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 cellsInput: 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++ 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 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
# 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# 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
<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:
35
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.
