<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 {
public:
    int mod = 1000000007;

    int dfs(int n, int key, unordered_map<int, vector<int>>& mp, vector<vector<int>>& dp)
    {
        if(n == 0) // exhausted all moves found unique number
            return 1;
        
        if(dp[n][key] != -1)
            return dp[n][key];

        long ans = 0;
        
        for(int i = 0; i < mp[key].size(); i++)
            ans = ((ans % mod) + (dfs(n - 1, mp[key][i], mp, dp) % mod)) % mod;

        return dp[n][key] = (int)(ans % mod);
    }

    int knightDialer(int n)
    {
        unordered_map<int, vector<int>> mp;
        vector<vector<int>> dp(n + 1, vector<int> (10, -1));

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

        long ans = 0;

        for(int i = 0; i <= 9; i++)
            ans = (ans + dfs(n - 1, i, mp, dp)) % mod;

        return (int)(ans % mod);
    }
};

/*

knight on any of 0 - 9 initially
n - 1 jumps to form distinct numbers of len n

n = 1 => 10, as 10 unique starting pos

n = 2 => 20, 
0 -> 2
1 -> 2
2 -> 2
3 -> 2
4 -> 3
5 -> 0
6 -> 3
7 -> 2
8 -> 2
9 -> 2

Maintain a map of possible jumps from nums then do dfs ?

*/
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