Please explain What is the worse-case performance of quicksort? Select the correct answer with the correct justification. Group of answer choices a. O(n*n), because during the partitioning stage it is possible that we select a 'bad' random pivot point that only sorts one item at a time. If we continuously do this over and over again, we are effectively only sorting 1 item each time. We then need to perform this pivot step n more times to sort the remaining n-1 items, thus giving us an O(n*n) complexity. b. O(n*n), because quicksort is a comparison-based sorting algorithm in which we have to compare every possible element against every other element in order to swap each element into the correct position. c. O(n*log(n)), because each time we are selecting a pivot point, on the average-case we are able to split our problem into roughly two collections, and sort half of our elements at the pivot poin

Please explain What is the worse-case performance of quicksort? Select the correct answer with the correct justification. Group of answer choices a. O(nn), because during the partitioning stage it is possible that we select a 'bad' random pivot point that only sorts one item at a time. If we continuously do this over and over again, we are effectively only sorting 1 item each time. We then need to perform this pivot step n more times to sort the remaining n-1 items, thus giving us an O(nn) complexity. b. O(nn), because quicksort is a comparison-based sorting algorithm in which we have to compare every possible element against every other element in order to swap each element into the correct position. c. O(nlog(n)), because each time we are selecting a pivot point, on the average-case we are able to split our problem into roughly two collections, and sort half of our elements at the pivot poin

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter8: Arrays And Strings

Section: Chapter Questions

Problem 24PE

See similar textbooks

Similar questions

Implement a quicksort based on partitioning on the median of a random sample of five items from the subarray. Put the items of the sample at the appropriate ends of the array so that only the median participates in partitioning. Run doubling tests to determine the effectiveness of the change, in comparison both to the standard algorithm and to median-of-3 partitioning (see the previous exercise). Extra credit : Devise a median-of-5 algorithm that uses fewer than seven compares on any input.
) Implement a quicksort based on using a sample of size 2k 1. First, sort the sample, then arrange to have the recursiveroutine partition on the median of the sample and to move the two halves of the rest ofthe sample to each subarray, such that they can be used in the subarrays, without havingto be sorted again. This algorithm is called samplesort.
You will analyze three algorithms to solve the maximum contiguous subsequence sum problem, and then evaluate the performance of instructor-supplied implementations of those three algorithms. You will compare your theoretical results to your actual results in a written report. What is the maximum contiguous subsequence sum problem? Given a sequence of integers A1, A2, ..., An (where the integers may be positive or negative), find a subsequence Aj, ... , Ak that has the maximum value of all possible subsequences. The maximum contiguous subsequence sum is defined to be zero if all of the integers in the sequence are negative. Consider the sequence shown below. A1: -2 A2: 11 A3: -4 A4: 13 A5: -5 A6: 2 The maximum contiguous subsequence sum is 20, representing the contiguous subsequence in positions 2, 3, and 4 (i.e. 11 + (-4) + 13 = 20). The sum of the values in all other contiguous subsequences is less than or equal to 20. Consider a second sequence, shown below. A1: 1…
Show, in the style of the trace given with the code, how the entropy-optimal sort first partitions the array B A B A B A B A C A D A B R A Use the form to answer the question as same on the picture. lt i gt 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 B A B A B A B A C A D A B R A.
When is O(n2) considered a costly algorithm? Is there a less costly alternative? Hint: Find the answer among algorithms with difficulty of O(n) , O(log n) , O(nn) , . . . Why does overlap testing have a degree of error? Does overlap testing work for small objects? Why or why not? When overlap testing is performed, does it always take five iterations of the bisection technique to detect the time right before collision?
Implement a quicksort with a sample size of 2k 1. Sort the sample first, then have the recursive procedure partition on the sample's median and shift the two halves of the rest of the sample to each subarray so that they may be utilised in the subarrays without having to be sorted again. This algorithm is known as samplesort. Put into practise a quicksort based on a 2k sample. The sample should be sorted first, after which you should set up the recursive procedure to split the sample based on its median and to shift the two halves of the remaining sample to each subarray so they can be utilised in the subarrays without needing to be sorted again. The name of this algorithm is samplesort.
We argued in class that if the input is not uniformally distributed, the worst-case running time of Bucket Sort (using Insertion Sort in each bucket) in O(n^2). The reason: all those numbers could fall in the same bucket, and then the time of Insertion sorting then dominates. How many out of the total n numbers could fall in one bucket and the algorithm still run in linear time if the rest are evenly distributed? (Choose one option, no points if you choose more than one.) A. O(1)B. O(√n)C. O(On)D. O(log n)E. None of the above
We talked about the trade-off between using sequential search on an unsorted list as opposed to sorting the list and then using binary search. If the list size is n = 9,000, about how many worst-case searches must be done before the second alternative is better in terms of number of comparisons? (Hint: Let p represent the number of searches done.) Use selection search to sort the binary search list.
Implement a quicksort based on partitioning on themedian of a random sample of five items from the subarray. Put the items of the sampleat the appropriate ends of the array so that only the median participates in partitioning.Run doubling tests to determine the effectiveness of the change, in comparison bothto the standard algorithm and to median-of-3 partitioning (see the previous exercise).Extra credit : Devise a median-of-5 algorithm that uses fewer than seven compares onany input.
Consider a divide-and-conquer algorithm that calculates the sum of all elements in a set of n numbers by dividing the set into two sets of n/2 numbers each, finding the sum of each of the two subsets recursively, and then adding the result. What is the recurrence relation for the number of operations required for this algorithm? Answer is f(n) = 2 f(n/2) + 1. Please show why this is the case.
We learned how to merge two ordered lists with the two-finger method, and we generalizedthis to merge k ordered lists, to obtain one sorted list with all the n items. k is a parameter that canbe set as needed, as long as 2 ≤ k ≤ n. However, for some values of k it is faster to simply sort all thenumbers ignoring the initial order. For which (big O) range of k is worth to merge the lists rather thansort?
Given a sorted array of n comparable items A, and a search value key, return the position (array index) of key in A if it is present, or -1 if it is not present. If key is present in A, your algorithm must run in order O(log k) time, where k is the location of key in A. Otherwise, if key is not present, your algorithm must run in O(log n) time.