2542.Maximum Subsequence Score

tags: Medium,Array,Greedy,Heap,Sorting

2542. Maximum Subsequence Score

題目描述

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, …, ik - 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[ik - 1]) * min(nums2[i0] , nums2[i1], ... ,nums2[ik - 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.

範例

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

解答

C++

class Solution {
public:
    long long maxScore(vector<int>& nums1, vector<int>& nums2, int k) {
        ios_base::sync_with_stdio(0); cin.tie(0);
        int n = nums1.size();
        vector<pair<int, int>> numPairs;
        for (int i = 0; i < n; i ++) {
            numPairs.push_back({nums1[i], nums2[i]});
        }
        sort(numPairs.begin(), numPairs.end(), [](const auto p1, const auto p2) {
            return p1.second > p2.second;
        });
        priority_queue<int, vector<int>, greater<int>> topkHeap;
        long long topkSum = 0;
        for (int i = 0; i < k; i ++) {
            topkSum += numPairs[i].first;
            topkHeap.push(numPairs[i].first);
        }
        long long ans = topkSum * numPairs[k - 1].second;
        for (int i = k; i < n; i ++) {
            topkSum += numPairs[i].first - topkHeap.top();
            topkHeap.pop();
            topkHeap.push(numPairs[i].first);
            ans = max(ans, topkSum * numPairs[i].second);
        }
        return ans;
    }
};

Jerry WuWed, May24, 2023

Python

class Solution:
    def maxScore(self, nums1: List[int], nums2: List[int], k: int) -> int:
        ans = total = 0
        heap = []
        data = sorted(zip(nums1, nums2), key=lambda x: -x[1])
        for n1, n2 in data:
            total += n1
            heappush(heap, n1)
            if len(heap) > k:
                total -= heappop(heap)
            if len(heap) == k:
                ans = max(ans, total * n2)
        return ans

Yen-Chi ChenWed, May 24, 2023

class Solution:
    def maxScore(self, nums1: List[int], nums2: List[int], k: int) -> int:
        pairs = sorted(zip(nums1, nums2), key=lambda x: x[1], reverse=True)
        heap = []
        n1Sum, res = 0, 0

        for n1, n2 in pairs:
            n1Sum += n1
            heapq.heappush(heap, n1)

            if len(heap) > k:
                n1Sum -= heappop(heap)

            if len(heap) == k:
                res = max(res, n1Sum * n2)

        return res

Ron ChenWed, May 24, 2023

Reference

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