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:
nums1 multiplied with the minimum of the selected elements from nums2.(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.length1 <= n <= 1050 <= nums1[i], nums2[j] <= 1051 <= k <= n
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;
}
};
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