@theexceptioncat I’ve been through it 4 times & don’t see how it works I would build a 2nd array where the value of a1 is the index of a2, & you stop as soon as you get a duplicate: for k=0..n a2[a1[k]]++ if a2[a1[k]] > 1 break (unless you’re looking for *more* than one duplicate)

I've been through it 4 times

& don't see how it works

I would build a 2nd array where the value of a1 is the index of a2, & you stop as soon as you get a duplicate:

for k=0..n
a2[a1[k]]++
if a2[a1[k]] > 1
break

(unless you're looking for *more* than one duplicate)

— Warren Myers ?? (@warrenmyers) May 22, 2020

from Twitter https://twitter.com/warrenmyers

May 22, 2020 at 10:46AM
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