Sorting

Question 1

Consider the following sequence of numbers

  92, 37, 52, 12, 11, 25  

Use bubble sort to arrange the sequence in ascending order. Give the sequence at the end of each of the first five passes.

A
Theory Explanation.
Question 2

Merge sort uses

A
Divide and conquer strategy
B
Backtracking approach
C
Heuristic search
D
Greedy approach
Question 2 Explanation: 
Merge sort uses the divide and conquer strategy.
Question 3

A two dimensional array A[1...n][1...n] of integers is partially sorted if

    ∀i, j ∈ [1...n−1],   A[i][j] < A[i][j+1] and 
                           A[i][j] < A[i+1][j] 

Fill in the blanks:
(a) The smallest item in the array is at A[i][j] where i=............and j=..............
(b) The smallest item is deleted. Complete the following O(n) procedure to insert item x (which is guaranteed to be smaller than any item in the last row or column) still keeping A partially sorted.

procedure  insert (x: integer);
var        i,j: integer;
begin
(1) i:=1; j:=1, A[i][j]:=x;
(2) while (x > ...... or x > ......) do
(3) if A[i+1][j] < A[i][j] ......... then begin
(4) A[i][j]:=A[i+1][j]; i:=i+1;
(5) end
(6) else begin
(7) ............
(8) end
(9) A[i][j]:= .............
    end  
A
Theory Explanation.
Question 4

If one uses straight two-way merge sort algorithm to sort the following elements in ascending order:

   20, 47, 15, 8, 9, 4, 40, 30, 12, 17 

then the order of these elements after second pass of the algorithm is:

A
8, 9, 15, 20, 47, 4, 12, 17, 30, 40
B
8, 15, 20, 47, 4, 9, 30, 40, 12, 17
C
15, 20, 47, 4, 8, 9, 12, 30, 40, 17
D
4, 8, 9, 15, 20, 47, 12, 17, 30, 40
Question 4 Explanation: 
Question 5

A sorting technique is called stable if

A
it takes O (nlog n) time
B
it maintains the relative order of occurrence of non-distinct elements
C
it uses divide and conquer paradigm
D
it takes O(n) space
Question 5 Explanation: 
Sorting techniques are said to be stable if it maintains the relative order of occurrence of non-distinct element.
Question 6

Let s be a sorted array of n integers. Let t(n) denote the time taken for the most efficient algorithm to determined if there are two elements with sum less than 1000 in s. Which of the following statements is true?

A
t(n) is O(1)
B
n ≤ t(n) ≤ n log2 n
C
n log2 n ≤ t(n) < (n/2)
D
t(n) = (n/2)
Question 6 Explanation: 
Since the array is sorted. Now just pick the first two minimum elements and check if their sum is less than 1000 or not. If it is less than 1000 then we found it and if not then it is not possible to get the two elements whose sum is less than 1000. Hence, it takes constant time. So, correct option is (A).
Question 7

Randomized quicksort is an extension of quicksort where the pivot is chosen randomly. What is the worst case complexity of sorting n numbers using randomized quicksort?

A
O(n)
B
O(n log n)
C
O(n2)
D
O(n!)
Question 7 Explanation: 
In worst case Randomized quicksort execution time complexity is same as quicksort.
Question 8

In a permutation a1...an of n distinct integers, an inversion is a pair (ai, aj) such that i < j and ai > aj.

If all permutations are equally likely, what is the expected number of inversions in a randomly chosen permutation of 1...n ?

A
n(n-1)/2
B
n(n-1)/4
C
n(n+1)/4
D
2n[log2n]
Question 8 Explanation: 
Probability of inverse (ai, aj(i Probability of expected no. of inversions = (1/2) × (n(n-1)/2) = n(n-1)/4
Question 9

In a permutation a1...an of n distinct integers, an inversion is a pair (ai, aj) such that ii > aj.

What would be the worst case time complexity of the Insertion Sort algorithm, if the inputs are restricted to permutations of 1...n with at most n inversions?

