[LeetCode] Shortest Distance from All Buildings
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):
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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.
Java solution:
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public class Solution { public int shortestDistance(int[][] grid) { int row = grid.length; if (row == 0) { return -1; } int col = grid[0].length; int[][] record1 = new int[row][col]; // visited num int[][] record2 = new int[row][col]; // distance int num1 = 0; for (int r = 0; r < row; r++) { for (int c = 0; c < col; c++) { if (grid[r][c] == 1) { num1 ++; boolean[][] visited = new boolean[row][col]; Queue<int[]> queue = new LinkedList<int[]>(); queue.offer(new int[]{r, c}); int dist = 0; while (!queue.isEmpty()) { int size = queue.size(); for (int i = 0; i < size; i++) { int[] node = queue.poll(); int x = node[0]; int y = node[1]; record2[x][y] += dist; record1[x][y] ++; if (x > 0 && grid[x - 1][y] == 0 && !visited[x - 1][y]) { queue.offer(new int[]{x - 1, y}); visited[x - 1][y] = true; } if (x + 1 < row && grid[x + 1][y] == 0 && !visited[x + 1][y]) { queue.offer(new int[]{x + 1, y}); visited[x + 1][y] = true; } if (y > 0 && grid[x][y - 1] == 0 && !visited[x][y - 1]) { queue.offer(new int[]{x, y - 1}); visited[x][y - 1] = true; } if (y + 1 < col && grid[x][y + 1] == 0 && !visited[x][y + 1]) { queue.offer(new int[]{x, y + 1}); visited[x][y + 1] = true; } } dist ++; } } } } int result = Integer.MAX_VALUE; for (int r = 0; r < row; r++) { for (int c = 0; c < col; c++) { if (grid[r][c] == 0 && record1[r][c] == num1 && record2[r][c] < result) { result = record2[r][c]; } } } return result == Integer.MAX_VALUE ? -1 : result; } |
Time complexity:
The time complexity for BFS/DFS is O(|V|+|E|), In this problem, every vertex has up to 4 edges (left, right, up, down), so |E| ~ 4|V|. Thus, you have overall O(|V|) = O(mn) for a BFS. This has been proven for all sparse graphs like this problem. Now, we do a BFS for each building, so the overall complexity is O(#buildings*(mn)). In worst case, every vertex is a building. So the number of buildings is also upper bounded by O(mn), and thus you have O((mn)(mn)) = O(m^2n^2). This is a very loose bound since when every vertex is a building, we don’t even need to do a BFS (nowhere to go).
https://discuss.leetcode.com/topic/31925/java-solution-with-explanation-and-time-complexity-analysis/5
