leetcode - 1293. Shortest Path in a Grid with Obstacles Elimination-Toy模板网

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Description

You are given an m x n integer matrix grid where each cell is either 0 (empty) or 1 (obstacle). You can move up, down, left, or right from and to an empty cell in one step.

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:
leetcode - 1293. Shortest Path in a Grid with Obstacles Elimination,OJ题目记录,leetcode,算法,职场和发展

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:
leetcode - 1293. Shortest Path in a Grid with Obstacles Elimination,OJ题目记录,leetcode,算法,职场和发展

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:

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

Solution

BFS, but use the remaining time for removing rocks as the visited label. It’s always better if we reach the same point, and the remaining chance to remove rocks is larger.

Time complexity: $o (m * n)$
Space complexity: $o (m * n)$ 文章来源地址https://www.toymoban.com/news/detail-719526.html

Code

class Solution:
    def shortestPath(self, grid: List[List[int]], k: int) -> int:
        queue = collections.deque([(0, 0, k, 0)])
        visited = {}
        m, n = len(grid), len(grid[0])
        while queue:
            x, y, r, step = queue.popleft()
            if visited.get((x, y), -1) >= r:
                continue
            if x == m - 1 and y == n - 1:
                return step
            visited[(x, y)] = r
            for dz in (-1, 1):
                if 0 <= x + dz < m:
                    if grid[x + dz][y] == 0:
                        queue.append((x + dz, y, r, step + 1))
                    elif r - 1 >= 0:
                        queue.append((x + dz, y, r - 1, step + 1))
                if 0 <= y + dz < n:
                    if grid[x][y + dz] == 0:
                        queue.append((x, y + dz, r, step + 1))
                    elif r - 1 >= 0:
                        queue.append((x, y + dz, r - 1, step + 1))
        return -1