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-rw-r--r--other/max_subseq_score.cpp47
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diff --git a/other/max_subseq_score.cpp b/other/max_subseq_score.cpp
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-// 2542. Maximum Subsequence Score
-#include "leetcode.h"
-
-/*
- * given two 0-indexed integer arrays 'num1' 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 '(num1[i0] + nums1[i1] + ... + nums1[ik - 1]) *
- * min(nums2[i0], nums2[i1], ..., nums2[ik - 1])' return the maximum possible
- * score. a subseqnce of indices of an array is a set that can be derived from
- * the set '{0, 1, ..., n - 1}' by deleting no elements
- */
-
-class Solution {
-public:
- long long maxScore(vector<int> &nums1, vector<int> &nums2, int k) {
- int n = nums1.size();
- vector<pair<int, int>> ess(n);
- for (int i = 0; i < n; ++i)
- ess[i] = {nums2[i], nums1[i]};
- sort(rbegin(ess), rend(ess));
- long long sum_s = 0, ans = 0;
- priority_queue<int, vector<int>, greater<int>> pq;
- for (auto &[e, s] : ess) {
- pq.emplace(s);
- sum_s += s;
- if (pq.size() > k) {
- sum_s -= pq.top();
- pq.pop();
- }
- if (pq.size() == k)
- ans = max(ans, sum_s * e);
- }
- return ans;
- }
-};
-
-int main() {
- Solution obj;
- vector<int> n11 = {1, 3, 3, 2}, n21 = {2, 1, 3, 4};
- vector<int> n12 = {4, 2, 3, 1, 1}, n22 = {7, 5, 10, 9, 6};
- printf("%lld\n", obj.maxScore(n11, n21, 3)); // expect: 12
- printf("%lld\n", obj.maxScore(n12, n22, 1)); // expect: 30
-}