<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 to implement the Bottom-up approach:
class Solution {
    const int MOD = 1e9+7;

    vector<vector<int>> directions = {{-2, -1}, {-1, -2}, {-2, 1}, {-1, 2}, {2, -1}, {1, -2}, {2, 1}, {1, 2}};
    vector<vector<int>> dialpad    = {{1, 2, 3},
                                      {4, 5, 6},
                                      {7, 8, 9},
                                      {-1, 0, -1}};
public:
    int knightDialer(int N) {
        // Stores the result value
        int resCount = 0;
        
        // 3D DP table
        vector<vector<vector<int>>> dp(N, vector<vector<int>>(4, vector<int>(3, 0)));

        // Initialize the second edge case: If you've correctly dialed a phone number of length N then you've one valid way
        for(int R = 0; R < 4; ++R)
            for(int C = 0; C < 3; ++C)
                dp[0][R][C] = 1;

        // Fill the rest of the table
        for(int length = 1; length < N; ++length) {
            for(int R = 3; R >= 0; --R) {
                for(int C = 2; C >= 0; --C) {
                    if(dialpad[R][C] != -1) {
                        int count = 0;
                        for(auto& dir : directions) {
                            int reachRow = R + dir[0];
                            int reachCol = C + dir[1];
                            if(reachRow >= 0 && reachCol >= 0 && reachRow < 4 && reachCol < 3 && dialpad[reachRow][reachCol] != -1) {
                                count = (count + dp[length - 1][reachRow][reachCol]) % MOD;
                            }
                        }
                        dp[length][R][C] = count % MOD;
                    }
                }
            }
        }

        // Start dialing from the cells which doesn't starts from '*' or '#'
        for(int R = 0; R < 4; ++R) 
            for(int C = 0; C < 3; ++C) 
                if(dialpad[R][C] != -1) 
                    resCount = (resCount + dp[N - 1][R][C]) % MOD;

        // Return the result value
        return resCount % MOD;
    }
};
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