A
Θ(n2)
B
Θ(n log n)
C
Θ(n1.5)
D
Θ(n)
Question 9 Explanation: 
Here the inputs are to be restricted to 1...n with atmost 'n' inversions. Then the worst case time complexity of inversion sort reduces to Θ(n).
Question 10

Consider the recursive algorithm given below:

 procedure bubblersort (n);
 var i,j: index; temp : item;
 begin
    for i:=1 to n-1 do
    if A[i] > A [i+1] then
 begin
    temp : A[i];
    A[i]:=A[i+1]; 
    A[i+1]:=temp
    end;
   bubblesort (n-1)
 end 

Let an be the number of times the ‘if…then….’ Statement gets executed when the algorithm is run with value n. Set up the recurrence relation by defining an in terms of an-1. Solve for an.

A
Theory Explanation.
Question 11

Following algorithm(s) can be used to sort n integers in the range [1...n3] in O(n) time

A
Heapsort
B
Quicksort
C
Mergesort
D
Radixsort
Question 11 Explanation: 
As no comparison based sort can ever do any better than nlogn. So option (A), (B), (C) are eliminated. O(nlogn) is lower bound for comparison based sorting.
As Radix sort is not comparison based sort (it is counting sort), so can be done in linear time.
Question 12

The minimum number of comparisons required to sort 5 elements is _____

A
7
Question 12 Explanation: 
Minimum no. of comparisons
= ⌈log(n!)⌉
= ⌈log(5!)⌉
= ⌈log(120)⌉
= 7
Question 13

Let a and b be two sorted arrays containing n integers each, in non-decreasing order. Let c be a sorted array containing 2n integers obtained by merging the two arrays a and b. Assuming the arrays are indexed starting from 0, consider the following four statements

1. a[i] ≥ b [i] => c[2i] ≥ a [i]
2. a[i] ≥ b [i] => c[2i] ≥ b [i]
3. a[i] ≥ b [i] => c[2i] ≤ a [i]
4. a[i] ≥ b [i] => c[2i] ≤ b [i] 
Which of the following is TRUE?

A
only I and II
B
only I and IV
C
only II and III
D
only III and IV
Question 13 Explanation: 
a[i] ≥ b[i]
Since both 'a' and 'b' are sorted in the beginning, there are 'i' elements than or equal to a[i] and similarly 'i' elements smaller than or equal to b[i]. So, a[i] ≥ b[i] means there are 2i elements smaller than or equal to a[i] and hence in the merged array, a[i] will come after these 2i elements. So, c[2i] ≤ a[i].
Similarly, a[i] ≥ b[i] says for b that, there are not more than 2i elements smaller than b[i] in the sorted array. So, b[i] ≤ c[2i].
So, option (C) is correct.
Question 14

If we use Radix Sort to sort n integers in the range [nk/2, nk], for some k>0 which is independent of n, the time taken would be?

A
Θ(n)
B
Θ(kn)
C
Θ(nlogn)
D
Θ(n2)
Question 14 Explanation: 
Time complexity for radix sort = Θ(wn)
where n = keys, w = word size
w = log2 (nk) = k × log2 (n)
Complexity Θ(wn) = Θ(k × log2(n) × n) = Θ(n log n) = Θ(n log n)
Question 15

Let P be a quicksort program to sort numbers in ascending order. Let t1 and t2 be the time taken by the program for the inputs [1 2 3 4] and [5 4 3 2 1] respectively. Which of the following holds?

A
t1 = t2
B
t1 > t2
C
t1 < t2
D
t1 = t2 + 5 log 5
Question 15 Explanation: 
Since both are in sorted order and no. of elements in second list is greater.
Question 16

Quicksort is ________ efficient than heapsort in the worst case.

