<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:
    const long long int MOD = 1'000'000'007;
    int row[8] = {2, 2, -2, -2, 1, -1, 1, -1};
    int col[8] = {1, -1, 1, -1, 2, 2, -2, -2};
    
    long long int f(int i, int j, int L, int n, vector<vector<vector<long long int>>>& dp, vector<vector<int>>& dial) {
        if (L == n) {
            return 1;
        }
        if (dp[i][j][L] != -1) {
            return dp[i][j][L];
        }
        long long int take = 0;
        for (int k = 0; k < 8; ++k) {
            int newr = i + row[k];
            int newc = j + col[k];
            if (newr >= 0 && newr < dial.size() && newc >= 0 && newc < dial[0].size() && dial[newr][newc] != -1) {
                take = (take + f(newr, newc, L + 1, n, dp, dial)) % MOD;
            }
        }
        return dp[i][j][L] = take;
    }
    
    int knightDialer(int n) {
        if (n == 0) return 0;

        vector<vector<int>> dial(4, vector<int>(3, -1));
        int c = 1;
        for (int i = 0; i < 4; ++i) {
            for (int j = 0; j < 3; ++j) {
                if (i == 3 && (j == 0 || j == 2)) {
                    continue;
                }
                dial[i][j] = c++;
            }
        }
        dial[3][1] = 0; // Center position (0)

        long long int maxi = 0;
        vector<vector<vector<long long int>>> dp(4, vector<vector<long long int>>(3, vector<long long int>(n, -1)));
        for (int i = 0; i < 4; ++i) {
            for (int j = 0; j <3; ++j) {
                if (dial[i][j] != -1) {
                    maxi = (maxi + f(i, j,1 , n, dp, dial)) % MOD;
                }
            }
        }
        return maxi;
    }
};

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