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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