2542. Maximum Subsequence Score

Description

You are given two 0-indexed integer arrays nums1 and nums2 of equal length n and a positive integer k. You must choose a subsequence of indices from nums1 of length k.

For chosen indices i₀, i₁, ..., i_{k - 1}, your score is defined as:

The sum of the selected elements from nums1 multiplied with the minimum of the selected elements from nums2.
It can defined simply as: (nums1[i₀] + nums1[i₁] +...+ nums1[i_{k - 1}]) * min(nums2[i₀] , nums2[i₁], ... ,nums2[i_{k - 1}]).

Return the maximum possible score.

A subsequence of indices of an array is a set that can be derived from the set {0, 1, ..., n-1} by deleting some or no elements.

Example 1:

Input: nums1 = [1,3,3,2], nums2 = [2,1,3,4], k = 3
Output: 12
Explanation: 
The four possible subsequence scores are:
- We choose the indices 0, 1, and 2 with score = (1+3+3) * min(2,1,3) = 7.
- We choose the indices 0, 1, and 3 with score = (1+3+2) * min(2,1,4) = 6. 
- We choose the indices 0, 2, and 3 with score = (1+3+2) * min(2,3,4) = 12. 
- We choose the indices 1, 2, and 3 with score = (3+3+2) * min(1,3,4) = 8.
Therefore, we return the max score, which is 12.

Example 2:

Input: nums1 = [4,2,3,1,1], nums2 = [7,5,10,9,6], k = 1
Output: 30
Explanation: 
Choosing index 2 is optimal: nums1[2] * nums2[2] = 3 * 10 = 30 is the maximum possible score.

Constraints:

n == nums1.length == nums2.length
1 <= n <= 10⁵
0 <= nums1[i], nums2[j] <= 10⁵
1 <= k <= n

Solution

maximum-subsequence-score.py

class Solution:
    def maxScore(self, nums1: List[int], nums2: List[int], k: int) -> int:
        N = len(nums1)
        A = sorted([(x, mmin) for x, mmin in zip(nums1, nums2)], key = lambda x : -x[1])
        heap = []
        res = 0
        total = 0

        for x, mmin in A:
            heappush(heap, x)
            total += x

            if len(heap) > k:
                total -= heappop(heap)

            if len(heap) == k:
                res = max(res, mmin * total)

        return res