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Number of blocks in a chessboard a knight can move to in exactly k moves

  • Last Updated : 09 Feb, 2021

Given integers i, j, k and n where (i, j) is the initial position of the Knight on a n * n chessboard, the task is to find the number of positions the Knight can move to in exactly k moves.
Examples: 
 

Input: i = 5, j = 5, k = 1, n = 10 
Output: 8
Input: i = 0, j = 0, k = 2, n = 10 
Output: 10 
The knight can see total 10 different positions in 2nd move. 
 

 

Approach: Use a recursive approach to solve the problem. 
First find all the possible positions where the knight can move to so if the initial position is i, j. Get to all valid locations in single move and recursively find all the possible positions where knight can move to in k – 1 steps from there. The base case of this recursion is when k == 0 (no move to make) then we will mark the position of the chessboard as visited if it is unmarked and increase the count. Finally, display the count .
Below is the implementation of the above approach:
 

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// function that will be called recursively
int recursive_solve(int i, int j, int steps, int n,
                      map<pair<int, int>, int> &m)
{
    // If there's no more move to make and
    // this position hasn't been visited before
    if (steps == 0 && m[make_pair(i, j)] == 0) {
 
        // mark the position
        m[make_pair(i, j)] = 1;
 
        // increase the count       
        return 1;
    }
     
    int res = 0;
    if (steps > 0) {
 
        // valid movements for the knight
        int dx[] = { -2, -1, 1, 2, -2, -1, 1, 2 };
        int dy[] = { -1, -2, -2, -1, 1, 2, 2, 1 };
 
        // find all the possible positions
        // where knight can move from i, j
        for (int k = 0; k < 8; k++) {
 
            // if the positions lies within the
            // chessboard
            if ((dx[k] + i) >= 0
                && (dx[k] + i) <= n - 1
                && (dy[k] + j) >= 0
                && (dy[k] + j) <= n - 1) {
 
                // call the function with k-1 moves left
                res += recursive_solve(dx[k] + i, dy[k] + j,
                                       steps - 1, n, m);
            }
        }
    }
    return res;
}
 
// find all the positions where the knight can
// move after k steps
int solve(int i, int j, int steps, int n)
{
    map<pair<int, int>, int> m;
    return recursive_solve(i, j, steps, n, m);
}
 
// driver code
int main()
{
    int i = 0, j = 0, k = 2, n = 10;
 
    cout << solve(i, j, k, n);
 
    return 0;
}

Java




// Java implementation of above approach
import java.util.*;
import java.awt.Point;
public class GFG
{
 
  // function that will be called recursively
  static int recursive_solve(int i, int j, int steps, int n,
                             HashMap<Point,Integer> m)
  {
 
    // If there's no more move to make and
    // this position hasn't been visited before
    if (steps == 0 && !m.containsKey(new Point(i, j)))
    {
 
      // mark the position
      m.put(new Point(i, j), 1);
 
      // increase the count       
      return 1;
    }
    int res = 0;
    if (steps > 0)
    {
 
      // valid movements for the knight
      int[] dx = { -2, -1, 1, 2, -2, -1, 1, 2 };
      int[] dy = { -1, -2, -2, -1, 1, 2, 2, 1 };
 
      // find all the possible positions
      // where knight can move from i, j
      for (int k = 0; k < 8; k++)
      {
 
        // if the positions lies within the
        // chessboard
        if ((dx[k] + i) >= 0
            && (dx[k] + i) <= n - 1
            && (dy[k] + j) >= 0
            && (dy[k] + j) <= n - 1)
        {
 
          // call the function with k-1 moves left
          res += recursive_solve(dx[k] + i, dy[k] + j,
                                 steps - 1, n, m);
        }
      }
    }
    return res;
  }
 
  // find all the positions where the knight can
  // move after k steps
  static int solve(int i, int j, int steps, int n)
  {
    HashMap<Point, Integer> m = new HashMap<>();
    return recursive_solve(i, j, steps, n, m);
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int i = 0, j = 0, k = 2, n = 10;
    System.out.print(solve(i, j, k, n));
  }
}
 
 
// This code is contributed by divyeshrabadiya07.

Python3




# Python3 implementation of above approach
from collections import defaultdict
 
# Function that will be called recursively
def recursive_solve(i, j, steps, n, m):
 
    # If there's no more move to make and
    # this position hasn't been visited before
    if steps == 0 and m[(i, j)] == 0:
 
        # mark the position
        m[(i, j)] = 1
 
        # increase the count        
        return 1
     
    res = 0
    if steps > 0:
 
        # valid movements for the knight
        dx = [-2, -1, 1, 2, -2, -1, 1, 2]
        dy = [-1, -2, -2, -1, 1, 2, 2, 1]
 
        # find all the possible positions
        # where knight can move from i, j
        for k in range(0, 8):
 
            # If the positions lies
            # within the chessboard
            if (dx[k] + i >= 0 and
                dx[k] + i <= n - 1 and
                dy[k] + j >= 0 and
                dy[k] + j <= n - 1):
 
                # call the function with k-1 moves left
                res += recursive_solve(dx[k] + i, dy[k] + j,
                                       steps - 1, n, m)
     
    return res
 
# Find all the positions where the
# knight can move after k steps
def solve(i, j, steps, n):
 
    m = defaultdict(lambda:0)
    return recursive_solve(i, j, steps, n, m)
 
# Driver code
if __name__ == "__main__":
 
    i, j, k, n = 0, 0, 2, 10
     
    print(solve(i, j, k, n))
 
# This code is contributed by Rituraj Jain

C#




// C# implementation of above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG
{
 
// function that will be called recursively
static int recursive_solve(int i, int j, int steps, int n,
                      Dictionary<Tuple<int,int>,int>m)
{
    // If there's no more move to make and
    // this position hasn't been visited before
    if (steps == 0 && !m.ContainsKey(new Tuple<int,int>(i, j))) {
 
        // mark the position
        m[new Tuple<int,int>(i, j)] = 1;
 
        // increase the count       
        return 1;
    }
     
    int res = 0;
 
    if (steps > 0) {
 
        // valid movements for the knight
        int []dx = { -2, -1, 1, 2, -2, -1, 1, 2 };
        int []dy = { -1, -2, -2, -1, 1, 2, 2, 1 };
 
        // find all the possible positions
        // where knight can move from i, j
        for (int k = 0; k < 8; k++) {
 
            // if the positions lies within the
            // chessboard
            if ((dx[k] + i) >= 0
                && (dx[k] + i) <= n - 1
                && (dy[k] + j) >= 0
                && (dy[k] + j) <= n - 1) {
 
                // call the function with k-1 moves left
                res += recursive_solve(dx[k] + i, dy[k] + j,
                                       steps - 1, n, m);
            }
        }
    }
    return res;
}
 
// find all the positions where the knight can
// move after k steps
static int solve(int i, int j, int steps, int n)
{
    Dictionary<Tuple<int,int>,int> m=new Dictionary<Tuple<int,int>,int>();
    return recursive_solve(i, j, steps, n, m);
}
 
// driver code
public static void Main(params string []args)
{
    int i = 0, j = 0, k = 2, n = 10;
 
    Console.Write(solve(i, j, k, n));
 
}
}
Output: 
10

 




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