Count all possible visited cells of a knight after N moves

Last Updated : 08 Mar, 2023

Given the current position of a Knight as (i, j), find the count of different possible positions visited by a knight after N moves (in a 10 x 10 board).

Examples:

Input: i = 3, j = 3, n = 1
Output: 9
The Knight is initially at position [3][3]. After one move it can visit 8 more cells

Input: i = 3, j = 3, n = 2
Output: 35

Approach: The idea is simple, we start from a given position, try all possible moves. After every move, recursively call for n-1 moves. We need to ensure that we never visit a cell again. We make a visited boolean matrix which will serve as a visited matrix so that the positions do not get repeated. When we visit a position, mark that position as true in the matrix.

Steps:-

Take a boolean visited matrix (10X10) and initialize all the cells as false (non-visited)
Create two vectors with all possible moves of a knight. We find that there are 8 possible moves of a knight.
Valid position = The knight is inside the boundaries of the board and the cell is non-visited.
Call the method for the next valid position with n = n-1.
If n == 0, return.

Below is the implementation of the above approach:

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
const int N = 10;
 
// All possible moves of the knight.
 
// In X axis.
vector<int> X = { 2, 1, -1, -2, -2, -1, 1, 2 };
 
// In Y axis.
vector<int> Y = { 1, 2, 2, 1, -1, -2, -2, -1 };
 
void getCountRec(vector<vector<bool> >& board,
                 int i, int j, int n)
{
    // if n=0, we have our result.
    if (n == 0)
        return;
 
    for (int k = 0; k < 8; k++) {
        int p = i + X[k];
        int q = j + Y[k];
 
        // Condition for valid cells.
        if (p >= 0 && q >= 0
            && p < 10 && q < N) {
            board[p][q] = true;
            getCountRec(board, p, q, n - 1);
        }
    }
}
 
int getCount(int i, int j, int n)
{
    vector<vector<bool> > board(N, vector<bool>(N));
    board[i][j] = true;
 
    // Call the recursive function to mark
    // visited cells.
    getCountRec(board, i, j, n);
 
    int cnt = 0;
    for (auto row : board) {
        for (auto cell : row) {
            if (cell)
                cnt++;
        }
    }
    return cnt;
}
 
// Driver Code
int main()
{
    int i = 3, j = 3, N = 2;
    cout << getCount(i, j, N) << endl;
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
   
static int N = 10;
 
// All possible moves of the knight.
 
// In X axis.
static int[] X = { 2, 1, -1, -2, -2, -1, 1, 2 };
 
// In Y axis.
static int[] Y = { 1, 2, 2, 1, -1, -2, -2, -1 };
 
static void getCountRec(boolean[][] board,
                        int i, int j, int n)
{
     
    // If n=0, we have our result.
    if (n == 0)
        return;
 
    for(int k = 0; k < 8; k++)
    {
        int p = i + X[k];
        int q = j + Y[k];
 
        // Condition for valid cells.
        if (p >= 0 && q >= 0 &&
            p < 10 && q < N) 
        {
            board[p][q] = true;
            getCountRec(board, p, q, n - 1);
        }
    }
}
 
static int getCount(int i, int j, int n)
{
    boolean[][] board = new boolean[N][N];
    board[i][j] = true;
 
    // Call the recursive function to mark
    // visited cells.
    getCountRec(board, i, j, n);
 
    int cnt = 0;
    for(boolean[] row : board) 
    {
        for(boolean cell : row)
        {
            if (cell != false)
                cnt++;
        }
    }
    return cnt;
}
 
// Driver code
public static void main(String[] args)
{
    int i = 3, j = 3, N = 2;
     
    System.out.println(getCount(i, j, N));
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python program for the above approach
SIZE = 10
 
# All possible moves of the knight.
# In X axis.
X = [2, 1, -1, -2, -2, -1, 1, 2] 
 
# In Y axis.
Y = [1, 2, 2, 1, -1, -2, -2, -1]
 
def getCountRec(board, i, j, n):
     
    # If n=0, we have our result.
    if n == 0:
        return
     
    for k in range(8):
        p = i + X[k]
        q = j + Y[k]
         
        # Condition for valid cells.
        if p >= 0 and q >= 0 and p < 10 and q < SIZE:
            board[p][q] = True
            getCountRec(board,p,q,n-1)
     
def getCount(i, j, n):
    board = [[False for i in range(SIZE)] for j in range(SIZE)]
    board[i][j] = True
     
    # Call the recursive function to mark
    # visited cells.
    getCountRec(board, i, j, n)
    cnt = 0
     
    for row in board:
        for cell in row:
            if cell != False:
                cnt += 1
             
    return cnt
 
# Driver code    
i = 3
j = 3
N = 2
print(getCount(i, j, N))
 
# This code is contributed by rdtank.

C#

// C# program for the above approach
using System;
class GFG
{
 
  static int N = 10;
 
  // All possible moves of the knight.
 
