human
What is a the Python code you requested given the following:
I found an interesting problem on https://leetcode.com/problems/max-dot-product-of-two-subsequences/:
Given two arrays nums1 and nums2.
Return the maximum dot product between non-empty subsequences of nums1 and nums2 with the same length.
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, [2,3,5] is a subsequence of [1,2,3,4,5] while [1,5,3] is not).
 
Example 1:
Input: nums1 = [2,1,-2,5], nums2 = [3,0,-6]
Output: 18
Explanation: Take subsequence [2,-2] from nums1 and subsequence [3,-6] from nums2.
Their dot product is (2*3 + (-2)*(-6)) = 18.
Example 2:
Input: nums1 = [3,-2], nums2 = [2,-6,7]
Output: 21
Explanation: Take subsequence [3] from nums1 and subsequence [7] from nums2.
Their dot product is (3*7) = 21.
Example 3:
Input: nums1 = [-1,-1], nums2 = [1,1]
Output: -1
Explanation: Take subsequence [-1] from nums1 and subsequence [1] from nums2.
Their dot product is -1.
 
Constraints:

1 <= nums1.length, nums2.length <= 500
-1000 <= nums1[i], nums2[i] <= 1000

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python code you requested.

import numpy as np
class Solution:
    def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:
        
        # DP(a=index of last, b=index of last) = max of:
        #   DP(a-1, b)
        #   DP(a-1, i) + nums1[a] * max_or_min(nums2[i+1:b+1])
        #   same for b
        
        INF = int(1e9)
        n, m = len(nums1), len(nums2)

        DP = np.ones((n, m), dtype=int) * -INF
        DP[0, 0] = nums1[0] * nums2[0]
        
        for a in range(n):
            for b in range(m):
                
                el = nums1[a] * nums2[b]
                curr = el
                
                if a > 0:
                    curr = max(curr, DP[a - 1, b])
                if b > 0:
                    curr = max(curr, DP[a, b - 1])
                if a > 0 and b > 0:
                    prev = DP[a - 1, b - 1]
                    curr = max(curr, prev, prev + el)
                    
                DP[a, b] = curr
                    
        return DP[-1, -1]