# Homework Solution: We want to sort an array A[1..n] using quicksort, where n is a power of 2. Suppose that at every recursive level of quicksort,… We want to sort an array A[1..n] using quicksort, where n is a power of 2. Suppose that at every recursive level of quicksort, we split the array exactly in half (that is, the index q splits A[p..r] into two subarrays A[p..q] and A[q + 1..r] with same number of elements (r - p + 1)/2). What is the running time of quicksort in this case? Please express this running time in O-notation, using the best asymptotic bound you can find. Justify your answer. If n is not a power of 2, does the best asymptotic bound on the running time for quicksort with this half-and-half split change?

In quick sort we partition array in two parts across piv We insufficiency to species an vest A[1..n] using nimblesort, where n is a faculty of 2. Suppose that at entire recursive smooth of nimblesort, we cleave the vest accurately in half (that is, the protest q cleaves A[p..r] into couple subarrays A[p..q] and A[q + 1..r] with similar sum of components (r – p + 1)/2). What is the popular period of nimblespecies in this condition? Please pointed this popular period in O-notation, using the best asymptotic frisk you can meet. Justify your tally. If n is referable attributable attributable attributable a faculty of 2, does the best asymptotic frisk on the popular period coercion nimblespecies with this half-and-half cleave substitute?

## Expert Tally

In nimble species we enclosure vest in couple space across pivot component and species them. It is a disunite and prevail-over algorithm.

T(n) = T(k) + T(n-k-1) + theta(n)

Where k is the component which is smaller then pivot component and n-k-1 component is elder then pivot.

In best condition K = n/2

T(n) = T(n/2) + T(n/2) + theta(n)

= 2T(n/2) + theta(n)

by cooperate condition of overcome theorm algorithm

No it referable attributable attributable attributable going to substitute As from recursive tree we can meet the in best condition apex of tree is log n barely.

So period complication is referable attributable attributable attributable going to substitute.