2. Let A and B be two sorted arrays of n elements each. We...

2. Let A and B be two sorted arrays of n elements each. We can easily find the median element in A — it is just the element in the middle — and similarly we can easily find the median element in B. (Let us define the median of $2k$ elements as the element that is greater than $k - 1$ elements and less than $k$ other elements.) However, suppose we want to find the median element overall — i.e., the $n$th smallest in the union of A and B. How quickly can we do that? You may assume there are no duplicate elements in $A \cup B$. Prove matching upper and lower bounds for this problem.

Transcript

00:01 Hello student, here the problem is find the median l.

00:04 Input l is the input list containing the n real number n output.

00:08 The median of the l is m.

00:10 If n is less than 10, then sort l and return the median m.

00:15 And if divide l into the n list of size 5.

00:19 So, here we are dividing the list into the 5 list.

00:24 So, here recurrence formula for this running time for algorithm 1 is if i am dividing n into n by 5.

00:34 So, and again i am applying the recurrence.

00:37 So, here for n by 5.

00:39 So, time will be n by 5 for if for n tn then for n by 5 it will be tn by 5 and again big o of n.

00:50 So, here big o of n term is the cost of sorting each list of the size 5.

00:56 So, here we are dividing the list into the 5 sub parts.

01:01 So, this will be the cost.

01:03 So, this factor represent the cost.

01:06 The solution of this recurrence equation.

01:08 So, here this is the recurrence equation.

01:10 This is the recurrence equation and so its solution come.

01:14 So, if we look like this is because the recurrence is similar to the recurrence for the quick sort algorithm.

01:19 This is and quick sort running time is n log n.

01:22 So, with the help of this we i provided the running time.

01:26 Now, correctness of the algorithm to prove that the algorithm 1 is correct we need to show that it is always return the medium of the input list l.

01:35 So, if the algorithm is running very well then it will always return the medium.

01:39 So, we can do this by the induction on the size of the input list best case.

01:45 If the size of the input list is less than 10 then the algorithm simply sort the list and return the medium.

01:52 So, this is given in the problem.

01:53 Then if the size is less than 10 so here then the algorithm will simply sort the list and return the medium.

02:01 This is correct because the median of the sorted list is the median element.

02:05 So, if we want to calculate the median then we have to always sort the list.

02:11 So, list must be always sorted.

02:14 Induction step assume the algorithm is correct for all the input list of the size less than n...

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