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Home/ Questions/Q 1076577
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T21:26:53+00:00 2026-05-16T21:26:53+00:00

I have a list of items that is somewhat like this: [ [orange, 9],

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I have a list of items that is somewhat like this:

[
  ["orange", 9],
  ["watermelon", 3],
  ["grapefruit", 6],
  ["peach", 8],
  ["durian", 2],
  ["apricot", 6]
]

I would like to split this list into… say two groups so that the sum of the weights of the items in each group are as equal as possible, i.e.:

List 1:
  orange: 9
  durian: 2
  apricot: 6
  TOTAL: 17

List 2:
  watermelon: 3
  grapefruit: 6
  peach: 8
  TOTAL: 17

Currently I’m solving this by traversing the ordered list in a zigzag’esque way. Assigning the items with more weight in the first pass to each group, assiging the items with less weight on the second pass, and so on.

This works ok, but it has it flaws. I’m thinking that a second pass on the groups in which I exchange items between them would result in more equally distributed groups, but the code involved could become too complicated.

Does someone know of a more efficient or intelligent way to do this?

Thanks!

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  1. Editorial Team
    Editorial Team
    2026-05-16T21:26:54+00:00Added an answer on May 16, 2026 at 9:26 pm

    This is the optimization version of the partition problem, which is NP-complete, although, according to that article, “there are heuristics that solve the problem in many instances, either optimally or approximately.”

    The methods section of that article contains a number of ways to do approximate or pseudo-polynomial solutions. Specifically, if you can live with an approximate answer, you could try the greedy approach:

    One approach to the problem, imitating the way children choose teams for a game, is the greedy algorithm, which goes through the numbers in descending order, assigning each of them to whichever subset has the smaller sum.

    The running time of this approach is O(n^2) and is guaranteed to get you to within a factor of 4/3 of the exact solution.

    If you must have an exact solution and your data set is small enough, you could always take a brute force approach of generating every possibility (this is exponential, O(2^n)). Depending on your performance needs, this might be sufficient.

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