(a) (Nothing to prove here) Suppose that p is the (k+1)-th smallest element in A, so that the pivotappears in B at B[k]. In this case, the length of Ao is k and the length of A₁ is n - k- 1, and sothe expected runtime of the recursive calls made in Step 4 is Tk + Tn-k-1. Because the pivot waschosen randomly, the probability that p is indeed the (k+1)-th smallest element in A is 1/n forall k = 0,..., n - 1. Therefore, the expected runtime of the recursive calls is 1/2 h=(Tk + Tn-k-1).Finally, choosing the pivot takes time O(1) and the filter step takes time O(n). Thus, Tr satisfies:nk=0n-11Tn = cn +=(Tk + Tn_k_1).k=0This will be our starting point. Also, note T₁ = 1.(b) Prove n Tn = cn² +2=Tk.(c) Deduce n. Tn = c(2n-1) + (n +1) · Tn-1.(d) DeducenTnn+12cn+1+Tn-1,n(e) Now iterate this bound to get 1 ≤2c(+1+ + ... + 1/3 ) + 1/2 .n+1n(f) Finally use the fact from calculus: 1+ ½ + 3 + ··· +/= O(log n) to get In=O(n log n).

The objective of the question is to derive the time complexity of a randomized algorithm for finding…

Answered: (a) (Nothing to prove here) Suppose…

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