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Apr 10, 2020 at 1:18 vote accept thecommexokid
Apr 10, 2020 at 1:17 comment added thecommexokid Basing a solution around IntegerPartitions was absolutely the way to go for me, so this one gets the Accept for my personal use-case.
Apr 9, 2020 at 23:59 comment added ciao @thecommexokid - basically.
Apr 9, 2020 at 18:20 comment added thecommexokid Just to make sure I'm understanding the approach: You compute all possible unique multisets of the distinct elements, then reject the ones that exceed the actual multiplicities?
Apr 9, 2020 at 16:53 comment added Marius Ladegård Meyer If you want speed, ciao usually provides! +1
Apr 9, 2020 at 16:52 history answered ciao CC BY-SA 4.0