{"id":13080,"date":"2020-05-22T15:42:35","date_gmt":"2020-05-22T15:42:35","guid":{"rendered":"https:\/\/merikebi.warrenmyers.com\/?p=13080"},"modified":"2020-05-22T15:42:35","modified_gmt":"2020-05-22T15:42:35","slug":"theexceptioncat-ive-been-through-it-4-times-dont-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-k0-n-a2a1","status":"publish","type":"post","link":"https:\/\/merikebi.warrenmyers.com\/?p=13080","title":{"rendered":"@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)"},"content":{"rendered":"
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I've been through it 4 times<\/p>\n

& don't see how it works<\/p>\n

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:<\/p>\n

for k=0..n
a2[a1[k]]++
if a2[a1[k]] > 1
break<\/p>\n

(unless you're looking for *more* than one duplicate)<\/p>\n

— Warren Myers ?? (@warrenmyers) May 22, 2020<\/a><\/p><\/blockquote>\n