Question 1

Which one of the following options arranges the functions in the increasing order of asymptotic growth rate?

f2, f3, f1

f3, f2, f1
f2, f1, f3
f1, f2, f3
Question 1 Explanation: 

The asymptotic notation order should be 

Constant < logarithmic < linear < polynomial < exponential < factorial 

F2 and F3 → Polynomial 

F1 → Exponential

By the order of asymptotic notations F1 is definitely larger. 


Consider n=100

F1 : 100 ^100 ⇒ 1.e+200

F2 : N^log(100) base 2 ⇒  100 ^ log(100) base 2 ⇒ 100^6.6438561897747 = 1,93,96,00,91,15,564.181300016469223466

F3 : N^Sqrt(n) ====> 100^Sqrt(100) ⇒ 100^10 ⇒ 10,00,00,00,00,00,00,00,00,000


We can apply "log" on both sides. 

log(F1)=nlog10 (base 2)

log(F2)=(logn)^2 = logn * logn (base 2)

log(F3)=sqrt(n)logn (base 2) 

Answer: F2< F3< F1

Question 2
 Define Rn< >to be the maximum amount earned by cutting a rod of length n metres into one or more pieces of integer length and selling them. For i > 0, let p[i] denote the selling price of a rod whose length is i meters. Consider the array of prices:
      p[1] = 1, p[2] = 5, p[3] = 8, p[4] = 9, p[5] = 10, p[6] = 17, p[7] = 18
Which of the following statements is/are correct about R7?
R7is achieved by three different solutions.
R7cannot be achieved by a solution consisting of three pieces.
Question 2 Explanation: 
The rod length of R7 is 18.

There are 3 different possible ways to get the maximum amount.
P[6] + P[1] → 17+1 = 18
P[2] + P[2] + P[3] → 5+5+8 = 18
P[7] → 18 = 18
Question 3
Consider the following undirected graph with edge weights as shown:

The number of minimum-weight spanning trees of the graph is _______
Question 3 Explanation: 

To find the number of spanning trees using 2 methods:

  1. If graph is complete, use n^n-2 formula
  2. If graph is not complete, then use kirchhoff theorem.

Steps in Kirchoff’s Approach:

(i) Make an Adjacency matrix.

(ii) Replace all non-diagonal is by – 1.

(iii) Replace all diagonal zero’s by the degree of the corresponding vertex.

(iv) Co-factors of any element will give the number of spanning trees.  

Using the Kirchhoff theorem will take lot of time because the number of vertices are 9. 

So, we use brute force technique to solve the problem with the help of Kruskal's algorithm.

Question 4
Consider the following recurrence relation.

Which one of the following options is correct?
Question 4 Explanation: 
Question 5
Consider the following array.

Which algorithm out of the following options uses the least number of comparisons (among the array elements) to sort the above array in ascending order?
Quicksort using the last element as pivot
Insertion sort
Selection sort
Question 5 Explanation: 

Quick sort(with last element as pivot) → will give the worst case time complexity as O(n^2).

Merge Sort → The merge sort will not depend upon the input order and will give O(nlogn) time complexity. 

Insertion Sort → Insertion sort will perform best case time complexity when the input array is in sorted order. If the array is already sorted then the inversion count is 0 and If the array is sorted in reverse order that inversion count is the maximum. 

Note: Inversion formal definition is two elements A[i] and A[j] form an inversion if A[i] > A[j] and i < j. 

The number of comparisons will not take more than “n” for the given input array. 

Selection Sort → Selection sort will not depend upon the input order and will give O(n^2) time complexity. 

Question 6
Let P be an array containing n integers. Let t be the lowest upper bound on the number of comparisons of the array elements, required to find the minimum and maximum values in an arbitrary array of n elements. Which one of the following choices is correct?
Question 6 Explanation: 

Let assume n=512


Using standard recursive algorithm:

MaxMin is a recursive algorithm that finds the maximum and minimum of the set of elements {a(i), a(i+1), ..., a(j)}. The situation of set sizes one (i=j) and two (i=j-1) are handled separately. For sets containing more than two elements, the midpoint is determined ( just as in binary search) and two new subproblems are generated. When the maxima and minima of these subproblems are determined, the two maxima are compared and the two minima are compared to achieve the solution for the entire set. 

