<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:
        # n = 1 => return 10
        # we know the exact numbers each cell can move towards
        # 1 => 6, 8 = 2 jumps
        # numbers(1, n) = numbers(6, n-1) + numbers(8, n-1)
        # sum this across all start points
        # base case => numbers(start, len) if len = 0 => 0 and len = 1 => 1
        next_nums = {
            0: [4,6],
            1: [6,8],
            2: [7,9],
            3: [4,8],
            4: [3,9,0],
            5: [],
            6: [7,0,1],
            7: [2,6],
            8: [1,3],
            9: [2,4]
        }
        num_cache = {}
        def helper(start, l):
            if (start, l) in num_cache:
                return num_cache[(start, l)]
            mod = (10**9) + 7
            if l <= 1:
                return l
            else:
                next_n = next_nums[start]
                if not next_n:
                    return 0
                else:
                    num_cache[(start, l)] = sum([helper(nxt, l-1)%mod for nxt in next_n])
                    return num_cache[(start, l)]
        mod = (10**9) + 7
        total = sum([helper(i, n)%mod for i in range(10)])
        return total % mod
39218