Binary search

Question

The minimum number of comparisons required to determine if an integer appears more than n/2 times in a sorted array of n integers is
(A) (n)
(B) (logn)
(C) (log*n)
(D) (1)

Answer 1

The Best way to find out whether an integer appears more than n/2 times in a sorted array(Ascending Order) of n integers, would be binary search approach.

1. The First occurrence of an element can be found out in O(log(n)) time using divide and conquer technique,lets say it is i.

2. The Last occurrence of an element can be found out in O(log(n)) time using divide and conquer technique,lets say it is j.

3. Now number of occurrence of that element(count) is (j-i+1). Overall time complexity = log n +log n +1 = O(logn).

Hence answer is Option B