  // In X axis.
  static int [] X = { 2, 1, -1, -2, -2, -1, 1, 2 };
 
  // In Y axis.
  static int [] Y = { 1, 2, 2, 1, -1, -2, -2, -1 };
 
  static void getCountRec(bool[,] board,
                          int i, int j, int n)
  {
 
      // If n=0, we have our result.
      if (n == 0)
          return;
 
      for(int k = 0; k < 8; k++)
      {
          int p = i + X[k];
          int q = j + Y[k];
 
          // Condition for valid cells.
          if (p >= 0 && q >= 0 &&
              p < 10 && q < N) 
          {
              board[p, q] = true;
              getCountRec(board, p, q, n - 1);
          }
      }
  }
 
  static int getCount(int i, int j, int n)
  {
      bool [, ] board = new bool[N, N];
      board[i, j] = true;
 
      // Call the recursive function to mark
      // visited cells.
      getCountRec(board, i, j, n);
 
      int cnt = 0;
      foreach(bool cell in board) 
      {
          if(cell != false)
            cnt++;
      }
      return cnt;
  }
 
  // Driver code
  public static void Main()
  {
      int i = 3, j = 3, N = 2;
 
      Console.WriteLine(getCount(i, j, N));
  }
}
 
// This code is contributed by ihritik

Javascript

<script>
 
// JavaScript implementation of the approach
const SIZE = 10;
 
// All possible moves of the knight.
 
// In X axis.
let X = [ 2, 1, -1, -2, -2, -1, 1, 2 ];
 
// In Y axis.
let Y = [ 1, 2, 2, 1, -1, -2, -2, -1 ];
 
function getCountRec(board,i,j,n)
{
    // if n=0, we have our result.
    if (n == 0)
        return;
 
    for (let k = 0; k < 8; k++) {
        let p = i + X[k];
        let q = j + Y[k];
 
        // Condition for valid cells.
        if (p >= 0 && q >= 0
            && p < 10 && q < SIZE) {
            board[p][q] = true;
            getCountRec(board, p, q, n - 1);
        }
    }
}
 
function getCount(i, j, n)
{
    let board = new Array(SIZE).fill(0).map(()=>new Array(N));
    board[i][j] = true;
 
    // Call the recursive function to mark
    // visited cells.
    getCountRec(board, i, j, n);
 
    let cnt = 0;
    for (let row of board) {
        for (let cell of row) {
            if (cell)
                cnt++;
        }
    }
    return cnt;
}
 
// Driver Code
 
let i = 3, j = 3,N = 2;
document.write(getCount(i, j, N),"</br>");
 
 
// This code is contributed by shinjanpatra
 
</script>// JavaScript implementation of the approach
 
const SIZE = 10;
 
// All possible moves of the knight.
 
// In X axis.
let X = [ 2, 1, -1, -2, -2, -1, 1, 2 ];
 
// In Y axis.
let Y = [ 1, 2, 2, 1, -1, -2, -2, -1 ];
 
function getCountRec(board,i,j,n)
{
    // if n=0, we have our result.
    if (n == 0)
        return;
 
    for (let k = 0; k < 8; k++) {
        let p = i + X[k];
        let q = j + Y[k];
 
        // Condition for valid cells.
        if (p >= 0 && q >= 0
            && p < 10 && q < SIZE) {
            board[p][q] = true;
            getCountRec(board, p, q, n - 1);
        }
    }
}
 
function getCount(i, j, n)
{
    let board = new Array(SIZE).fill(0).map(()=>new Array(N));
    board[i][j] = true;
 
    // Call the recursive function to mark
    // visited cells.
    getCountRec(board, i, j, n);
 
    let cnt = 0;
    for (let row of board) {
        for (let cell of row) {
            if (cell)
                cnt++;
        }
    }
    return cnt;
}
 
// Driver Code
 
let i = 3, j = 3,N = 2;
document.write(getCount(i, j, N),"</br>");
 
// This code is contributed by shinjanpatra
</script>

Output:

Time Complexity: O(8^N) where N is the number of moves the knight can make.

Space Complexity: O(N^2), where N is the size of the chessboard. This is because we are using a 2D vector to store the visited cells on the board. The size of this vector is N x N.

Suggest improvement

Count of all possible ways to reach a target by a Knight

Share your thoughts in the comments

Count all possible visited cells of a knight after N moves

C++14

Java

Python3

C#

Javascript

Please Login to comment...

Similar Reads

What kind of Experience do you want to share?