<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:
        invalid_options = {(3,0),(3,2)}
        knight_moves = [ (2,1), (1,2) , (-1,2), (-2,1), (2,-1), (1,-2), (-1,-2), (-2,-1)]
        dX = [-2, -1, 1, 2, -2, 1, -1, 2]
        dY = [-1, -2, 2, 1, 1, -2, 2, -1]

        cellpad = [
                    [1,2,3],
                    [4,5,6],
                    [7,8,9],
                    [-1,0,-1]
                ]

        def is_valid(x: int, y: int) -> bool:
            return x >= 0 and y >= 0 and x < 4 and y < 3 and cellpad[x][y] >= 0
        
        cellpad = [
            [1,2,3],
            [4,5,6],
            [7,8,9],
            [-1,0,-1]
        ]

        mem = {}
        def rec(i,j,size=0):
            key = (i,j,size)
            if key in mem: return mem[key]
            if size == n: return 1
            # find next move
            combinations = 0
            for move in zip(dX,dY):
                r, c = move[0] + i, move[1] + j
                if not is_valid(r,c): continue
                combinations += rec(r,c,size+1)
            mem[key] = combinations
            return combinations


        options = 0
        mod = 10**9 + 7

        for i,j in zip(range(4),range(3)):
            print(i,j)

        for i in range(4):
            for j in range(3):
                if not is_valid(i,j): continue
                options += rec(i, j,1)
        return options % mod

     
65281