You have an array of n elements. Suppose you implementquicksortby always choosing the central element of the array as the pivot. Then the tightest upper bound for the worst case performance isa)O(n2)b)O(nLogn)c)Theta(nLogn)d)O(n3)Correct answer is option 'A'. Can you explain this answer?

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Answer

Understanding Quicksort's Central Pivot Strategy
In quicksort, the choice of pivot significantly influences performance. When consistently selecting the central element as the pivot, the algorithm can face challenging scenarios.
Worst Case Scenario
- Choosing the central element does not guarantee balanced partitions, especially in already sorted or nearly sorted arrays.
- If the chosen pivot is the median, the partitioning may still yield unbalanced splits, leading to:
- Left sub-array with n/2 elements.
- Right sub-array with n/2 elements.
- This behavior can lead to a recursive depth of O(n), where each recursive call does not effectively reduce the problem size.
Time Complexity Analysis
- Each level of recursion processes n elements, resulting in a direct time complexity of:
- T(n) = T(n/2) + O(n)
- Solving this relation using the Master Theorem leads to:
- The depth of recursion is O(n), and at each level, we are doing O(n) work.
- Therefore, the total time complexity becomes O(n^2).
Conclusion
- The tightest upper bound for the worst-case performance of this quicksort implementation, where the central pivot is chosen, is O(n^2).
- This is due to the potentially inefficient partitioning leading to deep recursion without significant reduction of the problem size.
Final Note
- It is crucial to select the pivot wisely to avoid the worst-case scenario and achieve optimal performance, typically O(n log n) in average cases.

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