Snake and Ladder Problem

Posted 2020-08-16 鱼与海洋

Given a snake and ladder board, find the minimum number of dice throws required to reach the destination or last cell from source or 1st cell. Basically, the player has total control over outcome of dice throw and wants to find out minimum number of throws required to reach last cell.

If the player reaches a cell which is base of a ladder, the player has to climb up that ladder and if reaches a cell is mouth of the snake, has to go down to the tail of snake without a dice throw.

O(n) BFS

/**
 * Given a snake and ladder board, find the minimum number of dice throws required to reach the destination or last cell from source or 1st cell. Basically, the player has total control over outcome of dice throw and wants to find out minimum number of throws required to reach last cell.
 * If the player reaches a cell which is base of a ladder, the player has to climb up that ladder and if reaches a cell is mouth of the snake, has to go down to the tail of snake without a dice throw.
 * @author
 */
import java.util.*;
class Grid{
    int index;
    int distance;
    public Grid(int index, int distance){
        this.index = index;
        this.distance = distance;
    }
}
public class SnakeAndLadderProblem {

    public static int miniStep(int[] board){
        int min = 0;
        boolean visited[] = new boolean[board.length];
        Queue<Grid> queue = new LinkedList<Grid>();
        Grid root = new Grid(0,0);
        visited[0] = true;
        queue.offer(root);
        while(!queue.isEmpty()){
            Grid top = queue.poll();
            if(top.index == board.length -1){
                min = top.distance;
                break;
            }
            for(int i = top.index + 1; i <= top.index + 6 && i < board.length; i++){
                if(visited[i] == false){
                    Grid newGrid  = null;
                    visited[i] = true;                 
                    if(board[i] !=-1)
                         newGrid = new Grid(board[i], top.distance + 1);
                    else
                         newGrid = new Grid(i, top.distance + 1);
                    queue.add(newGrid);
                }
            }    
        }
        return min;
    }
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int board[] = new int[30];
        for(int i = 0; i < 30; i++){
            board[i] = -1;
        }
        // Ladders
                board[2] = 21;
                board[4] = 7;
                board[10] = 25;
                board[19] = 28;

        // Snakes
                board[26] = 0;
                board[20] = 8;
                board[16] = 3;
                board[18] = 6;
       System.out.println("min step is "+ miniStep(board));
    }

}