<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 knightDialer(int n) {
        int M = (int)1e9 + 7;
        unordered_map<int, vector<int>> m;
        m[0] = {4, 6};
        m[1] = {6,8};
        m[2] = {7,9};
        m[3] = {4,8};
        m[4] = {0, 3, 9};
        m[6] = {0,1,7};
        m[7] = {2,6};
        m[8] = {1,3};
        m[9] = {2,4};

        vector<vector<int>> dp(2, vector<int>(10, 0));
        for(int i=0;i<=9;i++) {
            dp[0][i] = dp[1][i] = 1;
        }

        for(int i=2;i<=n;i++) {
            for(int j=0;j<=9;j++) {
                dp[1][j] = 0;
                for(auto k: m[j]) {
                    dp[1][j] = (dp[1][j]%M + dp[0][k])%M;
                }
            }
            dp[0] = dp[1]; 
        }

        int total = 0;
        for(int i=0;i<=9;i++) {
            total = (total%M + dp[1][i])%M;
        }

        return total;
    }
};
9842