<p>The chess knight has a <strong>unique movement</strong>,&nbsp;it may move two squares vertically and one square horizontally, or two squares horizontally and one square vertically (with both forming the shape of an <strong>L</strong>). The possible movements of chess knight are shown in this diagram:</p>

<p>A chess knight can move as indicated in the chess diagram below:</p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/18/chess.jpg" style="width: 402px; height: 402px;" />
<p>We have a chess knight and a phone pad as shown below, the knight <strong>can only stand on a numeric cell</strong>&nbsp;(i.e. blue cell).</p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/18/phone.jpg" style="width: 242px; height: 322px;" />
<p>Given an integer <code>n</code>, return how many distinct phone numbers of length <code>n</code> we can dial.</p>

<p>You are allowed to place the knight <strong>on any numeric cell</strong> initially and then you should perform <code>n - 1</code> jumps to dial a number of length <code>n</code>. All jumps should be <strong>valid</strong> knight jumps.</p>

<p>As the answer may be very large, <strong>return the answer modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 10
<strong>Explanation:</strong> We need to dial a number of length 1, so placing the knight over any numeric cell of the 10 cells is sufficient.
</pre>

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

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 20
<strong>Explanation:</strong> All the valid number we can dial are [04, 06, 16, 18, 27, 29, 34, 38, 40, 43, 49, 60, 61, 67, 72, 76, 81, 83, 92, 94]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 3131
<strong>Output:</strong> 136006598
<strong>Explanation:</strong> Please take care of the mod.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 5000</code></li>
</ul>

from functools import cache


class Solution:
    def knightDialer(self, n: int) -> int:
        phone = [[1,2,3],[4,5,6],[7,8,9],["*",0,"#"]]

        #all directions a piece can move based on knight rules
        directions = [[1,2],[2,1],[-2,1],[1,-2],[-1,2],[2,-1],[-2,-1],[-1,-2]]
        #function to determine valid bounds
        def in_bounds(i,j):
            if i == 3 and j == 1:
                return True
            elif i >= 3 or i < 0:
                return False
            elif j >=3 or j < 0:
                return False
            return True
        #caching method, caching can only be used with return values, essentially stores the return values of dfs at specific values of i,j,renaining this allows us to reuse computations like dynamic programming. We cannoy use a self.count var in this case
        @cache
        def dfs(i, j, remaining):
            #return 1 path
            if remaining == 0:
                return 1  # Valid path found
            #count total paths at this point 
            total_count = 0
            #for each possible move
            for dx, dy in directions:
                new_i, new_j = i + dx, j + dy  
                #if in bound count its valid moves
                if in_bounds(new_i, new_j):
                    total_count += dfs(new_i, new_j, remaining - 1)
            
            return total_count

        total = 0
        #all starting points
        for i in range(3):
            for j in range(3):
                #start with n-1 moves using caching we can efficinetly compute maximum moves!
                total += dfs(i, j, n - 1)
        #0 starting point
        total += dfs(3, 1, n - 1)  # Starting from '0'
        return total % (10**9 + 7)
76965