LeetCode 1293. Shortest Path in a Grid with Obstacles Elimination 下一層坑

Given a m * n grid, where each cell is either 0 (empty) or 1 (obstacle). In one step, you can move up, down, left or right from and to an empty cell.

Return the minimum number of steps to walk from the upper left corner (0, 0) to the lower right corner (m-1, n-1) given that you can eliminate at most k obstacles. If it is not possible to find such walk return -1.

 

Example 1:

Input: 
grid = 
[[0,0,0],
 [1,1,0],
 [0,0,0],
 [0,1,1],
 [0,0,0]], 
k = 1
Output: 6
Explanation: 
The shortest path without eliminating any obstacle is 10. 
The shortest path with one obstacle elimination at position (3,2) is 6. Such path is (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (3,2) -> (4,2).

 

Example 2:

Input: 
grid = 
[[0,1,1],
 [1,1,1],
 [1,0,0]], 
k = 1
Output: -1
Explanation: 
We need to eliminate at least two obstacles to find such a walk.

 

Constraints:

• grid.length == m
• grid[0].length == n
• 1 <= m, n <= 40
• 1 <= k <= m*n
• grid[i][j] == 0 or 1
• grid[0][0] == grid[m-1][n-1] == 0

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BFS的時候下一輪條件剛開始沒有想明白，寫了個死循環的bug，題目並不難：

from collections import defaultdict

class Solution:
    def shortestPath(self, grid, k: int) -> int:
        m,n,step = len(grid),len(grid[0]),0
        layers,dic = [[(0,0,0)],[]],defaultdict(lambda:1600)
        dic[(0,0)] = 0
        cur,nxt = 0,1
        if (m == 1 and n == 1):
            return 0
        while (layers[cur] and step <= m*n):
            step += 1
            for x,y,ob in layers[cur]:
                for dx,dy in [[0,1],[0,-1],[-1,0],[1,0]]:
                    nx,ny = x+dx,y+dy
                    if (nx>=0 and nx<m and ny>=0 and ny<n):
                        nob = (ob+1) if grid[nx][ny] == 1 else ob
                        #bug1: if (nob<=k and (((nx,ny) not in dic) or dic[(nx,ny)]>=nob)): 而且只有((nx,ny) not in dic)才刷dic
                        if (nob<=k and dic[(nx,ny)]>nob):
                            if (nx == m-1 and ny == n-1):
                                return step
                            layers[nxt].append((nx,ny,nob))
                            dic[(nx,ny)] = nob
            layers[cur].clear()
            cur,nxt = nxt,cur
        return -1