LeetCode - Shortest Distance from All Buildings

ShortestDistancefromAllBuildings.java

// 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.

public int shortestDistance(int[][] grid) {
    if(grid == null || grid.length == 0 || grid[0].length == 0) return 0;
    int totalBuildings = countBuildings(grid);
    int minSteps = Integer.MAX_VALUE;
    for(int i = 0; i < grid.length; i++){
        for(int j = 0; j < grid[0].length; j++){
            if(grid[i][j] == 0){
                minSteps = Math.min(minSteps, bfs(grid, i, j, totalBuildings)); 
            }
        }
    }
    return minSteps == Integer.MAX_VALUE ? -1 : minSteps;
}

private int countBuildings(int[][] grid){
    int res = 0;
    for(int i = 0; i < grid.length; i++){
        for(int j = 0; j < grid[0].length; j++){
            if(grid[i][j] == 1) res++;
        }
    }
    return res;
}

private int bfs(int[][] grid, int i, int j, int totalBuildings){
    if(totalBuildings == 0) return Integer.MAX_VALUE;
    int m = grid.length, n = grid[0].length;
    int[][] visited = new int[m][n];
    Queue<Integer> queue = new LinkedList<Integer>();
    queue.add(i * n + j);
    int step = 0, foundBuildings = 0, totalSteps = 0, size = queue.size();
    while(!queue.isEmpty()){
        int val = queue.poll();
        size--;
        int x = val / n, y = val % n;
        if(visited[x][y] == 0){
            visited[x][y] = 1;
            if(grid[x][y] == 0){
                if(x - 1 >= 0) queue.offer((x-1) * n + y);
                if(y - 1 >= 0) queue.offer(x * n + y - 1);
                if(x + 1 < m) queue.offer((x+1) * n + y);
                if(y + 1 < n) queue.offer(x * n + y + 1);
                
            } else if(grid[x][y] == 1){
                foundBuildings++;
                totalSteps += step;
                if(foundBuildings == totalBuildings) return totalSteps;
            }
        }
        
        if(size == 0){
            step++;
            size = queue.size();
        }
    }
    return Integer.MAX_VALUE;
}