<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 mod = 1000000000 + 7;
    vector<pair<int, int>> directions = {{-1, -2}, {-2, -1}, {-1, 2}, {-2, 1},
                                        {1, -2}, {2, -1}, {1, 2}, {2, 1}};
    int rows = 4, cols = 3;

    bool valid(int row, int col) {
        if (row >= 0 && row < rows - 1 && col >= 0 && col < rows - 1) return true;
        if (row == rows - 1 && col == 1) return true;
        return false;
    }

    // int helper(int n, int row, int col) {
        
    //     int sum = 0;
    //     for (int i = 0; i < 4; i++) {
    //         for (int j = )
    //     }
    // }
    
    int knightDialer(int n) {
        vector<vector<vector<int>>> dp(n + 1, vector<vector<int>>(rows, vector<int>(cols, 0)));
        
        int res = 0;
        for (int k = 1; k <= n; k++) {
            for (int i = 0; i < rows; i++) {
                for (int j = 0; j < cols; j++) {
                    if (!valid(i, j)) continue;
                    int sum = 0;
                    if (k == 1) {
                        dp[k][i][j] = valid(i, j) ? 1 : 0;
                        sum += dp[k][i][j];
                    }
                    else {
                        for (const auto& dir : directions) {
                            int row = i + dir.first;
                            int col = j + dir.second;
                            if (valid(row, col)) sum = (sum + dp[k - 1][row][col]) % mod;
                        }
                        dp[k][i][j] = sum;
                    }
                    if (k == n) res = (res + sum) % mod;;
                }
            }
        }
        return res;
    }
};
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