<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 dp[5001][10];
    const int mo = 1e9 + 7;
    int sol(int i, int num, int n, vector<vector<int>>& mp) {
        if (i >= n)
            return 1;
        if(dp[i][num] != -1)
            return dp[i][num];
        long long ans = 0;
        vector<int> temp = mp[num];
        // cout << num << " ";
        
        for (int j = 0; j < temp.size(); j++) {
            int steps = sol(i + 1, temp[j], n, mp) % mo;
            ans += steps;
            ans = ans % mo;
        }
        return dp[i][num] = ans % mo;
    }
    int knightDialer(int n) {
        vector<vector<int>> mp = {
            {4, 6}, {8, 6},    {7, 9}, {4, 8}, {3, 9, 0},
            {},     {0, 1, 7}, {2, 6}, {1, 3}, {2, 4}};

        int ans = 0;
        memset(dp, -1, sizeof(dp));
        for(int i=0; i<=9; i++){
            ans += sol(1, i, n, mp);
            ans = ans % mo;
        }
        return ans;
    }
};
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