<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:
        moves = [(2,1), (1,2), (1,-2),(2,-1),(-1,-2),(-2,-1),(-2,1),(-1,2)]

        def isValid(nextSpot):
            new_r, new_c = nextSpot
            return new_r >= 0 and new_r < 4 and new_c >= 0 and new_c < 3 and (new_r, new_c) != (3,0) and (new_r, new_c) != (3,2)

        adj = {}
        print('hohoo')
        for r in range(4):
            for c in range(3):
                if (r,c) != (3,0) and (r,c) != (3, 2):
                    # print('finding adjacent for')
                    # print((r,c))
                    # generate all valid moves
                    valid_moves = []
                    for mr, mc in moves:
                        new_r = r + mr
                        new_c = c + mc
                        if isValid((new_r, new_c)):
                            valid_moves.append((new_r, new_c))
                    adj[(r,c)] = valid_moves
                    # print(valid_moves)
 
                    

        dialer = [[1,2,3],
                [4,5,6],
                [7,8,9],
                ['*',0,'#']]

        dp = {} # holds the digit : []. array of size n where arr[i] holds the number of numbers can generate at that point if  
        # we can only enter in i more numbers
        # so starting at i = 0 means we are done, and i = 1 means can enter one more number.
        for i in range(10):
            dp[i] = [1 for j in range(n)]

        for i in range(1, n):
            for r in range(4):
                for c in range(3):
                    if (r,c) != (3,0) and (r,c) != (3, 2):
                        # sum up all adjacent
                        total = 0
                        for mr, mc in adj[(r,c)]:
                            total += dp[dialer[mr][mc]][i-1]
                        #     print((mr, mc, dp[dialer[mr][mc]][i-1]))
                        dp[dialer[r][c]][i] = total
        ret = 0
        # print(adj)
        for i in range(10):
            ret += dp[i][n-1]
        return ret % (10**9 + 7)
36521