Leetcode: Shortest Distance from All Buildings

Question

1. 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]