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Diffstat (limited to 'marbles_in_bags.cpp')
-rw-r--r-- | marbles_in_bags.cpp | 48 |
1 files changed, 48 insertions, 0 deletions
diff --git a/marbles_in_bags.cpp b/marbles_in_bags.cpp new file mode 100644 index 0000000..4640eb9 --- /dev/null +++ b/marbles_in_bags.cpp @@ -0,0 +1,48 @@ +// 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 +} |