<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:
        // recursion + memoization
    int mod = 1000000007;

    int f(vector<vector<int>>& mat, int n, int count, int i, int j, vector<vector<vector<int>>>& dp) {
        if (count == n - 1) {
            return 1;
        }
        if (dp[count][i][j] != -1) return dp[count][i][j];

        int path = 0;

        if ((i - 2) >= 0 && (j - 1) >= 0 && mat[i - 2][j - 1] != -1) {
            path = (path + f(mat, n, count + 1, i - 2, j - 1, dp)) % mod;
        }
        if ((i - 2) >= 0 && (j + 1) < 3 && mat[i - 2][j + 1] != -1) {
            path = (path + f(mat, n, count + 1, i - 2, j + 1, dp)) % mod;
        }
        if ((i - 1) >= 0 && (j - 2) >= 0 && mat[i - 1][j - 2] != -1) {
            path = (path + f(mat, n, count + 1, i - 1, j - 2, dp)) % mod;
        }
        if ((i + 1) < 4 && (j - 2) >= 0 && mat[i + 1][j - 2] != -1) {
            path = (path + f(mat, n, count + 1, i + 1, j - 2, dp)) % mod;
        }
        if ((i - 1) >= 0 && (j + 2) < 3 && mat[i - 1][j + 2] != -1) {
            path = (path + f(mat, n, count + 1, i - 1, j + 2, dp)) % mod;
        }
        if ((i + 1) < 4 && (j + 2) < 3 && mat[i + 1][j + 2] != -1) {
            path = (path + f(mat, n, count + 1, i + 1, j + 2, dp)) % mod;
        }
        if ((i + 2) < 4 && (j - 1) >= 0 && mat[i + 2][j - 1] != -1) {
            path = (path + f(mat, n, count + 1, i + 2, j - 1, dp)) % mod;
        }
        if ((i + 2) < 4 && (j + 1) < 3 && mat[i + 2][j + 1] != -1) {
            path = (path + f(mat, n, count + 1, i + 2, j + 1, dp)) % mod;
        }

        return dp[count][i][j] = path;
    }

    int knightDialer(int n) {
        vector<vector<int>> mat(4, vector<int>(3, -1));

        int temp = 1;
        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                mat[i][j] = temp;
                temp++;
            }
        }
        vector<vector<vector<int>>> dp(n, vector<vector<int>>(4, vector<int>(3, -1)));
        mat[3][1] = 0;
        int result = 0;
        for (int i = 0; i < 4; i++) {
            for (int j = 0; j < 3; j++) {
                if (mat[i][j] != -1) {
                    result = (result + f(mat, n, 0, i, j, dp)) % mod;
                }
            }
        }

        return result % mod;
    }
};

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