Andrew Welch schrieb:
2008/11/18 Michael Kay <mike(_at_)saxonica(_dot_)com>:
for $d in distinct-values($seq) return $d[count($seq[. eq $d]) ge $i]
They are both O(n^2).
$vSeq[index-of($vSeq,.)[$i]]
...would be O(n^2) for both best and worst cases - right?
With $vSeq being immutable, wouldn't the expression
index-of( $vSeq, $i) be memoized after the first evaluation?
Isn't that one of the advantages of the functional paradigm?
(Not that I'm qualified enough to know this - just asking.)
Would you then still call it O(n^2) or O(n*m) ?
Michael Ludwig
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