Adds DFS, BFS and Dijkstra's algorithm solutions for "The Maze II" #283
Conversation
|
Issue: #56 |
altay9
changed the title
As we use PQ to find the closest node, this method is obsolete now.
|
Hello @ErdemT09, I have done some improvements in Dijkstra implementation.
|
|
|
||
| if(maze[x][y]==-1) continue; | ||
|
|
||
| maze[x][y] = -1; |
There was a problem hiding this comment.
I reccommend moving -1 to a constant called "VISITED" for readability.
There was a problem hiding this comment.
I still suggest doing this.
There was a problem hiding this comment.
Oke, it is done. Thanks for the suggestion.
| if (startPoint[0] < 0) | ||
| break; |
There was a problem hiding this comment.
I don't understand the function of this statement here. Can you please explain that?
There was a problem hiding this comment.
It is also a resideue. I just removed it.
| return (nextX >= 0 && nextY >= 0 | ||
| && nextX < maze.length | ||
| && nextY < maze[0].length | ||
| && maze[nextX][nextY] != 1); |
There was a problem hiding this comment.
Why don't we check for this condition unlike in the previous two solutions?
maze[nextX][nextY] == 0
There was a problem hiding this comment.
I have understood the reason why we don't do this.
In the previous solution, the tiles that we come across are either empty or a wall. If empty, we move through them. If wall, we stop. Here, we have a three valued logic: A tile is either empty but visited, empty or a wall. We should be able to move through an empty tile even if it is visited. At the end, we can't stop and branch off there, but whilst moving to another valid tile, we should be able to pass them.
After all, we start the canPass loop from an already visited tile. This means we have to consider -1 as a valid passable tile value.
There was a problem hiding this comment.
Thanks for the explanation Erdem. Appreciated.
|
I have accidentally deleted the branch. I should have been more careful. |
ErdemT09
left a comment
ErdemT09
left a comment
There was a problem hiding this comment.
Approved. A good question in terms of graph theory.
resolves #56
This question is a version of #282, but this time we are asked to find the shortest path to the target.
Dijkstra's algorithm appears to be the fastest with ~5ms and the way it searches is almost the same as the BFS.
But Dijkstra starts with detecting the node which has the least distance from the starting point.
Priority Queueis a perfect match to poll the closest node.