#1368

Minimum Cost to Make at Least One Valid Path in a Grid

hard · verified · 70.9% accepted · 2,593 likes · top 80%

array · breadth-first search · graph theory · heap (priority queue) · matrix · shortest path

Description

Given an m x n grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of grid[i][j] can be:

- 1 which means go to the cell to the right. (i.e go from grid[i][j] to grid[i][j + 1])

- 2 which means go to the cell to the left. (i.e go from grid[i][j] to grid[i][j - 1])

- 3 which means go to the lower cell. (i.e go from grid[i][j] to grid[i + 1][j])

- 4 which means go to the upper cell. (i.e go from grid[i][j] to grid[i - 1][j])

Notice that there could be some signs on the cells of the grid that point outside the grid.

You will initially start at the upper left cell (0, 0). A valid path in the grid is a path that starts from the upper left cell (0, 0) and ends at the bottom-right cell (m - 1, n - 1) following the signs on the grid. The valid path does not have to be the shortest.

You can modify the sign on a cell with cost = 1. You can modify the sign on a cell one time only.

Return the minimum cost to make the grid have at least one valid path.

Example 1:

Input: grid = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]]

Output: 3

Explanation: You will start at point (0, 0).

The path to (3, 3) is as follows. (0, 0) --> (0, 1) --> (0, 2) --> (0, 3) change the arrow to down with cost = 1 --> (1, 3) --> (1, 2) --> (1, 1) --> (1, 0) change the arrow to down with cost = 1 --> (2, 0) --> (2, 1) --> (2, 2) --> (2, 3) change the arrow to down with cost = 1 --> (3, 3)

The total cost = 3.

Example 2:

Input: grid = [[1,1,3],[3,2,2],[1,1,4]]

Output: 0

Explanation: You can follow the path from (0, 0) to (2, 2).

Example 3:

Input: grid = [[1,2],[4,3]]

Output: 1

Solution