_317.java

package com.fishercoder.solutions;

import java.util.LinkedList;
import java.util.Queue;

/**
 * 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.
 */
public class _317 {

    public int shortestDistance(int[][] grid) {
        int m = grid.length;
        if(m == 0) return -1;
        int n = grid[0].length;
        int[][] reach = new int[m][n];
        int[][] distance = new int[m][n];
        int[] shift = new int[]{0, 1, 0, -1, 0};//how these five elements is ordered is important since it denotes the neighbor of the current node
        int numBuilding = 0;

        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 1){
                    numBuilding++;
                    int dist = 1;
                    boolean[][] visited = new boolean[m][n];

                    Queue<int[]> q = new LinkedList<int[]>();
                    q.offer(new int[]{i, j});
                    while(!q.isEmpty()){
                        int size = q.size();
                        for(int l = 0; l < size; l++){
                            int[] current = q.poll();
                            for(int k = 0; k < 4; k++){
                                int nextRow = current[0] + shift[k];
                                int nextCol = current[1] + shift[k+1];
                                if(nextRow >= 0 && nextRow < m && nextCol >= 0
                                        && nextCol < n && !visited[nextRow][nextCol] && grid[nextRow][nextCol] == 0){
                                    distance[nextRow][nextCol] += dist;
                                    visited[nextRow][nextCol] = true;
                                    reach[nextRow][nextCol]++;
                                    q.offer(new int[]{nextRow, nextCol});
                                }
                            }
                        }
                        dist++;
                    }
                }
            }
        }

        int result = 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] == numBuilding && distance[i][j] < result){
                    result = distance[i][j];
                }
            }
        }
        return result == Integer.MAX_VALUE ? -1 : result;
    }

}