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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 | 0 | 2 | 0 | 1 |
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 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;
}
};Input
[[1,0,2,0,1],[0,0,0,0,0],[0,0,1,0,0]]
Output
7
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