<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>

#include<bits/stdc++.h>
class Solution {
public:
int mod = 1e9+7;
int ver[8] = {2, 2, -2, -2, 1, -1, 1, -1};
 int hor[8] = {1, -1, 1, -1, 2, 2, -2, -2};
long long func(int i, int j,vector<vector<vector<long long>>> &dp, int n){
     if(i>3 || i<0) return 0;
    if(j>2 || j<0) return 0;
 if(i==3 && j==2) return 0;
 if(i==3 && j==0) return 0;
    if(n==0) return 1;
    long long ans=0;
 

    if(dp[n][i][j]!=-1) return dp[n][i][j];
   for(int k=0;k<8;k++){
        int newrow  = i+ hor[k];
        int newcol = j+ ver[k];
ans = (ans + func(newrow,newcol,dp,n-1))%mod;
    };
   return dp[n][i][j]= ans;

}
    int knightDialer(int n) {
        vector<vector<vector<long long>>> dp(n,vector<vector<long long>> (4,vector<long long>(4,-1)));
        long long ans=0;
        for(int i=0;i<4;i++){
            for(int j=0;j<3;j++){
                if(i==3 && j==0) continue;
                if(i==3 && j==2) continue;
               
            ans= (ans + func(i,j,dp,n-1))%mod;
            };
        };
        return ans;
    };
};
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