Knight on Chess Board - Shortest Path

Question

I'm trying to solve this problem: https://www.interviewbit.com/problems/knight-on-chess-board/#

Basically, you're given a board, a start point and an end point and have to find the shortest path. I'm trying to do BFS on the the board using the 8 possible moves a knight can make and returning the number of moves it took, or -1 if there was no solution. I'm getting a run time out of memory error. I'm not sure where the error (or potential errors) are occurring.

Edit: Previously I was getting an error because I forgot got to mark nodes as visited. I've added that in but I'm still not getting the right answer.

public class Solution {
    
    private class Node {
        int row;
        int col;
        int count;
        public Node() {
            this.row = 0;
            this.col = 0;
            this.count = 0;
        }
        public Node(int row, int col, int count) {
            this.row = row;
            this.col = col;
            this.count = count;
        }
    }
    
    public int knight(int A, int B, int sr, int sc, int er, int ec) {
        int[][] matrix = new int[A][B];
        Queue<Node> q = new LinkedList<>(); //linkedlist??
        
        Node n = new Node(sr, sc, 0);
        q.add(n);
        matrix[sr][sc] = -1; 
        final int[][] SHIFTS = {
            {-2,1},
            {-2,-1},
            {2,1},
            {2,-1},
            {-1,2},
            {-1,-2},
            {1,2},
            {1,-2}
        };
        int count = 0;
        while(!q.isEmpty()) {
            Node cur = q.remove();
            if(cur.row == er && cur.col == ec) {
                return cur.count;
            }
            for(int[] i : SHIFTS) {  
                if(canTraverse(matrix, cur.row + i[0], cur.col + i[1])) {
                    matrix[cur.row + i[0]][cur.col + i[1]] = -1;
                    q.add(new Node(cur.row + i[0], cur.col + i[1], cur.count + 1));
                }
            }
        
        }
        return -1;
    }
    
    public static boolean canTraverse(int[][] matrix, int sr, int sc) {
        if(sr < 0 || sr >= matrix.length || sc < 0 || sc >= matrix[sr].length || matrix[sr][sc] == -1) {
            return false;
        }
        return true;
    }
}

I don't see where you're marking spaces as **visited**. – user3386109 Oct 04 '20 at 06:19 — user3386109, Oct 04 '20 at 06:19

score 0 · Answer 1 · answered Oct 04 '20 at 06:19

0

BFS algorithm needs to mark every visited position (node) to work properly. Else, such code could cause (almost certainly) runtime error or memory limit exceded (in short terms: A calls B and B calls A).

Solution: Create a boolean array and mark the nodes at the time they enter to the queue and you are done.

answered Oct 04 '20 at 06:19

Eloy Pérez Torres

1,050
4
14

Instead of a boolean array I changed the matrix value for a node that has been visited to -1 and made that a condition for when to not traverse the node. However, I'm not getting the right answer. – DisplayName Oct 04 '20 at 06:42
@DisplayName please, don't change the code as you are getting closer to the solution because that isn't helpful for others since they will have no context at all. To get the right answer you have to add an additional line "matrix[row][col] = new_count" everywhere you use "q.add". That includes matrix[sr][sc] = 0. That's because you need to successfully mark every visited node while expanding BFS tree. – Eloy Pérez Torres Oct 04 '20 at 06:51
Am I not making every node I visit -1 with the first line in the for loop? I don't understand where else to add that. Thanks for the tips on using stack overflow. – DisplayName Oct 04 '20 at 18:55
@DisplayName Yes, you initialize every node as -1 but you need to also update the calculated count for every node you encounter. Review BFS pseudocode at wikipedia https://en.wikipedia.org/wiki/Breadth-first_search to note that your code is not labeling discovered nodes. – Eloy Pérez Torres Oct 05 '20 at 03:07
Could you please tell me what line I need to change in the code to fix it. I've read over the wiki page and still am not sure. I thought I was updating the calculated count for every node by initializing the new Node value to cur.count + 1. Thanks again. – DisplayName Oct 14 '20 at 05:12

Knight on Chess Board - Shortest Path

1 Answers1