Assume that we are given 'n' objects and a subroutine that takes two inputs and says if they are equivalent or not (e.g. it can give output as 1 if they are equal).
I need to come up with an algorithm that calls the above function O(n log n) times and decides if the input has more than 'n/2' items that are equivalent to each other.
Here is a classical divide and conquer solution which gives O(n log n)
split into two subsets, A1 and A2, ..., and show T(n) is O(n log n). If A has a majority element v, v must also be a majority element of A1 or A2 or both. The equivalent contra-positive restatement is immediate: (If v is <= half in each, it is <= half in the total.) If both parts have the same majority element, it is automatically the majority element for A. If one of the parts has a majority element, count the number of repetitions of that element in both parts (in O(n) time) to see if it is a majority element. If both parts have a majority, you may need to do this count for each of the two candidates, still O(n). This splitting can be done recursively. The base case is when n = 1. A recurrence relation is T(n) = 2T(n/2) + O(n), so T(n) is O(n log n) by the Master Theorem.