<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:
        n1, n2 = len(nums1), len(nums2)

        # Handle special cases
        max_1, min_1 = max(nums1), min(nums1)
        max_2, min_2 = max(nums2), min(nums2)
        # When all the values in 1 are negative and all the ones in the
        # other are positive, then we will return a negative value by
        # minimizing the product of just 2 values
        if max_1 < 0 and min_2 > 0: # All nums1 < 0 and all nums2 > 0
            return max_1 * min_2 # The least negative nums1 and smallest nums2
        if max_2 < 0 and min_1 > 0: # All nums2 < 0 and all nums1 > 0
            return max_2 * min_1 # The least negative nums2 and the smallest nums1

        dp_cache = {}

        def dp(i:int, j:int) -> int:
            ans = dp_cache.get((i, j), None)
            if i >= n1 or j >= n2:
                return -float("inf")
            
            if ans is None: # Then compute on-the-fly
                # The decision is whether to use the (i, j) pairing or not
                # This problem is very similar to the Longest Common Subseq
                # problem we say before.

                # We can either use it or not. If we do, then our best
                # result would be nums1[i]*nums2[j] + dp[i, j] the product
                # of these 2 elements and the best of what comes next. We
                # will wrap this second part is a min(0, *) wrapper since
                # we need a non-empty subseq and since we are using the pair
                # (i, j) then it is non-empty and don't need to use any future
                # pairings. For the base-cases, we return -inf since they
                # are not options that we should choose.

                # We can also consider the best options that we have by not
                # using one or the other. We consider skipping index i or
                # index j and starting from the next index instead
                ans = max(max(0, dp(i+1, j+1)) + nums1[i]*nums2[j], dp(i+1, j), dp(i, j+1))
                dp_cache[(i, j)] = ans
            return ans

        return dp(0, 0)

        
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