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;
            }
        }
    }
}

4 thoughts on “LeetCode – Shortest Distance from All Buildings (Java)”

  1. if (grid[i][j] == 0 && reach[i][j] == numBuilding) {
    on this line, should grid[i][j] be compared with -1 and we have updated all 0s to -1s in helper method

Leave a Comment