Re: Re: Array lookup

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Re: Re: Re: Array lookup by Limbic~Region (Chancellor) on Apr 13, 2003 at 21:26 UTC
CountZero, I am not sure I follow. The node said all elements (to include duplicates). If add a duplicate in @FirstArray, then there has to be a duplicate in @SecondArray to be valid (which works). If I add an duplicate element in the second array, the substring will just shift. Can you give me an example of a data set where it will not work? I am not saying there isn't one (I am fairly certain there is), but your logic doesn't match what my tests have shown. Cheers - L~R Update: As I was going to bed last night, I figured out when this wouldn't work. If the second array has duplicates that are in the middle of the list (not on either edge).	[reply]
Re: Re: Re: Array lookup by CountZero (Bishop) on Apr 14, 2003 at 19:56 UTC
I was thinking about an element being duplicated a number of times in one of the arrays but not in the other array but I now see that this will depend on how you define "all" in the original question. I understood it as "all duplicates are alike", but if you understand it as "all duplicates are different" then your program works (but all hash-based solutions are broken unless they keep an explicit tally of the number of each individual element). My solution suffers from the same "defect" as yours depending on whether or not you increment the index of the second array after you have found a match. If you don't increment the index, then duplicates are not relevant. Update: Fixed some typos. Thanks L~R for this insight. CountZero "If you have four groups working on a compiler, you'll get a 4-pass compiler." - Conway's Law	[reply]