<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) {
        if (n == 1) return 10;
        vector<vector<long>> dp(9, vector<long>(n, 0));
        int a; // current pad
        int i=0; // current round
        int res = 0;
        for (a = 0; a < 9; a++) {
            dp[a][0] = 1;
        }
        for (i = 1; i < n; i++) {
            for (a = 0; a < 9; a++) {
                if (a < 2) { // as pad 4  6
                    dp[a][i] = (dp[2][i-1]*2 + dp[8][i-1]) % 1000000007;
                }
                else if (a < 6) { // as pad 1 3 7 9
                    dp[a][i] = (dp[0][i-1] + dp[6][i-1])%1000000007;
                }
                else if (a < 8) {
                    dp[a][i] = (dp[2][i-1]*2)%1000000007;
                }
                else {
                    dp[a][i] = (dp[0][i-1]*2)%1000000007;
                }
            }
        }
        for (a = 0; a < 9; a++) {
            res = (res + dp[a][n-1])%1000000007;
            //cout<< dp[a][n-1] << " ";
        }
        return res;
    }
    // 9 * n
    // dp[a][i] = dp[ap1][i-1] + dp[ap2][i-1] + ...
    // 4 << 3 9 0 == 0
    // 6 << 1 7 0 == 1

    // 1 << 8 6 == 2
    // 3 << 4 8 == 3
    // 7 << 2 6 == 4
    // 9 << 2 4 == 5

    // 2 << 7 9 == 6
    // 8 << 1 3 == 7

    // 0 << 4 6 == 8
};
33572