You have an array of n elements. Suppose you implement quicksort by 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?

Question

Answer

Explanation:
In quicksort, the choice of pivot plays a crucial role in determining the performance of the algorithm. If a bad pivot is chosen, it can lead to worst-case time complexity.

Choosing the central element as the pivot:
In this case, the algorithm always chooses the element at the center of the array as the pivot. Let's analyze the worst-case scenario when using this pivot selection strategy.

Worst-case scenario:
The worst-case scenario occurs when the chosen pivot is always the smallest or largest element in the array. This leads to highly unbalanced partitions, where one partition contains n-1 elements and the other partition is empty.

Recurrence relation:
In each step, the algorithm partitions the array into two subarrays based on the pivot, and recursively performs the same operation on each subarray. The recurrence relation for the worst-case scenario can be given as:

T(n) = T(n-1) + T(0) + O(n)

Here, T(n) represents the time taken to sort an array of size n.

Explanation of the recurrence relation:
- T(n-1) represents the time taken to sort the larger partition with (n-1) elements.
- T(0) represents the time taken to sort the smaller partition with 0 elements.
- O(n) represents the time taken to partition the array.

Worst-case time complexity:
In the worst-case scenario, the larger partition has (n-1) elements. Therefore, the recurrence relation can be simplified as:

T(n) = T(n-1) + O(n)

Expanding the recurrence relation:

T(n) = T(n-2) + O(n-1) + O(n)
= T(n-3) + O(n-2) + O(n-1) + O(n)
...
= O(1) + O(2) + O(3) + ... + O(n-1) + O(n)

Using the sum of arithmetic progression formula, the worst-case time complexity can be calculated as:

T(n) = O(n^2)

Hence, the tightest upper bound for the worst-case performance when implementing quicksort by always choosing the central element as the pivot is O(n^2). Therefore, option 'A' is the correct answer.

Answer

For any input, there are some permutations for which worst case will be O(n²). In some case, choosing the middle element minimizes the chances of encountering O(n²), but in worst case it can go to O(n²). Whichever element we take as Pivot, either first or middle, worst case will be O(n²) since Pivot is fixed in position. While choosing a random pivot minimizes the chances of encountering worst case i.e. O(n²). Refer this article on Quick Sort.

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