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Diffstat (limited to 'other/max_subseq_score.cpp')
-rw-r--r-- | other/max_subseq_score.cpp | 47 |
1 files changed, 0 insertions, 47 deletions
diff --git a/other/max_subseq_score.cpp b/other/max_subseq_score.cpp deleted file mode 100644 index a739d77..0000000 --- a/other/max_subseq_score.cpp +++ /dev/null @@ -1,47 +0,0 @@ -// 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 -} |