从小于或等于某个 N 的数字列表中最小化或找到理想的子集和

Question

Given a list of numbers, lst = [1, 1, 2, 4, 3, 2, 1, 1, 1, 2, 1, 4, 3, 1] how would I find an ideal number of lists which are less than or equal to 4?

There are many possibilities here. The goal is to minimize the number of possible lists. The program would need to do create subset lists like the following: {4}, {4}, {3, 1}, ... , {1, 1} .

Notice how the last list subset does not equal to four, but it less. This problem is difficult for the following reasons:

• The program must be able to find subset-sums which are less than or equal to a sum
• The program will needs to begin by removing values out of the original list into subset lists by first looking at the largest values

Answer 1

Here's my attempt. The general idea is to sort the list, and go from left and right and picking the biggest subset every iteration greedily. The time complexity is O(n) .

#include <iostream>
#include <vector>
#include <algorithm>

int main()
{
    std::vector<int> lst{1, 1, 2, 4, 3, 2, 1, 1, 1, 2, 1, 4, 3, 1};
    std::sort(lst.begin(),lst.end()); //sort the list
    int target = 4;

    int left = 0;
    int right = lst.size()-1;

    std::vector<std::vector<int>> solutions;
    while (left<right ){
        if(lst[left] > target)   // break if no solutions
            break;

        if(lst[right] > target) // ignore larger right values
            right--;


        if(lst[right]<=target){   // while the total sum is less than target, keep adding the elements
            std::vector<int> subset;
            subset.push_back(lst[right]);
            int sum = lst[right];
            while(left<right && lst[left]+sum<=target){
                sum+=lst[left];
                subset.push_back({lst[left]});
                left++;
            }
            solutions.push_back(subset);
            right--;
        }

    }

    for(auto& ss : solutions){
        std::cout<<'{';
        for(auto n:ss){
            std::cout<<n<<',';
        }
        std::cout<<"\b}";
        std::cout<<std::endl;
    }

}