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The median of $n$ elements can be found in $O(n)$ time. Which one of the following is correct about the complexity of quick sort, in which median is selected as pivot?

1. $\Theta (n)$
2. $\Theta (n \log n)$
3. $\Theta (n^{2})$
4. $\Theta (n^{3})$
edited | 2.7k views

As we choose the pivot a median element ... so, every time we are going to have good splits guaranteed so the best case $O(n \log n)$.
edited
+17

every time we are finding median in O(n) times then solving the 2 half array recurcively...

therefore recurrence relation:

T(n) = 2T( n / 2) + O(n)..

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How is middle and median diiferent ?
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What if the array is already sorted? or All the elements in the array are same?

In that case the complexity will be O(n^2) ?
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@Dulqar middle means (array size/2)th element whereas median depends upon values in the array.

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Here input will be like this

7 6 2 1 3 5 4

So, median element we will get at last.

median means middle element of sorted list
+1
@Srestha this example will take O(n^2). I think O(n^2) will be the right.

This e.g break the equation T(n)=T(n-1)+T(0)+ O(n)
+3

See this question

https://gateoverflow.in/2048/gate2014-3-14

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Thanks @srestha
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@ parikshit If it was the case that all elements in the list are same then in question the word 'median' would not have been used.
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I think answer should be O(n^2)  please read the last statement by choosing hoare method we are trying to reduce probability  of getting worst but worst case can still happen. Please explain me this.!!

The answer depends on strategy for choosing pivot. In early versions of Quick Sort where leftmost (or rightmost) element is chosen as pivot, the worst occurs in following cases.

1) Array is already sorted in same order.
2) Array is already sorted in reverse order.
3) All elements are same (special case of case 1 and 2)

Since these cases are very common use cases, the problem was easily solved by choosing either a random index for the pivot, choosing the middle index of the partition or (especially for longer partitions) choosing the median of the first, middle and last element of the partition for the pivot. With these modifications, the worst case of Quick sort has less chances to occur, but worst case can still occur if the input array is such that the maximum (or minimum) element is always chosen as pivot.

To people having doubt on what will be the time complexity when all input values are equal. I had the same doubt for few days and after spending some time on this question here is what i found:

1.There are two standard partition procedure of quicksort, one is lomuto partition which is given in topic of quicksort in cormen in which last element is chosen as pivot and both i and j starts from starting index, second one is hoare's partition which is the original quicksort partition procedure given by hoare who gave quicksort algorithm, this partition procedure is given in problems section of cormen, in this first element is chosen as pivot but i and j starts from opposite ends.

2.Now if u consider lomuto partition then no matter what pivot choosing strategy u follow, be it median or middle or anything the worst case will always be O(n^2) because of input type being all elements with equal values. but if we consider hoare partition and we choose good pivot like median then if all elements are same then we get equal partition so in this worst case can be guaranteed O(nlogn).

here is the reference to geeks for geeks comparing hoare vs lomuto partition: http://www.geeksforgeeks.org/hoares-vs-lomuto-partition-scheme-quicksort/

what i think is in two gate questions asking time complexity when pivot is median and in other one pivot is 4th smallest element they might be considering hoare partition only.

+2
good observation , and yes your assumption is correct .

They took Hoare partition as Hoare is who invented the quick sort method so they follow this technique only .
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Thanks a lot ! Even I had this doubt.
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thanks vineet ,,,this doubt was very irritating
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I think even in Hoares partition, one situation might lead to a O(n^2) complexity.

If all elements are same then the starting index would travel upto ending index trying to find an element greater than pivot(which it obviously won't find).

So the division would be T(n) = T(n-1) + T(0) + cn. [One group will contain (n-1) elements and the other 1 element].So run-time would be O(n^2).

Correct me if I'm wrong here