25 votes 25 votes The worst case running times of Insertion sort , Merge sort and Quick sort, respectively are: $\Theta (n \log n)$, $\Theta (n \log n)$ and $\Theta(n^2)$ $\Theta (n^2)$, $\Theta (n^2)$ and $\Theta(n \log n)$ $\Theta (n^2)$, $\Theta (n \log n)$ and $\Theta (n \log n)$ $\Theta (n^2)$, $\Theta (n \log n)$ and $\Theta (n^2)$ Algorithms gatecse-2016-set1 algorithms sorting easy + – Sandeep Singh asked Feb 12, 2016 • edited Nov 12, 2017 by kenzou Sandeep Singh 13.0k views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply Show 4 previous comments Prince Sindhiya commented Aug 1, 2018 reply Follow Share Yes quick sort is not stable 0 votes 0 votes Aakash_ commented Sep 20, 2018 reply Follow Share This Table should not be referenced blindly, it says Randomized Quick Sort take O(nlogn) but it's not True, Worst Case Time is O(n^2) even in Randomized Quick Sort. 3 votes 3 votes shashankrustagi commented Feb 7, 2021 reply Follow Share THis table is incorrect BUCKET sort worst case time complexity is $O(n^{2})$ 0 votes 0 votes Please log in or register to add a comment.
Best answer 35 votes 35 votes Answer is D. Insertion sort: $= \Theta(n^2)$ Merge sort: $= \Theta(n\log n)$ Quick sort: $= \Theta (n^2)$ Note : here $\Theta$ is not average case since question asked worst case so $\Theta$ represent worst case only abhilashpanicker29 answered Feb 12, 2016 • edited Jun 24, 2018 by Milicevic3306 abhilashpanicker29 comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes .Hence the given answer is D varunrajarathnam answered Sep 22, 2020 • edited Sep 27, 2020 by varunrajarathnam varunrajarathnam comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Insertion Sort Worst-Case time complexity=O(n^2) Merge Sort Worst-Case time complexity=O(nlogn) Quick Sort Worst-Case time complexity=O(n^2) Option D is correct himanshu dhawan answered Apr 9, 2021 himanshu dhawan comment Share Follow See all 0 reply Please log in or register to add a comment.