HackMD
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
