[LintCode] Knight Shortest Path

 Given a knight in a chessboard (a binary matrix with 0 as empty and 1 as barrier) with a sourceposition, find the shortest path to a destination position, return the length of the route. 

Return -1 if knight can not reached.

source and destination must be empty.
Knight can not enter the barrier.

 

Clarification

If the knight is at (x, y), he can get to the following positions in one step:

(x + 1, y + 2)
(x + 1, y - 2)
(x - 1, y + 2)
(x - 1, y - 2)
(x + 2, y + 1)
(x + 2, y - 1)
(x - 2, y + 1)
(x - 2, y - 1)

Example

[[0,0,0],
 [0,0,0],
 [0,0,0]]
source = [2, 0] destination = [2, 2] return 2

[[0,1,0],
 [0,0,0],
 [0,0,0]]
source = [2, 0] destination = [2, 2] return 6

[[0,1,0],
 [0,0,1],
 [0,0,0]]
source = [2, 0] destination = [2, 2] return -1

 

Solution. BFS, O(m * n) runtime, O(m * n) space 

Instead of using an extra 2D boolean array to track if a cell has been visited or not, this solution does this tracking by marking visited cells as barriers. 

The upside of doing this is to save some memory usage. The downside is the original input will be changed. This will be a problem if the shortestPath method needs to be called more than once.

 

/**
 * Definition for a point.
 * public class Point {
 *     publoc int x, y;
 *     public Point() { x = 0; y = 0; }
 *     public Point(int a, int b) { x = a; y = b; }
 * }
 */
public class Solution {
    /**
     * @param grid a chessboard included 0 (false) and 1 (true)
     * @param source, destination a point
     * @return the shortest path 
     */
    protected int[] deltaX = {1, 1, -1, -1, 2, 2, -2, -2};
    protected int[] deltaY = {2, -2, 2, -2, 1, -1, 1, -1};
    public int shortestPath(boolean[][] grid, Point source, Point destination) {
        if(grid == null || grid.length == 0 || grid[0].length == 0)
        {
            return -1;
        }
        int rowLen = grid.length;
        int colLen = grid[0].length;
        
        Queue<Point> queue = new LinkedList<Point>();
        queue.offer(source);
        //mark visited cell as barrier
        grid[source.x][source.y] = true;
        int steps = 0;
        
        while(queue.isEmpty() == false)
        {
            int size = queue.size();
            for(int i = 0; i < size; i++)
            {
                Point curr = queue.poll();
                if(curr.x == destination.x && curr.y == destination.y)
                {
                    return steps;
                }
                //make all 8 possible moves
                for(int direction = 0; direction < 8; direction++)
                {
                    //check if the current direction is valid and has not been visited
                    if(isValidDirection(grid, curr.x + deltaX[direction], curr.y + deltaY[direction]))
                    {
                        queue.offer(new Point(curr.x + deltaX[direction], curr.y + deltaY[direction]));
                        //mark visited cell as barrier
                        grid[curr.x + deltaX[direction]][curr.y + deltaY[direction]] = true;
                    }
                }
            }
            steps++;
        }
        return -1;
    }
    private boolean isValidDirection(boolean[][] grid, int row, int col)
    {
        int rowLen = grid.length;
        int colLen = grid[0].length;
        //check if in boundary
        if(row < 0 || row >= rowLen || col < 0 || col >= colLen)
        {
            return false;
        }
        //check if barrier
        return !grid[row][col];
    }
}

 

 

 

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