<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>

class Solution:
    def knightDialer(self, n: int) -> int:
        mat = [[1,2,3],[4,5,6],[7,8,9],["*",0,"#"]]
        m = len(mat)
        n1 = len(mat[0])
        # dp = [[-1] * n1 for _ in range(m)]
        cache={}
        pos = [(-1, -2), (-1, 2), (1, -2), (1, 2), (-2, 1), (-2, -1), (2, 1), (2, -1)]
        cnt = 0
        mod=10**9+7
        def isInvalid(r, c):
            if r >= 4 or c >= 3 or r < 0 or c < 0:
                return True
            if (r, c) in [(3, 0), (3, 2)]:
                return True
            return False
        def f(x, y, l):
            if isInvalid(x,y): return 0
            if l == n:
                return 1
            
            # if dp[x][y] != -1:
            #     return dp[x][y]
            if (x,y,l) in cache:
                return cache[(x,y,l)]
            s=0
            for i,j in pos:
                r=x+i
                c=y+j
    
                s += f(r,c,l+1)
            # dp[x][y] = s%mod
            # return dp[x][y]
            cache[(x,y,l)]=s
            return cache[(x,y,l)]%mod
            
        for i in range(m):
            for j in range(n1):
                if mat[i][j] != "*" and mat[i][j] != "#":
                    cnt += f(i, j, 1)

        return cnt%mod

        

48205