Knight Remains - Amazon Top Interview Questions


Problem Statement :


You are given four integers n, x, y, and k. n represents an n by n chessboard and x, y represents a knight positioned at (x, y). The knight has to take exactly k steps, where at each step it chooses any of the 8 directions uniformly at random.

Return the percentage chance rounded down to the nearest integer that the knight remains in the chessboard after taking k steps, with the condition that it can’t enter the board again once it leaves it.

Constraints

1 ≤ n ≤ 25
0 ≤ k ≤ 100

Example 1

Input

n = 8
x = 0
y = 0
k = 1

Output

25

Explanation

This is an 8x8 chessboard and the initial position of the knight is (0, 0). It can take k = 1 step. After taking one step it will lie inside the board only at 2 out of 8 positions, and will lie outside at other positions.

So, the probability is 2/8 = 0.25


Solution :



title-img 




                        Solution in C++ :

int x[8] = {2, 1, -1, -2, -2, -1, 1, 2};
int y[8] = {1, 2, 2, 1, -1, -2, -2, -1};
double dp[26][26][101];
double s1(int n, int i, int j, int k) {
    if (i < 0 || j < 0 || i >= n || j >= n) {
        return 0;
    }

    if (k == 0) return 1;
    if (dp[i][j][k] != 0) return dp[i][j][k];
    double temp = 0;
    for (int t = 0; t < 8; t++) {
        temp += 0.125 * s1(n, i + x[t], j + y[t], k - 1);
    }
    return dp[i][j][k] = temp;
}
int solve(int n, int x, int y, int k) {
    memset(dp, 0, sizeof(dp));
    double k1 = s1(n, x, y, k);
    k1 = k1 * 100;
    // cout<<k1;
    return (int)k1;
}
                    




                        Solution in Python : 
                            
class Solution:
    def solve(self, n, x, y, K):
        def isvalid(i, j):
            return 0 <= i < n and 0 <= j < n

        movement = [[2, 1], [2, -1], [-2, 1], [-2, -1], [1, 2], [-1, 2], [1, -2], [-1, -2]]

        @lru_cache(None)
        def dp(x, y, k):
            if not isvalid(x, y):
                return 0
            if k == 0:
                return 1
            res = 0
            for mov in movement:
                res += dp(x + mov[0], y + mov[1], k - 1)
            return res

        return dp(x, y, K) * 100 // (8 ** K)


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