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) Θ(n)
(B) Θ(logn)
(C) Θ(log*n)
(D) Θ(1)


Answer: (B)

Explanation: If you answered Theta(1), then think of examples {1, 2, 2, 2, 4, 4}, {1, 1, 2, 2, 2, 2, 3, 3}

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 occurence 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 occuerence of that element(count) is (j-i+1). Overall time complexity = log n +log n +1 = O(logn)

See Check for Majority Element in a sorted array

This solution is contributed by Nirmal Bharadwaj

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