I have N objects, each of which has a weight. I need to form combinations of the objects to maximize how many sets of objects add up to at least x total weight. Combinations can consist of any number of objects.

1). What is a fast and correct algorithm to solve this?
2). Is this NP-complete, hard, or neither?

Ideally, I'd want the left over set of objects that does not add up to x to be maximal as well, but would settle for an understanding of how to solve the problem without this.

  • $\begingroup$ If I understand correctly the input is a collection of $N$ (natural?) numbers $x_1, x_2, \dots, x_N$, plus an additional (natural?) number $X$. Can you clarify what the expected output is? $\endgroup$
    – Steven
    Nov 28, 2020 at 23:57
  • $\begingroup$ Yes, apologies. So lets say I have input (2, 7, 8, 3, 11) and X = 10. A possible output is (7, 3), (8, 2), (11). This solution has 3 subsets that achieve a weight >=x, and the goal is to have a solution that maximizes that number--the number of such subsets. So, (8,3) (11) would be suboptimal because (7,2) is not >= x so I'm one subset shy of the optimal number. $\endgroup$
    – wbarts
    Nov 29, 2020 at 0:05
  • $\begingroup$ So, just you are looking for a partition of \{x_1, \dots, x_n\} into subsets $S_1, S_2, \dots$ such that the number of subsets $S_i$ satisfying $\sum_{x \in S_i} x \ge X$ is maximized. Is this correct? $\endgroup$
    – Steven
    Nov 29, 2020 at 0:07
  • $\begingroup$ yes! That's exactly right. And I'd like for the size of the remainder to be maximal as well if that's possible. $\endgroup$
    – wbarts
    Nov 29, 2020 at 0:09
  • 1
    $\begingroup$ Please edit the question to clarify it based on the feedback. Don't just put clarifications in the comments: we want the question to be understandable to future readers without having to read the comments. Part of our mission is to build up an archive of high-quality questions and answers that will be useful not only to you but also to others in the future. Thank you! $\endgroup$
    – D.W.
    Nov 29, 2020 at 0:47

1 Answer 1


This problem is NP-hard as it can be seen by a reduction from $3$-partition.

In the $3$-partition problem we are given a (multi-)set $\mathcal{S}$ containing $3n$ positive integers $x_1, \dots, x_{3n}$ and the goal is that of deciding whether there exists a partition of $\mathcal{S}$ into $n$ sets $S_1, \dots, S_n$ of $3$ elements each, such that, for each $i$, $\sum_{x \in S_i} x = T$, where $T = \frac{1}{n} \sum_{i=1}^{3n} x_i$.

It is known that this problem remains NP-hard even when each $x_i$ is strictly between $T/4$ and $T/2$. We will therefore restrict to such instances.

To obtain an instance of your problem, it suffices to use the same set of integers $\{ x_1, \dots, x_{3n} \}$ (i.e., $N=3n$) with $X = T$. Clearly, since the sum of any two input elements can be at most $2 \cdot (\frac{T}{2}-1) < T = X$, we have that, in any partition $\mathcal{P}$ of $\{ x_1, \dots, x_{3n} \}$, the number of sets $S \in \mathcal{P}$ such that $\sum_{x \in S} x \ge X$ can be at most $N/3 = n$.

If the answer to the 3-partition instance is "yes", then the same partition $S_1, \dots, S_n$ exactly matches this upper bound in the instance of your problem. On the other hand, if the answer to the 3-partition instance is "no", then in any partition of $\{ x_1, \dots, x_{3n} \}$ there is at most one set with sum smaller than $T$, i.e., there are at most $n-1$ sets with sum at least $T$.

To summarize: the answer to the instance of $3$-partition is yes if and only if the optimal solution to the corresponding instance of your problem has measure at least $n$.

Your problem is not NP-complete since it does not belong to NP (since it is not a decision problem). Unless P=NP there is no efficient (polynomial-time) algorithm for your problem.


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