<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:
    int solve(int n,int jumps,unordered_map<int,vector<int>>ump,vector<vector<int>>&dp){
        if(jumps==1){return 1;}
        int mod=1e9+7;
        if(dp[n][jumps]!=-1){return dp[n][jumps];}
        int x=0;
        for(auto y:ump[n]){
            x+=solve(y,jumps-1,ump,dp)%mod;
            x%=mod;
        }
        return dp[n][jumps]=x;
    }
    int knightDialer(int n) {
        int mod=1e9+7;
        unordered_map<int,vector<int>>ump;
        ump[0]={4,6};
        ump[1]={6,8};
        ump[2]={7,9};
        ump[3]={4,8};
        ump[4]={0,3,9};
        ump[5]={};
        ump[6]={0,1,7};
        ump[7]={2,6};
        ump[8]={1,3};
        ump[9]={2,4};
        vector<vector<int>>dp(5001,vector<int>(10,0));
        //int ans=0;
        // for(int i=0;i<10;i++){
        //     ans+=solve(i,n,ump,dp)%mod;
        //     ans%=mod;
        // }
        //return ans;
        for(int i=0;i<10;i++){
            dp[1][i]=1;
        }
        //jumps==n
        int res=0;
        for(int j=1;j<=n;j++){
            for(int i=0;i<10;i++){
                for(auto x:ump[i]){
                    dp[j][i]+=dp[j-1][x]%mod;
                    dp[j][i]%=mod;
                }
                if(j==n){
                    res+=dp[j][i];
                    res%=mod;
                }
            }
        }
        return res;
    }
};
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