<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>

/*
    Tabulation:
    Build from 0 - n
    -   Base cases:
        >   n = 0: There are 0 phone numbers with length 0
        >   n = 1: There are 10 phone numbers with length 1
    -   For each i from 2 to n:
        >   Consider adding a digit
*/

class Solution {
public:
    int MOD = 1000000007;
    vector<vector<int>> moves = {
        {5, 7}, {6, 8}, {3, 7}, {2, 8, 10}, {}, {0, 6, 10}, {1, 5}, {0, 2}, {1, 3}, {}, {3, 5}, {}
    };

    int knightDialer(int n) {
        vector<vector<int>> dp(n + 1);
        dp[1] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0};
        for (int i = 2; i <= n; i++) {
            vector<int> prev = dp[i - 1];
            vector<int> next(12, 0);
            for (int d = 0; d < 12; d++) {
                for (int nd : moves[d]) {
                    next[nd] = (next[nd] + prev[d]) % MOD;
                }
            }
            dp[i] = next;
        }

        int ans = 0;
        for (int i = 0; i < 12; i++) {
            // cout << (i + 1) % 11 << ": " << dp[n][i] << endl;
            ans = (ans + dp[n][i]) % MOD;
        }

        return ans;
    }
};
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