<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:
    def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:
        dp = {}

        def dfs(p1, p2):
            if (p1, p2) in dp:
                return dp[(p1, p2)]

            if p1 == len(nums1) or p2 == len(nums2):
                return 0

            best = max(
                dfs(p1 + 1, p2), 
                dfs(p1, p2 + 1),
                dfs(p1 + 1, p2 + 1) + nums1[p1] * nums2[p2]
            )

            dp[(p1, p2)] = best
            return best

        best = -float('inf')
        for i in range(len(nums1)):
                for j in range(len(nums2)):
                    best = max(best, dfs(i + 1, j + 1) + nums1[i] * nums2[j])
        return best
93400