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Diffstat (limited to 'marbles_in_bags.cpp')
-rw-r--r-- | marbles_in_bags.cpp | 48 |
1 files changed, 0 insertions, 48 deletions
diff --git a/marbles_in_bags.cpp b/marbles_in_bags.cpp deleted file mode 100644 index 4640eb9..0000000 --- a/marbles_in_bags.cpp +++ /dev/null @@ -1,48 +0,0 @@ -// 2551. Put Marbles in Bags -#include "leetcode.h" - -/* - * you have 'k' bags. you are given a 0-indexed integer array 'weights' where - * 'weights[i]' is the weight of the i'th marble. you are also given the integer - * 'k'. divide the marbles into the 'k' bags according to the following rules - * - no bag is empty - * - if the i'th marble and the j'th marble are in a bag, then all the marbles - * with an index between the i'th and j'th indices should also be in that same - * bag. - * - if a bag onsists of all the marbles with an index from 'i' to 'j' - * inclusively, then the cost of the bag is 'weights[i] + weights[j]' the score - * after distributing the marbles is the sum of the costs of all the 'k' bags. - * return the difference between the maximum and minimum scores among marble - * distributions - */ - -class Solution { -public: - long long putMarbles(vector<int> &weights, int k) { - priority_queue<int> pq1; - priority_queue<int, vector<int>, greater<int>> pq2; - for (int i = 0; i < weights.size() - 1; i++) { - pq1.push(weights[i] + weights[i + 1]); - pq2.push(weights[i] + weights[i + 1]); - if (pq1.size() > k - 1) { - pq1.pop(); - pq2.pop(); - } - } - long long ans1 = 0, ans2 = 0; - while (!pq1.empty()) { - ans1 += pq1.top(); - ans2 += pq2.top(); - pq1.pop(); - pq2.pop(); - } - return ans2 - ans1; - } -}; - -int main() { - Solution obj; - vector<int> w1 = {1, 3, 5, 1}, w2 = {1, 3}; - printf("%lld\n", obj.putMarbles(w1, 2)); // expect: 4 - printf("%lld\n", obj.putMarbles(w2, 2)); // expect: 0 -} |