Shortest Distance from All Buildings in C++

Suppose we want to make a house on an empty land which reaches all buildings in the shortest amount of distance. We can only move four directions like up, down, left and right. We have a 2D grid of values 0, 1 or 2, where −

0 represents an empty land which we can pass by freely.
1 represents a building which we cannot pass through.
2 represents an obstacle which we cannot pass through.

So, if the input is like

1	2	1
0	0	0
0	1	0

then the output will be 7 as Given three buildings are present at (0,0), (0,4), (2,2), and an obstacle is at (0,2) so the point (1,2) is an ideal empty land to build a house, as the total travel distance is 3+3+1=7 is minimum.

To solve this, we will follow these steps −

ret := inf
n := row size of grid
m := col size of grid
numberOfOnes := 0
Define one 2D array dist of size n x m
Define one 2D array reach of size n x m
for initialize i := 0, when i < n, update (increase i by 1), do −
- for initialize j := 0, when j < m, update (increase j by 1), do −
  - if grid[i, j] is same as 1, then −
    - (increase numberOfOnes by 1)
    - Define one queue q
    - insert {i, j} into q
    - Define one set visited
    - for initialize lvl := 1, when not q is empty, update (increase lvl by 1), do −
      - sz := size of q
      - while sz is non-zero, decrease sz in each iteration, do −
        curr := first element of q
        delete element from q
        x := curr.first
        y := curr.second
        for initialize k := 0, when k < 4, update (increase k by 1), do −
        nx := x + dir[k, 0]
        ny := y + dir[k, 1]
        if nx and ny are in range of grid or grid[nx,ny] is not 0, then
        Ignore following part, skip to the next iteration
        insert {nx, ny} into visited
        dist[nx, ny] := dist[nx, ny] + lvl
        (increase reach[nx, ny] by 1)
        insert {nx, ny} into q
for initialize i := 0, when i < n, update (increase i by 1), do −
- for initialize j := 0, when j < m, update (increase j by 1), do −
  - if grid[i, j] is same as 0 and reach[i, j] is same as numberOfOnes, then −
    - ret := minimum of ret and dist[i, j]
return (if ret is same as inf, then -1, otherwise ret)

Example

Let us see the following implementation to get better understanding −

int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
class Solution {
public:
   int shortestDistance(vector<vector<int>>& grid) {
      int ret = INT_MAX;
      int n = grid.size();
      int m = grid[0].size();
      int numberOfOnes = 0;
      vector < vector <int> > dist(n, vector <int>(m));
      vector < vector <int> > reach(n, vector <int>(m));
      for(int i = 0; i < n; i++){
         for(int j = 0; j < m; j++){
            if(grid[i][j] == 1){
               numberOfOnes++;
               queue < pair <int, int> > q;
               q.push({i, j});
               set < pair <int, int> > visited;
               for(int lvl = 1; !q.empty(); lvl++){
                  int sz = q.size();
                  while(sz--){
                     pair <int, int> curr = q.front();
                     q.pop();
                     int x = curr.first;
                     int y = curr.second;
                     for(int k = 0; k < 4; k++){
                        int nx = x + dir[k][0];
                        int ny = y + dir[k][1];
                        if(nx < 0 || ny < 0 || nx >= n || ny >= m || visited.count({nx, ny}) || grid[nx][ny] != 0) continue;
                        visited.insert({nx, ny});
                        dist[nx][ny] += lvl;
                        reach[nx][ny]++;
                        q.push({nx, ny});
                     }
                  }
               }
            }
         }
      }
      for(int i = 0; i < n; i++){
         for(int j = 0; j < m; j++){
            if(grid[i][j] == 0 && reach[i][j] == numberOfOnes){
               ret = min(ret, dist[i][j]);
            }
         }
      }
      return ret == INT_MAX ? -1 : ret;
   }
};