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

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `using` `namespace` `std; ` ` `  `// function that will be called recursively ` `int` `recursive_solve(``int` `i, ``int` `j, ``int` `steps, ``int` `n,  ` `                      ``map, ``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, ``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; ` `} `

## 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 `

Output:

```10
``` My Personal Notes arrow_drop_up Third year Department of Information Technology Jadavpur University

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Improved By : rituraj_jain