Sponsored
In this approach, we will use a greedy strategy to partition the weights such that we achieve both the minimum and maximum possible scores for the marble distribution. The key observation here is to leverage sorting of the weights to maximize or minimize the cost of each bag.
Time Complexity: O(n log n) due to sorting the weights array.
Space Complexity: O(1) as we only use a fixed amount of extra space.
1#include <iostream>
2#include <vector>
3#include <algorithm>
4
5int putMarbles(std::vector<int>& weights, int k) {
6 std::sort(weights.begin(), weights.end());
7 int total_min = 0, total_max = 0;
8 for(int i = 0; i < k; ++i) {
9 total_min += weights[i] + weights[weights.size() - 1 - i];
10 total_max += weights[weights.size() - 1 - i] + weights[i];
11 }
12 return total_max - total_min;
13}
14
15int main() {
16 std::vector<int> weights = {1, 3, 5, 1};
17 std::cout << putMarbles(weights, 2) << std::endl;
18 return 0;
19}This C++ solution uses the STL sort function to order the weights and then calculates the min and max scores using the first and last k elements.
The dynamic programming route considers evaluating partial solutions and utilizing them to form the global optimum. Here, DP arrays keep track of the best possible score up to each marble in the sequence. This methodology, while more complex, offers an exhaustive evaluation and solution-guarantee mechanism for larger constraints when direct greedy methods might struggle.
Time Complexity: O(k * n^2) where we analyze subarray splits and their costs.
Space Complexity: O(k * n) to store subproblem solutions.
1
JavaScript's exploratory DP conditionality combines pre-computed data to furnish problem's resultant greatest resolve.