For the improvement of efficiency of quick sort the pivot can be Select one: O a. The first element O b. The mean element O c. The last element O d. None of the above Straight selection sort is basically a method of repeated Select one: 0 a. interchange O b. searching O c. Position adjustment O d. None of the above To analyze the sequential algorithm for the average-case efficiency all of the following are correct except one. Select one: O a. We must make some assumptions about possible input of size n. O b. The probability of a successful search is p (0 spsi). O c. Possibility of the first match in the ith position of the list is the same for every O d. The best case do not exceed n/2 when the element is not exist. The queue after each operation of the following sequence that starts with the empty queue: enqueue (8), enqueue (7), dequeu, enqueue (3), enqueue (9), dequeue, enqueue (5), enqueue (2), dequeue Select one: O a. 5 9 2 O b. 2 5 9 O c.9 52 O d. 9 25 The solution of the following Recurrence relation: ao = 8; an = nan-1 is: Select one: a. an = 8n(n-1)! O b. an= 5n! O c. an = 8n! d. an= 5n(n-1)(n-2) By using which method sorting is not possible Select one: O a. Insertion O b. Selection O c. Deletion O d. Exchange In the following linear search alg. that returns the index location of a value in an array. The average number of lines executed /*a is the array of size n we are searching through */ i = 0; while ((i 1 Select one: O a. T(n) = 2n+2, O(n) O b. T(n) = 2n, O(2n) O c. T(n) = 2n, O(n) d. T(n) = n, on) Giving the following list 42 6718 95 using merge sort the third step of merging will produce the following list: Select one: O a. 2,4,6,7,1,8,5,9 O b. 1,2,4,5,6,7,8,9 O c. there is no merging step O d. 2,4,6,7,1,5,8,9 In quick sort, the number of partitions into which the file of size n is divided by a selected record is Select one: O a. 4 O b. n/2(n-1) O c.2 0 d. 2(n-1) T(n), is the running time of a particular algorithm on input of size n, In the following algorithm T(n) is: 1. n = read input from user 2. sum = = 0 3. i= 0 4. while i<><>