<p>On an <code>n x n</code> chessboard, a knight starts at the cell <code>(row, column)</code> and attempts to make exactly <code>k</code> moves. The rows and columns are <strong>0-indexed</strong>, so the top-left cell is <code>(0, 0)</code>, and the bottom-right cell is <code>(n - 1, n - 1)</code>.</p>

<p>A chess knight has eight possible moves it can make, as illustrated below. Each move is two cells in a cardinal direction, then one cell in an orthogonal direction.</p>
<img src="https://assets.leetcode.com/uploads/2018/10/12/knight.png" style="width: 300px; height: 300px;" />
<p>Each time the knight is to move, it chooses one of eight possible moves uniformly at random (even if the piece would go off the chessboard) and moves there.</p>

<p>The knight continues moving until it has made exactly <code>k</code> moves or has moved off the chessboard.</p>

<p>Return <em>the probability that the knight remains on the board after it has stopped moving</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3, k = 2, row = 0, column = 0
<strong>Output:</strong> 0.06250
<strong>Explanation:</strong> There are two moves (to (1,2), (2,1)) that will keep the knight on the board.
From each of those positions, there are also two moves that will keep the knight on the board.
The total probability the knight stays on the board is 0.0625.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1, k = 0, row = 0, column = 0
<strong>Output:</strong> 1.00000
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 25</code></li>
	<li><code>0 &lt;= k &lt;= 100</code></li>
	<li><code>0 &lt;= row, column &lt;= n - 1</code></li>
</ul>

class Solution:
    def knightProbability(self, n: int, k: int, row: int, col: int) -> float:
        # 8 possible knight moves
        dirs = [(2, 1), (1, 2), (-1, 2), (-2, 1), (-2, -1), (-1, -2), (1, -2), (2, -1)]
        
        # Memoization cache to store already computed results
        memo = {}
        
        # DFS function to calculate the probability
        def dfs(i, j, remaining_moves):
            # If out of bounds, return 0 (knight is off the board)
            if i < 0 or i >= n or j < 0 or j >= n:
                return 0
            
            # If no moves are left, return 1 (knight is on the board)
            if remaining_moves == 0:
                return 1
            
            # If this state has been computed before, return the cached result
            if (i, j, remaining_moves) in memo:
                return memo[(i, j, remaining_moves)]
            
            # Calculate probability by trying all 8 possible knight moves
            prob = 0
            for di, dj in dirs:
                ni, nj = i + di, j + dj
                prob += dfs(ni, nj, remaining_moves - 1) / 8.0
            
            # Store the result in the memo table
            memo[(i, j, remaining_moves)] = prob
            return prob
        
        # Start the DFS from the initial position
        return dfs(row, col, k)

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