<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 int mod=1e9+7;
    int ans(int i,int j,int cnt,vector<vector<vector<int>>>&dp,vector<vector<char>>&grid,int n){
        if((grid[i][j]=='*')||(grid[i][j]=='#')){
            return 0;
        }
        if(cnt==n){
            return 1;
        }
        
        if(dp[i][j][cnt]!=-1){
            return dp[i][j][cnt];
        }
        int del_row[]={-2,-1,1,2,2,1,-1,-2};
        int del_col[]={1,2,2,1,-1,-2,-2,-1};
        int f=0;
        for(int k=0;k<8;k++){
           int nrow=i+del_row[k];
           int ncol=j+del_col[k];
           if((nrow>=0)&&(ncol>=0)&&(nrow<4)&&(ncol<3)){
             f=((f%mod)+(ans(nrow,ncol,cnt+1,dp,grid,n)%mod))%mod;
           }
        }
        return dp[i][j][cnt]=f%mod;
    }
    int knightDialer(int n) {
        vector<vector<char>>mat(4,vector<char>(3,'0'));
        vector<vector<vector<int>>>dp(4+1,vector<vector<int>>(3+1,vector<int>(n+1,-1)));
        for(int i=0;i<3;i++){
            for(int j=0;j<3;j++){
              int val=j+1+(i*3);
              mat[i][j]=val+'0';
            }
        }
        mat[3][0]='*';
        mat[3][1]='0';
        mat[3][2]='#';
        int count=0;
        for(int i=0;i<4;i++){
            for(int j=0;j<3;j++){
                count=(count+ans(i,j,1,dp,mat,n))%mod;
            }
        }
        return count;
    }
};
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