@Anish: Does it work on 1,2,3,1?

Dave

On Aug 7, 2:04 am, Ashish Goel <[email protected]> wrote:
> I thought again on this, i think we have made a simple problem quite
> complicated
>
> if a number is half times in a list, that  implies that either it is
> repeated alternately otherwise adjacent somewhere else
>
> so if this is repeating alternatively that in first 4 occurrence, we should
> have this at index 0,2 or 1,3 otherwise it is adjacent somewhere else
>
> additional case(1,2,3,3,4,3,6,3) also covered
>
> therefore
>
> if ((a[0] ==a[1]) || (a[0]==a[2])) { printf("%d",a[0]);  return;}
> if ((a[1] ==a[2]) || (a[1]==a[3])) { printf("%d",a[1]);  return;}
>
> for (int i=3;i<n;i++)
>
> {
>     if a[i]==a[i-1] printf("%d", a[i]); return;
>
> }
>
> Best Regards
> Ashish Goel
> "Think positive and find fuel in failure"
> +919985813081
> +919966006652
>
>
>
> On Sat, Aug 7, 2010 at 11:50 AM, Ashish Goel <[email protected]> wrote:
> > 1,6,3,4,6,5,6,2,6,6
> > or
> > 1,2,3,6,4,6,5,6,6,6
>
> > lovely...
>
> > Use a two-step process. First, check for a repeated number in the
> >> first 4 elements. If none is found, then there are at least n/2-1
> >> occurrences of the repeated elements in the last n-3 elements, meaning
> >> that there must be at least two repeated elements in adjacent
> >> positions. So second, check for equal adjacent numbers in the last n-3
> >> elements.
>
> >> Dave
>
> >> On Aug 5, 8:36 am, AlgoBoy <[email protected]> wrote:
> >> > an array in which n/2 elements are unique...and the remaning n/2 have
> >> > the same elements but reapeated n/2 times. can anyone suggest a linear
> >> > solution with constant space/...
>
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