Problem
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.
Example:
Input: [[1,0,2,0,1],[0,0,0,0,0],[0,0,1,0,0]]
1 - 0 - 2 - 0 - 1
| | | | |
0 - 0 - 0 - 0 - 0
| | | | |
0 - 0 - 1 - 0 - 0
Output: 7
Explanation: Given three buildings at (0,0), (0,4), (2,2), and an obstacle at (0,2),
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.
Solution
class Solution {
public int shortestDistance(int[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) return -1;
int m = grid.length, n= grid[0].length;
int count = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 1) count++;
}
}
int[][] dist = new int[m][n];
int[][] conn = new int[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 1) {
if (!bfs(grid, i, j, count, dist, conn)) return -1;
}
}
}
int min = Integer.MAX_VALUE;
boolean found = false;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 0) {
if (conn[i][j] == count) {
found = true;
min = Math.min(min, dist[i][j]);
}
}
}
}
return found ? min : -1;
}
class Point {
int x;
int y;
int distance;
Point(int x, int y, int dist) {
this.x = x;
this.y = y;
this.distance = dist;
}
}
private boolean bfs(int[][] grid, int xstart, int ystart, int count, int[][] dist, int[][] conn) {
boolean[][] spot = new boolean[grid.length][grid[0].length];
Deque<Point> queue = new ArrayDeque<>();
queue.offer(new Point(xstart, ystart, 0));
spot[xstart][ystart] = true;
int buildings = 1;
int[] dx = {-1, 1, 0, 0};
int[] dy = {0, 0, -1, 1};
while (!queue.isEmpty()) {
Point cur = queue.poll();
for (int i = 0; i < 4; i++) {
int x = cur.x + dx[i];
int y = cur.y + dy[i];
if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length) continue;
//connect to a building
if (grid[x][y] == 1 && !spot[x][y]) {
spot[x][y] = true;
buildings++;
}
//connect to a spot
if (grid[x][y] == 0 && !spot[x][y]) {
spot[x][y] = true;
queue.offer(new Point(x, y, cur.distance+1));
dist[x][y] += cur.distance+1;
conn[x][y]++;
}
}
}
return buildings == count;
}
}