why have you divided it into three . I mean it can be divided into
many groups of consecutive .
Please write the statement properly !
On Jan 30, 11:13 am, snehal jain <learner....@gmail.com> wrote:
> @nishanth
> divide into groups ( not necessarily 2) in as many group as possible.. such
> that elements in each group is consecutive
>
> another example to clearify
>
> A= { 9,7, 13, 11,6,12,8,10,3, 4, 2, 16,14,17,13,15)
> ans
> <9,7,13,11,6,12,8,10>
> <3,4,2>
> <16,14,17,13,15>
>
>
>
> On Sun, Jan 30, 2011 at 11:32 AM, nishaanth <nishaant...@gmail.com> wrote:
> > @snehal..i guess you are missing something in the question...divide it into
> > 2 groups such that (there should be some other condition or criterion).....
>
> > On Sun, Jan 30, 2011 at 11:29 AM, snehal jain <learner....@gmail.com>wrote:
>
> >> my approach
>
> >> sort in nlogn and then while traversing the array put the elements in a
> >> group till they are consecutive . when a[i+1]!=a[i]+1 then put a[i+1] in
> >> different group.. now we need to rearrange elements in the group so that
> >> they are according to the given array.. but that will make it O(n^2) ..
> >> anyone with better ideas?
>
> >> On Fri, Jan 21, 2011 at 1:20 PM, snehal jain <learner....@gmail.com>wrote:
>
> >>> Divide a list of numbers into groups of consecutive numbers but their
> >>> original order should be preserved.
> >>> Example:
> >>> <8,2,4,7,1,0,3,6>
> >>> Two groups:
> >>> <2,4,1,0,3> <8,7,6>
> >>> Better than O(n^2) is expected.
>
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> > --
> > S.Nishaanth,
> > Computer Science and engineering,
> > IIT Madras.
>
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