<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:
        '''
        2 squares vertically and 1 hor.
        2 sq hor, 1 ver


        n-1 jumps;


        moves = [[2,-1],[2,1], [1,-2],[1,2],    [-2,-1],[-2,1], [1,-2],[-1,2],] 


        n-1 jumps





        2
        start:0 - 2
        2
         6
         4 

         06
         04

        def rec(start, n):   
            if n == 0:
                return 0      
            
            result = 0
            for next in hash_map[start]:         
                jumps = rec(next, n--)
                result += jumps

            return result 
        


        for i in range (0..10):
            rec( n )
       


        1 -  2 (6,8)
        2 -  2 (7,9)
        3 -  2 (4,8)
        4 -  3 (3,9,0)
        5 -  X
        6 - 3 (1,7,0)
        7 - 2 (6,2)
        8 - 2 (3,1)
        9- 2 (4,2)
        0 -2 (6,4)


        rec_fib

        return rec_fib(n-1)+ rec_fib(n-2)




        '''

        jumps_map = {
            0:[6,4],
            1: [6,8],
            2:[7,9],
            3:[4,8],
            4:[3,9,0],
            5:[],
            6:[1,7,0],
            7:[6,2],
            8:[3,1],
            9:[4,2],
        }


        def rec(start, n,dp):   
            if (start,n) in dp:
                return dp[(start,n)]

            if n == 0:
                return 1     

            result = 0

            for next in jumps_map[start]:  
                result_next = rec(next, n-1, dp)
                result += result_next
                result =  result % (10**9+7)

            dp[(start,n)] = result
            return dp[(start,n)] 
        

        dp = {}
        result = 0
        for i in range (0,10):
            result_i = rec(i, n-1,dp)
            result += result_i 
            

        return result % (10**9+7)
        
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