You have k bags. You are given a 0-indexed integer array weights where weights[i] is the weight of the ith marble. You are also given the integer k.
Divide the marbles into the k bags according to the following rules:
ith marble and jth marble are in a bag, then all marbles with an index between the ith and jth indices should also be in that same bag.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.
Example 1:
Input: weights = [1,3,5,1], k = 2 Output: 4 Explanation: The distribution [1],[3,5,1] results in the minimal score of (1+1) + (3+1) = 6. The distribution [1,3],[5,1], results in the maximal score of (1+3) + (5+1) = 10. Thus, we return their difference 10 - 6 = 4.
Example 2:
Input: weights = [1, 3], k = 2 Output: 0 Explanation: The only distribution possible is [1],[3]. Since both the maximal and minimal score are the same, we return 0.
Constraints:
1 <= k <= weights.length <= 1051 <= weights[i] <= 109class Solution {
public:
// placebo
long long putMarbles(vector<int>& weights, int k) {
int n = weights.size();
if (k == 1 || n == k) return 0;
vector<int> candidates;
for (int i = 0; i < n-1; i++)
{
candidates.push_back(weights[i] + weights[i+1]);
}
sort(candidates.begin(), candidates.end());
long long mins = 0, maxs = 0;
for (int i = 0; i < k-1; i++)
{
mins += candidates[i];
maxs += candidates[n-2-i];
}
return maxs - mins;
}
};