1368. Minimum Cost to Make at Least One Valid Path in a Grid

Given a 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 invalid signs on the cells of the grid which points outside the grid.

You will initially start at the upper left cell (0,0). A valid path in the grid is a path which 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 doesn 't 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

Example 4:

Input: grid = [[2,2,2],[2,2,2]]
Output: 3

Example 5:

Input: grid = [[4]]
Output: 0

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 100

# @lc code=start
using LeetCode

function min_cost1368(grid::Matrix)
    dx, dy = [1, -1, 0, 0], [0, 0, 1, -1]
    m, n = size(grid)
    dq = Deque{Tuple{Int,Int,Int}}()
    visited = fill(false, size(grid))
    pushfirst!(dq, (1, 1, 0))
    while !isempty(dq)
        x, y, w = popfirst!(dq)
        visited[x, y] && continue
        visited[x, y] = true
        (x == m && y == n) && return w
        for i in 1:4
            nx, ny = x + dx[i], y + dy[i]
            (1 <= nx <= m && 1 <= ny <= n) || continue
            grid[x, y] == i ? pushfirst!(dq, (nx, ny, w)) : push!(dq, (nx, ny, w + 1))
        end
    end
end
# @lc code=end

min_cost1368 (generic function with 1 method)

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