This is our solution and implementation to problem #1368: Minimum Cost to Make at Least One Valid Path in a Grid on LeetCode.
Our code is written in Java. If you want to code your solution in a different language, no worries, as the core concepts will carry over from language to language. This page includes our analysis of the problem, our implementation strategy a breakdown on the space and time complexity, and the full code implementation.
If you would like to view the original problem and solve it, please visit: Minimum Cost to Make at Least One Valid Path in a Grid on LeetCode.
Problem Analysis
The problem asks us to return the minimum cost to make the grid have at least one valid path. A valid path is a path from the top left to the bottom right cell where the arrows (as described in the problem) lead you from the start to the end. Notice that the problem doesn't ask which cells to change, only the count. Therefore, we don't actually need to determine which cells would need to change, and this simplifies the problem.
To figure out how many cells we need to change, we need to:
- Figure out which cells are in the initial path starting from the first position.
- Find all cells neighboring that are the cells in that path but aren't themselves in that path.
- For each of those cells, traverse its path and repeat. Each time we "jump" from a path to a neighboring cell's path, we increment the cost it would take to get to that cell, since a "jump" is the same as changing the arrow direction of a cell.
Implementation Strategy
The code will map cells [x,y]
to position x*width+y
so we can use one-dimensional arrays to store information. We'll keep track of cells that were visited using a boolean array cellIsVisited. First, we'll traverse the path starting from the initial position 0. The path traversal algorithm will be defined in its own method so we can re-use it. It will:
- Create a list to indicate cells that are a part of this path called visitedCells.
- Add the cursor, or current cell, to the visitedCells and set cellIsVisited[cursor] to true.
- If the cursor is already at the end cell, return a list of just that cell.
- While the cursor is in a valid position:
- Update the cursor based on the direction of the arrow in the position in the grid.
- If the position is out of bounds or points to a cell already visited, return the current list of visitedCells.
- Add the cursor, or current cell, to the visitedCells and set cellIsVisited[cursor] to true.
- If the cursor is already at the end cell, return a list of just that cell.
Once the first path traversal is done, return 0 if the path already reached the target cell. Otherwise, use a variable currentCost to represent the cost it would take to get to the end cell (so far). Then, for each cell in visitedCells:
- Make a new list newVisitedCells.
- For the current cell, calculate the cell to the left, right, up, and down. For each, if they're valid and haven't been visited, call traversePath with that cell and add all responses to newVisitedCells or return currentCost if we've hit the end cell.
- Once all cells are visited in visitedCells, increment currentCost and set visitedCells to newVisitedCells and repeat.
Space and Time Complexity
This has a space complexity of O(n * m) where n and m are the height and width of the grid. Each cell is visited at most twice: once when traversing its path for the first time, and once when iterating over visitedCells. The space complexity is O(n * m) since we have a boolean array of size n * m, and in the worst case, visitedCells will have almost every cell (i.e. if a path almost gets to the target cell but not quite).
Additional Resources
The links below outline some of the points discussed above in more detail.
Full Code
Our solution is given in the Java code below:
All of our solutions are hosted on GitHub. The code on this page was pulled from the repo.