**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 :

` ````
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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