<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:
    long long mod=1e9+7;
    bool isSafe(int i,int j,vector<vector<char>>&board)
    {
        return i>=0 && j>=0 && i<board.size() && j<board[0].size() && board[i][j]>='0' && board[i][j]<='9';
    }
    long long helper(int n,vector<vector<char>>&board,int i,int j,vector<vector<vector<long long>>>&m)
    {
        if(n==1)
        {
            return m[n][i][j]=1;
        }
        if(m[n][i][j]!=-1)
        {
            return m[n][i][j]%mod;
        }
        int dx[8]={-2,-2,-1,1,2,2,1,-1};
        int dy[8]={-1,1,2,2,1,-1,-2,-2};
        long long ans=0;
        for(int k=0;k<8;k++)
        {
            int nextX=i+dx[k],nextY=j+dy[k];
            if(isSafe(nextX,nextY,board))
            {
                if(m[n-1][nextX][nextY]==-1)
                {
                    m[n-1][nextX][nextY]=helper(n-1,board,nextX,nextY,m);
                }
                ans=(ans+m[n-1][nextX][nextY])%mod;
            }
        }
        return m[n][i][j]=ans;
    }
    int knightDialer(int n) 
    {
        vector<vector<char>>board={{'1','2','3'},{'4','5','6'},{'7','8','9'},{'*','0','#'}};
        long long ans=0;
        vector<vector<vector<long long>>>dp(n+1,vector<vector<long long>>(4,vector<long long>(3,-1)));
        for(int i=0;i<board.size();i++)
        {
            for(int j=0;j<board[0].size();j++)
            {
                if(board[i][j]>='0' && board[i][j]<='9')
                {
                    ans=(ans+helper(n,board,i,j,dp))%mod;
                }
            }
        }
        return ans;
    }
};
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