Find a number where it appears exactly N/2 times

Question

Here is one of my interview question. Given an array of N elements and where an element appears exactly N/2 times and the rest N/2 elements are unique

Answer 1

The answer is straightforward.. and can be achieved in worst case (n/2 + 1) comparisons

1. Compare pairwise first (n-2) numbers, that is, compare nos. at 0 and 1, then 2 and 3 and so on... total n/2 -1 comparisons. If we find identical numbers in any of the above comparisons.. we have the repeated number... else:

2. Take any one of the last two remaining numbers (say second last one I took) and compare it with the numbers in the second last pair.. if match occurs..second last no. is the repated one, else last one is the repeated one... in all 2 comparisons.

Total comparisons = n/2 - 1 + 2 =n/2 + 1 (worst case) I dont think there is any O(log n) method to achieve this