A
LESS.
Question 16 Explanation: 
As worst case time for quicksort is O(n2) and worst case for heap sort is O(n logn).
Question 17

You have an array of n elements. Suppose you implement quicksort by always choosing the central element of the array as the pivot. Then the tightest upper bound for the worst case performance is

A
O(n2)
B
O(n log n)
C
Θ(n log⁡n)
D
O(n3)
Question 17 Explanation: 
The Worst case time complexity of quick sort is O (n2). This will happen when the elements of the input array are already in order (ascending or descending), irrespective of position of pivot element in array.
Question 18
Assume that the algorithms considered here sort the input sequences in ascending order. If the input is already in ascending order, which of the following are TRUE?
I. Quicksort runs in Θ(n2) time
II. Bubblesort runs in Θ(n2) time
III. Mergesort runs in Θ(n) time
IV. Insertion sort runs in Θ(n) time
A
I and II only
B
I and III only
C
II and IV only
D
I and IV only
Question 18 Explanation: 
If input sequence is already sorted then the time complexity of Quick sort will take O(n2) and Bubble sort will take O(n) and Merge sort will takes O(nlogn) and insertion sort will takes O(n).
→ The recurrence relation for Quicksort, if elements are already sorted,
T(n) = T(n-1)+O(n) with the help of substitution method it will take O(n2).
→ The recurrence relation for Merge sort, is elements are already sorted,
T(n) = 2T(n/2) + O(n) with the help of substitution method it will take O(nlogn).
We can also use master's theorem [a=2, b=2, k=1, p=0] for above recurrence.
Question 19

The worst case running times of Insertion sort, Merge sort and Quick sort, respectively, are:

A
Θ(nlogn), Θ(nlogn), and Θ(n2)
B
Θ(n2 ), Θ(n2 ), and Θ(nlogn)
C
Θ(n2), Θ(nlogn), and Θ(nlogn)
D
Θ(n2), Θ(nlogn), and Θ(n2)
Question 19 Explanation: 
Question 20

There are n unsorted arrays: A1, A2, ..., An. Assume that n is odd. Each of A1, A2, ..., An contains n distinct elements. There are no common elements between any two arrays. The worst-case time complexity of computing the median of the medians of A1, A2, ..., An is

A
O(n)
B
O(n log n)
C
Ω(n2 log n)
D
O(n2)
Question 20 Explanation: 
Finding the median in an unsorted array is O(n).
But it is similar to quicksort but in quicksort, partitioning will take extra time.
→ Find the median will be (i+j)/2
1. If n is odd, the value is Ceil((i+j)/2)
2. If n is even, the value is floor((i+j)/2)
-> Here, total number of arrays are
⇒ O(n)*O(n)
⇒ O(n2)
Note:
They are clearly saying that all are distinct elements.
There is no common elements between any two arrays.
Question 21

An array of 25 distinct elements is to be sorted using quicksort. Assume that the pivot element is chosen uniformly at random. The probability that the pivot element gets placed in the worst possible location in the first round of partitioning (rounded off to 2 decimal places) is _____.

A
0.08
B
0.01
C
1
D
8
Question 21 Explanation: 
Step-1: Given, 25 distinct elements are to be sorted using quicksort.
Step-2: Pivot element = uniformly random.
Step-3: Worst case position in the pivot element is either first (or) last.
Step-4: So total 2 possibilities among 25 distinct elements
= 2/25
= 0.08
Question 22

Consider the Quicksort algorithm. Suppose there is a procedure for finding a pivot element which splits the list into two sub-lists each of which contains at least one-fifth of the elements. Let T(n) be the number of comparisons required to sort n elements. Then

A
T(n) ≤ 2T(n/5) + n
B
T(n) ≤ T(n/5) + T(4n/5) + n
C
T(n) ≤ 2T(4n/5) + n
D
T(n) ≤ 2T(n/2) + n
Question 22 Explanation: 
Consider the case where one subset of set of n elements contains n/5 elements and another subset of set contains 4n/5 elements.
So, T(n/5) comparisons are needed for the first subset and T(4n/5) comparisons needed for second subset.
Now, suppose that one subset contains more than n/5 elements then another subset will contain less than 4n/5 elements. Due to which time complexity will be less than
T(n/5) + T(4n/5) + n
Because recursion tree will be more balanced.
Question 23

Which one of the following in place sorting algorithms needs the minimum number of swaps?

