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

Problem 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.

Key Insights

• The grid represents a directed graph where each cell points to another cell.
• A valid path must start at (0, 0) and end at (m - 1, n - 1).
• Modifying a cell's direction incurs a cost of 1.
• Use a graph traversal algorithm to explore paths and track costs.

Space and Time Complexity

Time Complexity: O(m * n log(m * n)), where m is the number of rows and n is the number of columns, due to the priority queue operations. Space Complexity: O(m * n), for storing the cost to reach each cell.

Solution

The problem can be solved using a modified Dijkstra's algorithm or a priority queue (min-heap) approach. The idea is to explore each cell in the grid and calculate the minimum cost to reach the target cell (m - 1, n - 1). We maintain a priority queue that stores the current cost and the cell coordinates. We also keep a 2D array to track the minimum cost to reach each cell.

1. Initialize a priority queue and start from the top-left corner (0, 0) with a cost of 0.
2. For each cell, check the direction indicated by the sign.
3. If the direction leads to a valid cell within the grid, add it to the queue with the same cost.
4. If the direction is invalid (points outside the grid), consider changing the direction (incurring a cost of 1) and enqueue the new cell.
5. Continue until you reach the bottom-right corner or exhaust all possibilities.

Code Solutions