<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 MOD = 1e9 + 7;
    vector<int> getNext(int val){
        vector<int> res;
        switch(val) {
            case 0:
                res = {4,6};
                break;
            case 1:
                res = {6,8};
                break;
            case 2:
                res = {7,9};
                break;
            case 3:
                res = {4,8};
                break;
            case 4:
                res = {0,3,9};
                break;
            case 6:
                res = {0,1,7};
                break;
            case 7:
                res = {2,6};
                break;
            case 8:
                res = {1,3};
                break;
            case 9:
                res = {2,4};
                break;
            break;
                res = {};
        }
        return res;
    }
    int dp[10][5003];
    int solve(int index,int n){
        if(n==1){
            return 1;
        }
        if(dp[index][n]!=-1){
            return dp[index][n];
        }
        int ans=0;
        vector<int> nx = getNext(index);
        for(int j=0;j<nx.size();j++){
            ans=(ans+solve(nx[j],n-1)%MOD)%MOD;
        }

        return dp[index][n]=ans%MOD;
    }
public:
    int knightDialer_bruteforce(int n) {
    
        queue<string> st;
        for(int i=0;i<10;i++){
            st.push(to_string(i));
        }
        while(st.front().size() < n){
            int sz = st.size();
            for(int i=0;i<sz;i++){
                string cur = st.front();
                cout << cur  << " ";
                st.pop();
                int val = cur.back() - '0';
                cout << "val : " << val << endl;
                vector<int> v = getNext(val);
                for(int j=0;j<v.size();j++){
                    string temp = cur;
                    st.push(temp+to_string(v[j]));
                }   
            }
            cout << endl;
        }
        return st.size();
    }
    int knightDialer(int n) {
        int res=0;
        memset(dp,-1,sizeof(dp));
        for(int i=0;i<10;i++){
            res=(res+solve(i,n)%MOD)%MOD;
        }
        return res%MOD;
    }
};
120017