<p>Given two arrays <code>nums1</code>&nbsp;and <code><font face="monospace">nums2</font></code><font face="monospace">.</font></p>

<p>Return the maximum dot product&nbsp;between&nbsp;<strong>non-empty</strong> subsequences of nums1 and nums2 with the same length.</p>

<p>A subsequence of a array is a new array which is formed from the original array by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie,&nbsp;<code>[2,3,5]</code>&nbsp;is a subsequence of&nbsp;<code>[1,2,3,4,5]</code>&nbsp;while <code>[1,5,3]</code>&nbsp;is not).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [2,1,-2,5], nums2 = [3,0,-6]
<strong>Output:</strong> 18
<strong>Explanation:</strong> Take subsequence [2,-2] from nums1 and subsequence [3,-6] from nums2.
Their dot product is (2*3 + (-2)*(-6)) = 18.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [3,-2], nums2 = [2,-6,7]
<strong>Output:</strong> 21
<strong>Explanation:</strong> Take subsequence [3] from nums1 and subsequence [7] from nums2.
Their dot product is (3*7) = 21.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [-1,-1], nums2 = [1,1]
<strong>Output:</strong> -1
<strong>Explanation: </strong>Take subsequence [-1] from nums1 and subsequence [1] from nums2.
Their dot product is -1.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 500</code></li>
	<li><code>-1000 &lt;= nums1[i], nums2[i] &lt;= 1000</code></li>
</ul>

class Solution {
public:
    int dp[501][501][2];
    int help(int i, int j, vector<int>& v1, vector<int>& v2 , int l){
        if(i==v1.size() || j==v2.size()){
            if(l==1)return 0;
            else return -1;
        }
        if(dp[i][j][l]!= INT_MIN) return dp[i][j][l];
        else{
            int op1=v1[i]*v2[j]+help(i+1,j+1,v1,v2,1);
            int op2=help(i+1,j,v1,v2,l);
            int op3=help(i,j+1,v1,v2,l);
            return dp[i][j][l]= max(op1,max(op2,op3));
        }
    }
    int maxDotProduct(vector<int>& v1, vector<int>& v2) {
        for(int i=0;i<v1.size();i++)
        {
            for(int j=0;j<v2.size();j++)
            {
                dp[i][j][0]=INT_MIN;
                dp[i][j][1]=INT_MIN;
            }
        }
        int ans=help(0,0,v1,v2,0);
        if(ans == -1)
        {
            sort(v1.begin(),v1.end());
            sort(v2.begin(),v2.end());
            return max(v1.front()*v2.back(),v2.front()*v1.back());
        }
        else return ans;
    }
};
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