<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>

int m  = 1e9+7;
class Solution {
public:
int f(int cnt,int n,vector<int>&dr,vector<int>&dc,int r,int c, vector<vector<vector<int>>> &dp){
if (r < 0 || r >= 4 || c < 0 || c >= 3 || (r == 3 && c == 0) || (r == 3 && c == 2)) return 0;
if(cnt==n)return 1;
if(dp[r][c][cnt]!=-1)return dp[r][c][cnt];
int ans =0;
for(int i=0;i<8;i++){
    int nr = r + dr[i];
    int nc = c + dc[i];
   ans = (ans%m +  f(cnt+1,n,dr,dc,nr,nc,dp)%m)%m;
}
return dp[r][c][cnt]= ans;
}
    int knightDialer(int n) {
       vector<int>dr={-2,-2,-1,1,2,2,1,-1};
       vector<int>dc={-1,1,2,2,1,-1,-2,-2};
       vector<vector<vector<int>>> dp(6,vector<vector<int>>(5,vector<int>(n,-1)));
    int res=0;
    for(int r=0;r<4;r++){
        for(int c = 0;c<3;c++){
            if(r==3 && c==0)continue;
            if(r==3 && c==2)continue;
            res = (res%m + f(1,n,dr,dc,r,c,dp)%m)%m;
        }
    }
    return res;
    }
};
58997