[Jerrrry]

Your second example subtracts and adds 1 nearly arbitrarily, which wouldn't be needed if the convention of the 0-index wasn't so widespread.

[umanwizard]

You need the first three elements to go into the first group, the next three to go into the second group, and so on. How would you write it?

[binary132]

That’s not what that loop does, it puts one item into each next group and loops back over the groups after every three items. Really it ought to be a by-three stepped loop over items, inserting each into each group inline:

groups[1], groups[2], groups[3] = items[i], items[i+1], items[i+2]

If the group count is dynamic, you can just loop over groups instead, and then step through items by #groups, inserting.

[Benjamin_Dobell]

If it is dynamic one of the loops will also suffer from an off-by-one issue. You can't add 1-based indices together like you can zero-based indices.

It's also worth noting your solution exhibits similar off-by-one behaviour. The left hand side constants (integer values) do not match the right. It's error prone.

[Jerrrry]

  >You can't add 1-based indices together like you can zero-based indices.

I think you are right but I am unable to articulate why. But I think 0 based indexes are able to fruitfully capture "going nowhere" iteratively than 1 based indexes, which do require a decrement in that circumstance.

[binary132]

sorry, where is the off by one? the code offered is of course only a solution for the fixed-size groups

[umanwizard]

Fair enough, for some reason I thought it was i/3 and not i%3. Still, I think the point stands.