## Shortest Distance from All Buildings

Question

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.

Solution

BFS

``````Class Solution{
public:
int shortestDistance(vector<vector<int>> grid) {
int rows = grid.size();
if (rows == 0) {
return -1;
}
int cols = grid.size();

// sum of distances to all the buildings
vector<vector<int>> dist(rows, vector<int>(cols, 0));

// number of reachable buildings
vector<vector<int>> nums(rows, vector<int>(cols, 0));
int buildingNum = 0;

// BFS from each building
for (int i = 0; i < rows; ++i) {
for (int j = 0; j < cols; ++j) {
if (grid[i][j] == 1) {
buildingNum++;
bfs(grid, i, j, dist, nums);
}
}
}

int minval = INT_MAX;
for (int i = 0; i < rows; ++i) {
for (int j = 0; j < cols; ++j) {
if (grid[i][j] == 0 && dist[i][j] != 0 && nums[i][j] == buildingNum) {
minval = min(minval, dist[i][j]);
}
}
}
if (minval < INT_MIN) {
return minval;
}
return -1;
}

private:
void bfs(vector<vector<int>>& grid, int row, int col, vector<vector<int>>& dist, vector<vector<int>>& nums) {
int rows = grid.size();
int cols = grid.size();
queue<pair<int, int>> q;
q.push(make_pair(row, col));

vector<vector<int>> dirs = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};

// record visited node
vector<vector<bool>> visited(rows, vector<bool>(cols, false));
int level = 0;
while(!q.empty()) {
level++;
int size = q.size();
for (int i = 0; i < size; ++i) {
pair<int, int> coords = q.front();
q.pop();
for (int k = 0; k < dirs.size(); ++k) {
int x = coords.first + dirs[k];
int y = coords.second + dirs[k];
if (x >= 0 && x < rows && y >= 0 && y < cols && !visited[x][y] && grid[x][y] == 0) {
visited[x][y] = true;
dist[x][y] += level;
nums[x][y]++;
q.push(make_pair(x, y));
}
}
}
}
}
};
``````