1293. Shortest Path in a Grid with Obstacles Elimination

Shortest Path in a Grid with Obstacles Elimination - LeetCode

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

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).

Solution

Use bfs to explore from (0,0) with elimination k, which is state (0,0,k)

If the grid[x][y] is obstacle, eliminate it and push a new state to stack (x, y, k-1) if k>0

If it is a cell, push a new state to stack (x, y, k)

Maintain a visited set to record the visited sate (i, j, k)

If the x,y hits the destination (m-1, n-1), return depth.

class Solution:
    def shortestPath(self, grid: List[List[int]], k: int) -> int:
        directions = [(0,1),(-1,0),(1,0),(0,-1)]
        elimination = k
        m = len(grid)
        n = len(grid[0])
        q = [(0,0,elimination)]
        new_q = []
        visited = set()
        if(m+n-1<=k):
            return m+n-2
        depth = 1
        while(q):
            (i,j,k) = q.pop()
            for dx,dy in directions:
                x = i+dx
                y = j+dy
                if(x<0 or x>=m or y<0 or y>=n):
                    continue
                if(grid[x][y] == 1):
                    if(k>0 and (x,y,k-1) not in visited):
                        visited.add((x,y,k-1))
                        new_q.append((x,y,k-1))
                else:
                    if((x,y,k) not in visited):
                        if(x==m-1 and y == n-1):
                            return depth
                        visited.add((x,y,k))
                        new_q.append((x,y,k))
            if(not q):
                depth += 1
                q = new_q
                new_q = []

        return -1