this theorem is true for comparision sorts only! counting sort is not a
comparison sort.
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There is a theorem which states that a sorting algorithm must take o(nlogn) to
sort a list of any possible values. however there are certain sorts like
insertion sort where o(n) in the best case ie when a list is sorted.
On 02-Oct-2010, at 10:32 PM, "Harshal ..Bet oN iT!!" wrote:
> you are gi
but the range (k) is cubic in n, if we use counting sort also in radix sort,
k wont be of the order n, so how can solution be in linear time?
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use radix sort
On Sat, Oct 2, 2010 at 10:32 PM, Harshal ..Bet oN iT!! wrote:
> you are given n integers in the range 0 to n3-1 ((n to the power 3) - 1) .
> how can u sort the numbers in O(n)?
>
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you are given n integers in the range 0 to n3-1 ((n to the power 3) - 1) .
how can u sort the numbers in O(n)?
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