_317.java

package com.fishercoder.solutions;

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

/**
 * 317. 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.
 */

public class _317 {
    public static class Solution1 {
        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;
        }
    }

}