LeetCode – Shortest Distance from All Buildings (Java)

Java Solution

This problem can be solved by BFS. We define one matrix for tracking the distance from each building, and another matrix for tracking the number of buildings which can be reached.

public int shortestDistance(int[][] grid) {
    int[][] distance = new int[grid.length][grid[0].length];
    int[][] reach = new int[grid.length][grid[0].length];
 
 
    int numBuilding = 0;
    for (int i = 0; i < grid.length; i++) {
        for (int j = 0; j < grid[0].length; j++) {
            if (grid[i][j] == 1) {
                helper(grid, distance, reach, i, j);
                numBuilding++;
            }
        }
    }
 
    int result = 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 && reach[i][j] == numBuilding) {
                result = Math.min(result, distance[i][j]);
            }
        }
    }
 
    return result == Integer.MAX_VALUE ? -1 : result;
}
 
private void helper(int[][] grid, int[][] distance, int[][] reach, int i, int j) {
 
    int[] dx = {-1, 0, 1, 0};
    int[] dy = {0, 1, 0, -1};
 
    //two queue, one for direction, one for distance tracking
    LinkedList<int[]> q = new LinkedList<>();
    LinkedList<Integer> qDist = new LinkedList<>();
 
    q.offer(new int[]{i, j});
    qDist.offer(1);
 
    while (!q.isEmpty()) {
        int[] head = q.poll();
        int dis = qDist.poll();
 
        for (int k = 0; k < 4; k++) {
            int x = head[0] + dx[k];
            int y = head[1] + dy[k];
 
            if (x >= 0 && y >= 0 && x < grid.length && y < grid[0].length && grid[x][y] == 0) {
                grid[x][y] = -1;
 
                q.offer(new int[]{x, y});
                qDist.offer(dis + 1);
 
                distance[x][y] += dis;
                reach[x][y]++;
            }
        }
    }
 
    for (int m = 0; m < grid.length; m++) {
        for (int n = 0; n < grid[0].length; n++) {
            if (grid[m][n] == -1) {
                grid[m][n] = 0;
            }
        }
    }
}
Category >> Algorithms  
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  • ryanlr

    corrected.

  • mousepotato

    why you put dfs function name here? Should be bfs right?