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1569. Max Dot Product Of Two Subsequences

Array Dynamic Programming

Problem - Max Dot Product Of Two Subsequences

Hard

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

Solutions

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class Solution:
    def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:
        m, n = len(nums1), len(nums2)
        dp = [[float('-inf')] * (n + 1) for _ in range(m + 1)]
        for i, x in enumerate(nums1, 1):
            for j, y in enumerate(nums2, 1):
                v = x * y
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1], max(0, dp[i - 1][j - 1]) + v)
        return dp[-1][-1]

Submission Stats:

  • Runtime: 126 ms (95.48%)
  • Memory: 21.9 MB (30.65%)