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

Similar to https://stackoverflow.com/a/1191881/199556 explanation.

Let's compare 3 elements(3 comparison operation) in worse case "same" element will appear once. So we reduce the tail by 3 and reduce count of "same" elements by one.

In final step(after k iterations) our tail will contain (n/2) - k "same" elements. Let's compare the length of the tail.

On the one hand it will n-3k on the other hand (n/2) - k + 1. Last unsame elements may exist.

n-3k = (n/2) - k + 1

k = 1/4*(n-2)

After k iterations we'll surely get result.

Number of comparisons 3/4*(n-2)