flag=0;
for(int i=0;i<n-1;i+=2)
{
if(arr[0]!=arr[1])
{
flag=1;
break;
}
}
if(! flag)
cout<<arr[n-1];
else
cout<<arr[i];
T(N) : O(n/2)
More can be done using 2 pointer , 1st going forward , 2nd in reverse.
You can use xor operation in pairing too ..from moving back as well as
moving front,
Binary search is of no use !
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
