<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 solve(int i,int j,vector<int>& nums1, vector<int>& nums2,vector<vector<int>>& dp)
    {
        if(i>=nums1.size()||j>=nums2.size())
        return 0;
        if(dp[i][j]!=-1)
        return dp[i][j];
        int op1=0,op2=0,op3=0,op4=0;
        op1=0+solve(i+1,j,nums1,nums2,dp); 
        op2=0+solve(i,j+1,nums1,nums2,dp); 
        op3=0+solve(i+1,j+1,nums1,nums2,dp); 
        op4=nums1[i]*nums2[j]+solve(i+1,j+1,nums1,nums2,dp); 
        return dp[i][j]=max(op1,max(op2,max(op3,op4)));
    }
    int maxDotProduct(vector<int>& nums1, vector<int>& nums2) {
        int n=nums1.size();int m=nums2.size();
        vector<vector<int>>dp(n+1,vector<int>(m+1,-1));
        int max1=INT_MIN,max2=INT_MIN;
        int min1=INT_MAX,min2=INT_MAX; 
        for(int i=0;i<nums1.size();i++)
        {
            max1=max(nums1[i],max1);
            min1=min(nums1[i],min1);
        }
        
        for(int i=0;i<nums2.size();i++)
        {
            max2=max(nums2[i],max2);
            min2=min(nums2[i],min2);
        }
        
        int ans= solve(0,0,nums1,nums2,dp);
        if(ans!=0)
        return ans;
        return max(max1*max2,max(min1*min2,max(min1*max2,min2*max1)));
    }
};
16300