1458. Max Dot Product of Two Subsequences

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

对两个数组问从中各选一个子串，满足它俩积最大且长度一样（并>=1）时积是多少。优化 + subseq => DP。一共M * N步，每步action为1. 选i和j 2. 选i或选j。dp[i][j]表前i和前j元素组成的子串满足条件时最大积。都选时最大为dp[i][j]+=max(dp[i -1][j -1],0)，选一个时最大为dp[i -1][j])或dp[i][j -1])。

代码参考了这。

class Solution:
    def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:
        M, N = len(nums1), len(nums2)
        dp = [[0] * N for _ in range(M)]
        for i in range(M):
            for j in range(N):
                dp[i][j] = nums1[i] * nums2[j]
                if i and j: dp[i][j] += max(dp[i - 1][j - 1], 0)
                if i: dp[i][j] = max(dp[i][j], dp[i - 1][j])
                if j: dp[i][j] = max(dp[i][j], dp[i][j - 1])
        return dp[-1][-1]