<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 = 1e9 + 7;
    bool isValid(int row, int col, vector<vector<int>>& board) {
        int n = board.size();
        int m = board[0].size();
        if (row == 3) {
            if (col == 1)
                return true;
            return false;
        }
        return row >= 0 && row < n && col >= 0 && col < m;
    }
    int solve(int row, int col, int cnt, vector<vector<int>>& board,
              vector<vector<vector<int>>>& dp) {
        if (cnt == 0)
            return 1;

        if (dp[row][col][cnt] != -1)
            return dp[row][col][cnt];

        int mRow[] = {-2, 0, 2, 0};
        int mCol[] = {0, 2, 0, -2};
        int sum = 0;
        for (int i = 0; i < 4; i++) {
            int nRow = row + mRow[i];
            int nCol = col + mCol[i];
            if (i & 1) {
                if (isValid(nRow - 1, nCol, board))
                    sum =
                        (sum + solve(nRow - 1, nCol, cnt - 1, board, dp)) % mod;
                if (isValid(nRow + 1, nCol, board))
                    sum =
                        (sum + solve(nRow + 1, nCol, cnt - 1, board, dp)) % mod;
            } else {
                if (isValid(nRow, nCol - 1, board))
                    sum =
                        (sum + solve(nRow, nCol - 1, cnt - 1, board, dp)) % mod;
                if (isValid(nRow, nCol + 1, board))
                    sum =
                        (sum + solve(nRow, nCol + 1, cnt - 1, board, dp)) % mod;
            }
        }

        return dp[row][col][cnt] = sum;
    }
    int knightDialer(int n) {
        vector<vector<int>> board(4, vector<int>(3, 1));
        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                board[i][j] += i * 3 + j;
            }
        }
        board[3][0] = board[3][2] = -1;
        board[3][1] = 0;

        int sum = 0;
        vector<vector<vector<int>>> dp(
            4, vector<vector<int>>(3, vector<int>(5001, -1)));

        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                sum = (sum + solve(i, j, n - 1, board, dp)) % mod;
            }
        }
        sum = (sum + solve(3, 1, n - 1, board, dp)) % mod;

        return sum % mod;
    }
};
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