Test: Sorting- 1 - Computer Science Engineering (CSE) MCQ

• The worst case of QuickSort occurs when the picked pivot is always one of the corner elements in sorted array. In worst case, QuickSort recursively calls one subproblem with size 0 and other subproblem with size (n-1).
• So recurrence is T(n) = T(n-1) + T(0) + O(n).
• The above expression can be rewritten as T(n) = T(n-1) + O(n) 

void exchange(int *a, int *b) 
{ 
int temp; 
temp = *a; 
*a = *b; 
*b = temp; 
} 

int partition(int arr[], int si, int ei) 
{ 
int x = arr[ei]; 
int i = (si - 1); 
int j; 

for (j = si; j <= ei - 1; j++) 
{ 
    if(arr[j] <= x) 
    { 
    i++; 
    exchange(&arr[i], &arr[j]); 
    } 
} 

exchange (&arr[i + 1], &arr[ei]); 
return (i + 1); 
} 

/* Implementation of Quick Sort 
arr[] --> Array to be sorted 
si --> Starting index 
ei --> Ending index 
*/
void quickSort(int arr[], int si, int ei) 
{ 
int pi; /* Partitioning index */
if(si < ei) 
{ 
    pi = partition(arr, si, ei); 
    quickSort(arr, si, pi - 1); 
    quickSort(arr, pi + 1, ei); 
} 
} 

If we use median as a pivot element, then the recurrence for all cases becomes T(n) = 2T(n/2) + O(n) The above recurrence can be solved using Master Method. It falls in case 2 of master method.

Insertion sort takes linear time when input array is sorted or almost sorted (maximum 1 or 2 elements are misplaced). All other sorting algorithms mentioned above will take more than lienear time in their typical implementation.

7 and 9 both are at their correct positions (as in a sorted array). Also, all elements on left of 7 and 9 are smaller than 7 and 9 respectively and on right are greater than 7 and 9 respectively.

Selection sort makes O(n) swaps which is minimum among all sorting algorithms mentioned above.

In Heapsort, we first build a heap, then we do following operations till the heap size becomes 1. a) Swap the root with last element b) Call heapify for root c) reduce the heap size by 1. In this question, it is given that heapify has been called few times and we see that last two elements in given array are the 2 maximum elements in array. So situation is clear, it is maxheapify whic has been called 2 times.

The only significant advantage that bubble sort has over most other algorithms, even quicksort, but not insertion sort, is that the ability to detect that the list is sorted efficiently is built into the algorithm. When the list is already sorted (best-case), the complexity of bubble sort is only O(n).

The data can be sorted using external sorting which uses merging technique. This can be done as follows: 1. Divide the data into 10 groups each of size 100. 2. Sort each group and write them to disk. 3. Load 10 items from each group into main memory. 4. Output the smallest item from the main memory to disk. Load the next item from the group whose item was chosen. 5. Loop step #4 until all items are not outputted. The step 3-5 is called as merging technique.

1. To sort the array firstly create a min-heap with first k+1 elements and a separate array as resultant array.
2. Because elements are at most k distance apart from original position so, it is guranteed that the smallest element will be in this K+1 elements.
3. Remove the smallest element from the min-heap(extract min) and put it in the result array.
4. Now,insert another element from the unsorted array into the mean-heap, now,the second smallest element will be in this, perform extract min and continue this process until no more elements are in the unsorted array.finally, use simple heap sort for the remaining elements

Time Complexity

1. O(k) to build the initial min-heap.
2. O((n-k)logk) for remaining elements.
3. 0(1) for extract min.

So overall O(k) + O((n-k)logk) + 0(1) = O(nlogk)

Out of the given options quick sort is the only algorithm which is not stable. Merge sort is a stable sorting algorithm.

Any comparison based sorting algorithm can be made stable by using position as a criteria when two elements are compared. Counting Sort is not a comparison based sorting algortihm. Heap Sort is not a comparison based sorting algorithm.

Applying binary search to calculate the position of the data to be inserted doesn't reduce the time complexity of insertion sort. This is because insertion of a data at an appropriate position involves two steps: 1. Calculate the position. 2. Shift the data from the position calculated in step #1 one step right to create a gap where the data will be inserted. Using binary search reduces the time complexity in step #1 from O(N) to O(logN). But, the time complexity in step #2 still remains O(N). So, overall complexity remains O(N^2).

The number of comparisons that a comparison sort algorithm requires increases in proportion to Nlog(N), where N is the number of elements to sort. This bound is asymptotically tight: Given a list of distinct numbers (we can assume this because this is a worst-case analysis), there are N factorial permutations exactly one of which is the list in sorted order. The sort algorithm must gain enough information from the comparisons to identify the correct permutations. If the algorithm always completes after at most f(N) steps, it cannot distinguish more than 2^f(N) cases because the keys are distinct and each comparison has only two possible outcomes. Therefore, 2^f(N) >= N! or equivalently f(N) >= log(N!). Since log(N!) is Omega(NlogN), the answer is NlogN. For more details, read here

The time complexity is given by: T(N) = T(N/3) + T(2N/3) + N Solving the above recurrence relation gives, T(N) = N(logN base 3/2)

The insertion sort will take time when input array is already sorted.

The recurrence tree for merge sort will have height Log(n). And O(n²) work will be done at each level of the recurrence tree (Each level involves n comparisons and a comparison takes O(n) time in worst case). So time complexity of this Merge Sort will be

The recursion expression becomes: T(n) = T(n/4) + T(3n/4) + cn After solving the above recursion, we get

For the case where n/5 elements are in one subset, T(n/5) comparisons are needed for the first subset with n/5 elements, T(4n/5) is for the rest 4n/5 elements, and n is for finding the pivot. If there are more than n/5 elements in one set then other set will have less than 4n/5 elements and time complexity will be less than T(n/5) + T(4n/5) + n because recursion tree will be more balanced.

Worst case complexities for the above sorting algorithms are as follows: Merge Sort — nLogn Bubble Sort — n^2 Quick Sort — n^2 Selection Sort — n^2

Counting sort algorithm is efficient when range of data to be sorted is fixed. In the above question, the range is from 0 to 255(ASCII range). Counting sort uses an extra constant space proportional to range of data.