# 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.

### Solutions:

```java
public class Solution {
    public int shortestDistance(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return -1;
        }
        Queue<Integer> xs = new LinkedList<Integer>();
        Queue<Integer> ys = new LinkedList<Integer>();
        int[][] sums = new int[grid.length][grid[0].length];
        int[][] reach = new int[grid.length][grid[0].length];
        for (int i = 0; i < grid.length; i ++) {
            for (int j = 0; j < grid[0].length; j ++) {
                if (grid[i][j] == 1) {
                    xs.add(i);
                    ys.add(j);
                }
            }
        }
        int target = xs.size();
        int min = Integer.MAX_VALUE;
        boolean found = false;
        while (!xs.isEmpty()) {
            boolean[][] visited = new boolean[grid.length][grid[0].length];
            int x = xs.poll();
            int y = ys.poll();
            Queue<Integer> qx = new LinkedList<Integer>();
            Queue<Integer> qy = new LinkedList<Integer>();
            Queue<Integer> dis = new LinkedList<Integer>();
            qx.add(x);
            qy.add(y);
            dis.add(0);
            while (!qx.isEmpty()) {
                int i = qx.poll();
                int j = qy.poll();
                int d = dis.poll();
                if (i - 1 >= 0 && visited[i - 1][j] == false && grid[i - 1][j] == 0) {
                    qx.add(i - 1);
                    qy.add(j);
                    dis.add(d + 1);
                    visited[i - 1][j] = true;
                }
                if (i + 1 < grid.length && visited[i + 1][j] == false && grid[i + 1][j] == 0) {
                    qx.add(i + 1);
                    qy.add(j);
                    dis.add(d + 1);
                    visited[i + 1][j] = true;
                }
                if (j - 1 >= 0 && visited[i][j - 1] == false && grid[i][j - 1] == 0) {
                    qx.add(i);
                    qy.add(j - 1);
                    dis.add(d + 1);
                    visited[i][j - 1] = true;
                }
                if (j + 1 < grid[0].length && visited[i][j + 1] == false && grid[i][j + 1] == 0) {
                    qx.add(i);
                    qy.add(j + 1);
                    dis.add(d + 1);
                    visited[i][j + 1] = true;
                }
                if (d != 0) {
                    sums[i][j] += d;
                    reach[i][j] ++;
                    if (xs.isEmpty() && reach[i][j] == target) {
                        found = true;
                        min = Math.min(min, sums[i][j]);
                    }
                }
                
            }
           
           
        }
        if (found == false) {
            return -1;
        }
        return min;
    }
}
```

```java
public class Solution {
    public int shortestDistance(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return -1;
        }
        int result = Integer.MAX_VALUE, look = 0;
        int[][] sums = new int[grid.length][grid[0].length];
        int[][] dirs = new int[][]{{0,-1},{-1,0},{0,1},{1,0}};
        for (int i = 0; i < grid.length; i ++) {
            for (int j = 0; j < grid[0].length; j ++) {
                if (grid[i][j] == 1) {
                    result = Integer.MAX_VALUE;
                    Queue<Integer> xs = new LinkedList<Integer>();
                    Queue<Integer> ys = new LinkedList<Integer>();
                    Queue<Integer> dis = new LinkedList<Integer>();
                    xs.add(i);
                    ys.add(j);
                    dis.add(0);
                    while (!xs.isEmpty()) {
                        int x = xs.poll();
                        int y = ys.poll();
                        int d = dis.poll();
                        for (int k = 0; k < dirs.length; k ++) {
                            int nX = x + dirs[k][0];
                            int nY = y + dirs[k][1];
                            if (nX >= 0 && nX < grid.length && nY >= 0 && nY < grid[0].length && grid[nX][nY] == look) {
                                grid[nX][nY] --;
                                sums[nX][nY] += d + 1;
                                xs.add(nX);
                                ys.add(nY);
                                dis.add(d + 1);
                                result = Math.min(result, sums[nX][nY]);
                            }
                        }
                    }
                    look --;
                }
            }
        }
        
        
        return result == Integer.MAX_VALUE ? -1 : result;
    }
}
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://ugopireddy.gitbook.io/leet-code-solutions/solutions-301-350/317-shortest-distance-from-all-buildings.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.