LeetCode Challenge #2542. 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.

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`
Video Solution
C++ Solution
```				```
class Solution {
public:
long long maxScore(vector<int>& nums1, vector<int>& nums2, int k) {
vector<pair<int,int>> pairs;

int n=nums1.size();

for(int i=0; i<n; i++)
{
pairs.push_back({nums2[i],nums1[i]});
}
sort(pairs.begin(),pairs.end());
reverse(pairs.begin(),pairs.end());

priority_queue<int,vector<int>,greater<int>> pq;

long long sum=0;
long long ans=0;

for(int i=0; i<n; i++)
{
pq.push(pairs[i].second);
sum+=pairs[i].second;

if(pq.size()>k)
{
sum-=pq.top();
pq.pop();
}

if(pq.size()==k) ans= max(ans,(long long)sum*pairs[i].first);
}

return ans;

}
};
```
```