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: 8Input: 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; } |
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 |
10
