<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>

from typing import List

class Solution:
    def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:
        n1 = len(nums1)
        n2 = len(nums2)
        
        # Initialize the DP table
        dp = [[[-float('inf')] * 2 for _ in range(n2)] for _ in range(n1)]
        
        # Fill the DP table
        for i in range(n1):
            for j in range(n2):
                current_product = nums1[i] * nums2[j]
                # When k = 1 (including the sum of i, j)
                dp[i][j][1] = current_product
                if i > 0 and j > 0:
                    dp[i][j][1] = max(dp[i][j][1], dp[i-1][j-1][1] + current_product)
                if i > 0:
                    dp[i][j][1] = max(dp[i][j][1], dp[i-1][j][1])
                if j > 0:
                    dp[i][j][1] = max(dp[i][j][1], dp[i][j-1][1])
                
                # When k = 0 (not including the sum of i, j)
                if i > 0 and j > 0:
                    dp[i][j][0] = max(dp[i][j][0], dp[i-1][j-1][0])
                if i > 0:
                    dp[i][j][0] = max(dp[i][j][0], dp[i-1][j][0])
                if j > 0:
                    dp[i][j][0] = max(dp[i][j][0], dp[i][j-1][0])
                # Include the case where we start a new subarray
                dp[i][j][0] = max(dp[i][j][0], 0)
                dp[i][j][1] = max(dp[i][j][1], dp[i][j][0] + current_product)

        return dp[n1-1][n2-1][1]

38600