Q44.Marks: +2.0UGC NET Paper 2: Computer Science 2nd January 2026 Shift 1
How many number of comparison are required in worst case to find out second smallest element of n elements list?
1.n + ⌈log n⌉ - 1
2.n + ⌈log n⌉
3.⌈log n⌉
4.n + ⌈log n⌉ - 2✓ Correct
Solution
The correct answer is n + ⌈log n⌉ - 2.
Key Points
To find the second smallest element in a list of n elements, we need to conduct comparisons in a systematic way.
In the worst-case scenario, the comparisons involve building a tournament tree structure where each element competes to find the smallest and second smallest elements.
The number of comparisons required to determine the smallest element is n - 1, as each pair of elements is compared.
To find the second smallest element, additional comparisons are needed within the subset of elements that were compared against the smallest element.
In total, the number of comparisons required in the worst case is n + ⌈log n⌉ - 2.
Additional Information
Understanding the Process:
A tournament tree is constructed where elements are compared to find the smallest element.
After identifying the smallest element, the second smallest element is determined by comparing the elements that were defeated by the smallest element.
The height of the tournament tree is ⌈log n⌉, representing the number of levels of comparison.
This systematic approach ensures that the second smallest element is found efficiently.
Important Points:
The formula n + ⌈log n⌉ - 2 accurately represents the worst-case number of comparisons required.
The tournament tree method is commonly used in competitive programming and algorithm design for efficient comparison-based operations.
Understanding this concept is crucial for optimizing algorithms in scenarios involving element selection.