A
Quick sort
B
Insertion sort
C
Selection sort
D
Heap sort
Question 23 Explanation: 
Selection sort requires minimum number of swaps i.e O(n). The algorithm finds the minimum value, swaps it with the value in the first position, and repeats these steps for the remainder of the list. It does no more than n swaps, and thus is useful where swapping is very expensive.
Question 24

In quick sort, for sorting n elements, the (n/4)th smallest element is selected as pivot using an O(n) time algorithm. What is the worst case time complexity of the quick sort? 

A
θ(n)
B
θ(n log n)
C
θ(n2)
D
θ(n2 log n)
Question 24 Explanation: 

n→n/(4/3)→n/(4/3)2→n/(4/3)3-----n/(4/3)k=1
n/(4/3)k = 1
⇒n=(4/3)k ⇒ k = log4/3n [k=no. of levels]
In each level workdone = Cn
So, total workdone = Cn⋅log4/3n = (nlog4/3n)
Question 25

What is the number of swaps required to sort n elements using selection sort, in the worst case?

A
θ(n)
B
θ(n log n)
C
θ(n2)
D
θ(n2 logn)
Question 25 Explanation: 
Selection sort – There is no Worst case input for selection sort. Since it searches for the index of kth minimum element in kth iteration and then in one swap, it places that element into its correct position. For n-1 iterations of selection sort, it can have O(n) swaps. Selection Sort does a significant number of comparisons before moving each element directly to its final intended position. At most the algorithm requires N swaps. once we swap an element into place, you never go back again.So it is great for writes O(n) but not so great (at all) for reads — O(n2). It isn’t well-suited to generalized sorting, but might work well in specialized situations like EEPROM (where writes are inordinately expensive).
Question 26

The number of elements that can be sorted in Θ(log n) time using heap sort is

A
Θ(1)
B
Θ(√log⁡n)
C
Θ (log⁡n/log⁡log⁡n)
D
Θ(log n)
Question 26 Explanation: 
Using heap sort to n elements it will take O(n log n) time. Assume there are log n/ log log n elements in the heap.
So, Θ((logn/ log log n)log(logn/log log n))
= Θ(logn/log logn (log logn - log log logn))
= Θ((log n/log logn) × log log n)
= Θ(log n)
Hence, option (C) is correct answer.
Question 27

Which one of the following is the tightest upper bound that represents the number of swaps required to sort n numbers using selection sort?

A
O(log n)
B
O(n)
C
O(n log n)
D
O(n2)
Question 27 Explanation: 
Best, Average and worst case will take maximum O(n) swaps.
Selection sort time complexity O(n2) in terms of number of comparisons. Each of these scans requires one swap for n-1 elements (the final element is already in place).
Question 28
In the code below reverse(A,i,j) takes an array A, indices i and j with i ≤ j, and reverses the segment A[i], A[i+1],. . . ,A[j]. For instance if A = [0,1,2,3,4,5,6,7] then, after we apply reverse(A,3,6), the contents of the array will be A = [0,1,2,6,5,4,3,7].
function mystery (A[0..99]) {
int i,j,m;
for (i = 0; i < 100; i++) {
m = i;
for (j = i; j < 100, j++) {
if (A[j] > A[m]) {
m = j;
}
}
reverse(A,i,m);
}
return;
}
When the procedure terminates, the array A has been:
A
Sorted in descending order
B
Sorted in ascending order
C
Reversed
D
Left unaltered
Question 28 Explanation: 
This procedure is call pancake sort. Each iteration (of the outer for loop) moves the maximum value in A[i..99] to A[i].
Question 29
Which Sorting method is an external Sort?
A
Heap Sort
B
Quick Sort
C
Insertion Sort
D
None of the above
Question 29 Explanation: 
External sorting is a term for a class of sorting algorithms that can handle massive amounts of data. External sorting is required when the data being sorted do not fit into the main memory of a computing device (usually RAM) and instead they must reside in the slower external memory (usually a hard drive). External sorting typically uses a hybrid sort-merge strategy.
Example of external sorting is merge sort.
Question 30
In order to sort list of numbers using radix sort algorithm, we need to get the individual digits of each number ‘n’ of the list. If n is a positive decimal integer, then ith digit , from right , of the number n is:
A
B
C
D
Question 31
The algorithm design technique used in the quick sort algorithm is
A
Dynamic programming
B
Back tracking
C
Divide and conquer
D
Greedy method
Question 31 Explanation: 
The algorithm design technique used in the quick sort algorithm is divide and conquer.
Question 32

