<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>

const int delta_r[8] = {1, 2, 2, 1, -1, -2, -2, -1};
const int delta_c[8] = {2, 1, -1, -2, -2, -1, 1, 2};
const int mod = 1e9 + 7;

class Solution {
public:
    bool valid(int r, int c, int rows, int cols) {
        return r >= 0 && r < rows && c >= 0 && c < cols;
    }
    
    int dfs(vector<vector<int>> &grid, int rows, int cols, int r, int c, int n, vector<vector<vector<int>>> &dp) {
        if (n == 1)
            return 1;
        if (dp[n][r][c] != INT_MAX)
            return dp[n][r][c];
        
        int res = 0;
        
        for (int i = 0; i < 8; i++) {
            int rr = r + delta_r[i];
            int cc = c + delta_c[i];
            
            if (valid(rr, cc, rows, cols) && grid[rr][cc])
                res = (res + dfs(grid, rows, cols, rr, cc, n - 1, dp)) % mod;
        }
        
        return dp[n][r][c] = res;
    }
    
    int knightDialer(int n) {
        int res = 0;
        int rows = 4;
        int cols = 3;
        vector<vector<int>> grid(rows, vector<int>(cols, 1));
        vector<vector<vector<int>>> dp(n + 1, vector<vector<int>>(rows, vector<int>(cols, 0)));
        
        grid[rows - 1][0] = 0;
        grid[rows - 1][cols - 1] = 0;
        
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j])
                    dp[1][i][j] = 1;
            }
        }
        
        for (int i = 2; i <= n; i++) {
            for (int r = 0; r < rows; r++) {
                for (int c = 0; c < cols; c++) {
                    if (grid[r][c] != 1)
                        continue;
                    
                    for (int j = 0; j < 8; j++) {
                        int rr = r + delta_r[j];
                        int cc = c + delta_c[j];
                        
                        if (valid(rr, cc, rows, cols) && grid[rr][cc])
                            dp[i][r][c] = (dp[i][r][c] + dp[i - 1][rr][cc]) % mod;
                    }
                }
            }
        }
        
        for (int r = 0; r < rows; r++) {
            for (int c = 0; c < cols; c++) {
                res = (res + dp[n][r][c]) % mod;
            }
        }
        
        return res;
    }
};
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