<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 {
    int findMov( int i, int n , int& ans, vector<vector<int>>& next, vector<vector<int>>& dp){
        if(n==0){
            return 1 ;
        }
        if(dp[i][n]!=-1){
            return dp[i][n] ;            
        }

        int temp =0;
        for(auto el: next[i] ){
            temp = (temp+findMov(el,n-1,ans,next,dp)) % 1000000007;
        }
        return dp[i][n] = temp;
    }
public:
    int knightDialer(int n) {
        vector<vector<int>> next = { {4,6},{6,8},{7,9},{4,8},{0,3,9},{},{1,7,0},{2,6},{1,3},{2,4}} ;
        vector<vector<int>> dp(10,vector<int>(n,0));
        for( int i=0; i<10; i++ ){
            dp[i][0] = 1;
        }
        for( int j=1; j<n; j++ ){
            for (int i=0; i<10; i++){
                int temp =0;
                for(auto el: next[i] ){
                    temp = (temp+dp[el][j-1]) % 1000000007;
                }
                dp[i][j] = temp;
            }
        }

        int ans = 0;
        for ( int i=0; i<10; i++ ){
            ans = (ans+dp[i][n-1]) % 1000000007;
        }

        return ans;

        // memoiazation............
        // vector<vector<int>> dp(10,vector<int>(n+1,-1));
        // vector<vector<int>> next = { {4,6},{6,8},{7,9},{4,8},{0,3,9},{},{1,7,0},{2,6},{1,3},{2,4}} ;
        // int ans = 0 ;
        // for( int i=0; i<10; i++){
        //     ans = (ans+findMov(i,n-1,ans,next,dp)) % 1000000007; 
        // }
        // return ans;
    }
};
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