of open space can not be provided at home.
• Bfs simple search process, sun + = dist; point distance and the current implementation updates.
• now.dis + 1 each update of the current distance between two points is achieved.

class Coordinate {
    int x, y;
    public Coordinate(int x, int y) {
        this.x = x;
        this.y = y;
    }
}

public class Solution {
    public int EMPTY = 0;
    public int HOUSE = 1;
    public int WALL = 2;
    public int[][] grid;
    public int n, m;
    public int[] deltaX = {0, 1, -1, 0};
    public int[] deltaY = {1, 0, 0, -1};
    
    private List<Coordinate> getCoordinates(int type) {
        List<Coordinate> coordinates = new ArrayList<>();
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == type) {
                    coordinates.add(new Coordinate(i, j));
                }
            }
        }
        
        return coordinates;
    }
    
    private void setGrid(int[][] grid) {
        n = grid.length;
        m = grid[0].length;
        this.grid = grid;
    }
    
    private boolean inBound(Coordinate coor) {
        if (coor.x < 0 || coor.x >= n) {
            return false;
        }
        if (coor.y < 0 || coor.y >= m) {
            return false;
        }
        return grid[coor.x][coor.y] == EMPTY;
    }

    /**
     * @param grid a 2D grid
     * @return an integer
     */
    public int shortestDistance(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return -1;
        }
        
        // set n, m, grid
        setGrid(grid);
        
        List<Coordinate> houses = getCoordinates(HOUSE);
        int[][] distanceSum = new int[n][m];
        int[][] visitedTimes = new int[n][m];
        for (Coordinate house : houses) {
            bfs(house, distanceSum, visitedTimes);
        }
        
        int shortest = Integer.MAX_VALUE;
        List<Coordinate> empties = getCoordinates(EMPTY);
        for (Coordinate empty : empties) {
            if (visitedTimes[empty.x][empty.y] != houses.size()) {
                continue;
            }
            
            shortest = Math.min(shortest, distanceSum[empty.x][empty.y]);
        }
        
        if (shortest == Integer.MAX_VALUE) {
            return -1;
        }
        return shortest;
    }
    
    private void bfs(Coordinate start,
                     int[][] distanceSum,
                     int[][] visitedTimes) {
        Queue<Coordinate> queue = new LinkedList<>();
        boolean[][] hash = new boolean[n][m];
        
        queue.offer(start);
        hash[start.x][start.y] = true;
        
        int steps = 0;
        while (!queue.isEmpty()) {
            steps++;
            int size = queue.size();
            for (int temp = 0; temp < size; temp++) {
                Coordinate coor = queue.poll();
                for (int i = 0; i < 4; i++) {
                    Coordinate adj = new Coordinate(
                        coor.x + deltaX[i],
                        coor.y + deltaY [i]
                    );
                    if (!inBound(adj)) {
                        continue;
                    }
                    if (hash[adj.x][adj.y]) {
                        continue;
                    }
                    queue.offer(adj);
                    hash[adj.x][adj.y] = true;
                    distanceSum[adj.x][adj.y] += steps;
                    visitedTimes[adj.x][adj.y]++;
                } // direction
            } // for temp
        } // while
    }
}