<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;
    int mod = 1e9 + 7;
    vector<long long> dp1(n + 1);
    vector<long long> dp2(n + 1);
    vector<long long> dp3(n + 1);
    vector<long long> dp4(n + 1);
    
    dp1[1] = 1;
    dp2[1] = 2;
    dp3[1] = 2;
    dp4[1] = 4;
    
    for (int i = 2; i <= n; i++) {
        dp1[i] = dp2[i - 1] % mod;
        dp2[i] = (dp4[i - 1] + 2 * dp1[i - 1]) % mod;
        dp3[i] = dp4[i - 1] % mod;
        
        long long temp = (3 * dp2[i - 1]) % mod;
        temp = (temp + 2 * dp3[i - 1]) % mod;
        temp = (temp - dp2[i - 1] + mod) % mod; // thêm `mod` để tránh số âm
        dp4[i] = temp;
    }
    
    long long result = (dp1[n] + dp2[n]) % mod;
    result = (result + dp3[n] + dp4[n]) % mod;
    
    return result;
}
};
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