Quick sort is run on two inputs shown below to sort in ascending order:

I : 1,2,3,…n II : n, n-1,…, 2, 1

Let k1 and k2 be the number of comparisons made for the inputs I and II respectively. Then
A
k1 < k2
B
k1 = k2
C
k1 > k2
D
None
Question 32 Explanation: 
In quick sort whether the inputs are in descending order or ascending order, it will have the worst case comparisons and will be equal no. of comparisons.
Question 33
Which of the following sorting algorithms does not have a worst case running time of O(n2)?
A
Bubble Sort
B
Quick Sort
C
Merge Sort
D
Heap Sort
Question 33 Explanation: 
worst case time complexity of mergesort is O(nlogn).
Question 34
Which sorting method is best suited for external sorting?
A
Quick Sort
B
Heap Sort
C
Merge sort
D
All the above
Question 34 Explanation: 
External sorting is a term for a class of sorting algorithms that can handle massive amounts of data. External sorting is required when the data being sorted do not fit into the main memory of a computing device (usually RAM) and instead they must reside in the slower external memory (usually a hard drive). External sorting typically uses a hybrid sort-merge strategy. In the sorting phase, chunks of data small enough to fit in main memory are read, sorted, and written out to a temporary file. In the merge phase, the sorted sub-files are combined into a single larger file.
Question 35
You are given a sequence of n elements to sort. The input sequence consists of n/k subsequences,each containing k elements. The elements in a given subsequence are all smaller than the elements in the succeeding subsequence and larger than the elements in the preceding subsequence. Thus, all that is needed to sort the whole sequence of length n is to sort the k elements in each of the n/k subsequences. The lower bound on the number of comparisons needed to solve this variant of the sorting problem is
A
Ω(n)
B
Ω(n/k)
C
Ω(nlogk )
D
Ω(n/klogn/k)
Question 35 Explanation: 
There are n/k subsequences and each can be ordered in k! ways. This makes a (k!)n/k outputs. We can use the same reasoning:
(k!)n/k ≤ 2h
Taking the logarithm of both sides, we get:
h ≥ lg(k!)n/k
= (n/k)lg(k!)
≥ (n/k)(k/2)lg(k/2)
= (1/2)*(nlogk)-(1/2)*n
= Ω(nlogk)
Question 36
If the array A contains the items 10, 4, 7, 23, 67, 12 and 5 in that order, what will be the resultant array A after third pass of insertion sort?
A
67, 12, 10, 5, 4, 7, 23
B
4, 7, 10, 23, 67, 12, 5
C
4, 5, 7, 67, 10, 12, 23
D
10, 7, 4, 67, 23, 12, 5
Question 36 Explanation: 
Initial positions: 10, 4, 7, 23, 67, 12 , 5
Pass-1: 4, 10, 7, 23, 67, 12, 5
Pass-2: 4, 7, 10, 23, 67, 12, 5
Pass-3: 4, 7, 10, 23, 67, 12, 5 [ No change because the value 10 is placed in the same position ]
There are 36 questions to complete.

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