<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>

r'''
# Requirements

m = len(nums1), n = len(nums2)
1 <= m, n <= 500
-1000 <= nums1[i], nums2[i] <= 1000

# Edge cases

1. Max dot product is negative

# Solutions

1. Brute force, backtracking
    - O(3^(m + n)) time, O(m + n) auxiliary space
2. Memo / DP
    - O(mn) time, O(mn) auxiliary space
2. DP, optimized space
    - O(mn) time, O(min(m, n)) auxiliary space
'''
def max_dot_product_memo(nums1: list[int], nums2: list[int]) -> int:
    m, n = len(nums1), len(nums2)

    @lru_cache(maxsize = (m + 1) * (n + 1))
    def max_dot_product(l1: int, l2: int) -> int:
        if l1 == m or l2 == n:
            return -float('inf')
        
        return max(
            nums1[l1] * nums2[l2] + max(0, max_dot_product(l1 + 1, l2 + 1)),
            max_dot_product(l1 + 1, l2),
            max_dot_product(l1, l2 + 1)
        )
        
    return max_dot_product(0, 0)

def max_dot_product_dp(nums1: list[int], nums2: list[int]) -> int:
    m, n = len(nums1), len(nums2)

    # Optimize space complexity
    if m < n:
        return max_dot_product_dp(nums2, nums1)

    # DP
    dp1 = [-float('inf')] * (n + 1)

    for l1 in range(m - 1, -1, -1):
        dp0 = [0] * (n + 1)
        dp0[n] = -float('inf')

        for l2 in range(n - 1, -1, -1):
            dp0[l2] = max(
                nums1[l1] * nums2[l2] + max(0, dp1[l2 + 1]),
                dp1[l2],
                dp0[l2 + 1]
            )
        dp1 = dp0

    return dp1[0]

class Solution:
    def maxDotProduct(self, nums1: list[int], nums2: list[int]) -> int:
        return max_dot_product_dp(nums1, nums2)
16700