Today I have done leetcode's May LeetCoding Challenge with cpp.

May LeetCoding Challenge 24

Description

Maximum Subsequence Score

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.

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

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
  using pi = pair<int, int>;
public:
  long long maxScore(vector<int>& nums1, vector<int>& nums2, int k) {
    vector<pi> pairs;
    pairs.reserve(nums1.size());
    for(int i = 0; i < nums1.size(); ++i) {
      pairs.push_back({nums2[i], nums1[i]});
    }
    sort(pairs.begin(), pairs.end(), greater<pi>());
    priority_queue<int, vector<int>, greater<int>> pq;
    long long sum = 0;
    int count = 0;
    long long answer = 0;
    for(const auto &[factor, num] : pairs) {
      pq.push(num);
      count += 1;
      sum += num;
      if(count > k) {
        sum -= pq.top();
        pq.pop();
        count -= 1;
      }
      if(count == k) {
        answer = max(answer, sum * factor);
      }
    }
    return answer;
  }
};

// Accepted
// 28/28 cases passed (205 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 95.07 % of cpp submissions (87.2 MB)