Leetcode: Shortest Distance from All Buildings
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Question
- Shortest Distance from All Buildings My Submissions Question
Total Accepted: 919 Total Submissions: 3251 Difficulty: Hard
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.
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Solution
Get idea from here.
class Solution(object): def shortestDistance(self, grid): """ :type grid: List[List[int]] :rtype: int """ m, n, self.nth = len(grid), len(grid[0]), -1 distance = [ [0]*n for dummy in xrange(m) ] for i,row in enumerate(grid): for j, num in enumerate(row): if num==1: if self.helper(i,j,grid,distance)==False: return -1 return min([ distance[i][j] for i, row in enumerate(grid) for j, num in enumerate(row) if num == (self.nth+1) ] or [-1] ) def helper(self, i, j, grid, distance): m, n = len(grid), len(grid[0]) queue, level, nth = collections.deque( [(i,j)] ), 1, self.nth count = 0 while queue: for dummy in xrange(len(queue)): i, j = queue.popleft() for x, y in [(0,1), (0,-1), (1,0), (-1,0)]: nexti, nextj = i+x, j+y if 0<= nexti <m and 0<=nextj<n and grid[nexti][nextj]==(nth+1): count += 1 queue.append( (nexti,nextj) ) distance[nexti][nextj] += level grid[nexti][nextj] = nth level += 1 self.nth -= 1 return True if count!=0 else False
Take-home message
- 1.
print [1,2,3] or [-1]=> [1,2,3]print [] or [-1]=> [-1]
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