To find the number of element comparisons needed for Maxmin, T(n) represents this number, then the resulting recurrence relation is

When n is a power of two, n = 2k for some positive integer k, then








→ The given example n=512

Apply into 3n/2 -2

= (3*512)/2 -2

= 768-2

= 766


Find the minimum and maximum independently, using n-1 comparisons for each, for a total of 2n-2 comparisons. In fact, at most 3⌊n/2⌋ comparisons are sufficient to find both the minimum and the maximum. The strategy is to maintain the minimum and maximum elements seen thus far. Rather than processing each element of the input by comparing it against the current minimum and maximum, at a cost of 2 comparisons per element, we process elements in pairs. We compare pairs of elements from the input first with each other, and then we compare the smaller to the current minimum and the larger to the current maximum, at a cost of 3 comparisons for every 2 elements. 

Setting up initial values for the current minimum and maximum depends on whether n is odd or even. If n is odd, we set both the minimum and maximum to the value of the first element,and then we process the rest of the elements in pairs. If n is even, we perform 1 comparison on the first 2 elements to determine the initial values of the minimum and maximum, and then process the rest of the elements in pairs as in the case for odd n.

Let us analyze the total number of comparisons. If n is odd, then we perform 3⌊n/2⌋ comparisons. If n is even, we perform 1 initial comparison followed by 3(n-2)/2 comparisons, for a total of (3n/2)-2.  Thus, in either case, the total number of comparisons is at most 3⌊n/2⌋.

Given an even number of elements. So, 3n/2 -2 comparisons. 

= (3*512)/2 -2

= 768-2

= 766



By using Tournament Method: 

Step-1: To find the minimum element in the array will take n-1 comparisons. 

We are given 512 elements. So, to find the minimum element in the array will take 512-1= 511 

Step-2: To find the largest element in the array using the tournament method. 

  1. After the first round of Tournament , there will be exactly n/2 numbers =256 that will lose the round.
  2. The biggest loser (the largest number) should be among these 256 loosers.To find the largest number will take (n/2)−1 comparisons =256-1 = 255

Total 511+255= 766

Question 7

FORTRAN implementation do not permit recursion because

they use static allocation for variables
they use dynamic allocation for variables
stacks are not available on all machines
it is not possible to implement recursion on all machines
Question 7 Explanation: 
FORTRAN implementation do not permit recursion because they use the static allocation for variables.
→ Recursion requires dynamic allocation of data.
Question 8

The recurrence relation that arises in relation with the complexity of binary search is:

T(n) = T(n/2) + k, k a constant
T(n) = 2T(n/2) + k, k a constant
T(n) = T(n/2) + log n
T(n) = T(n/2) + n
Question 8 Explanation: 
In binary search, search for the half of the list and constant time for comparing. So,
∴ T(n) = 2T(n/2) + k, k a constant
Question 9

Which of the following algorithm design techniques is used in the quicksort algorithm?

Dynamic programming
Divide and conquer
Greedy method
Question 9 Explanation: 
In quick sort, we use divide and conquer technique.
Question 10

In which one of the following cases is it possible to obtain different results for call-by reference and call-by-name parameter passing methods?

Passing a constant value as a parameter
Passing the address of an array as a parameter
Passing an array element as a parameter
Passing an array following statements is true
Question 10 Explanation: 
Passing an array element as a parameter then it gives different output values for the call-by-reference and call-by-name parameters.
{ ........
a[ ] = {1, 2, 3, 4}
i = 0
print a[0];
fun(int x)
int i = 1;
x = 8;
Call-by-reference = 8
Call-by-value = 1
Question 11

Which one of the following statements is false?

Optimal binary search tree construction can be performed efficiently using dynamic programming.
Breadth-first search cannot be used to find connected components of a graph.
Given the prefix and postfix walks over a binary tree, the binary tree cannot be uniquely constructed.
Depth-first search can be used to find connected components of a graph.
Question 11 Explanation: 
In BFS algorithm, we can randomly select a source vertex and then run, after that whether we need to check distance to each and every vertex from source is still infinite (or) not. If we find any vertex having infinite distance then the graph is not connected.
Question 12

Consider the following two functions:

Which of the following is true?

g1(n) is O(g2(n))
g1 (n) is O(3)
g2 (n) is O(g1 (n))
g2 (n) is O(n)
Both A and B
Question 12 Explanation: 
In asymptotic complexity, we assume sufficiently large n. So, g1(n) = n2 and g2(n) = n3.
Growth rate of g1 is less than that of g2 i.e., g1(n) = O(g2(n)) = O(n).
Question 13

An array A contains n integers in locations A[0],A[1], …………… A[n-1]. It is required to shift the elements of the array cyclically to the left by K places, where 1≤K≤n-1. An incomplete algorithm for doing this in linear time, without using another is given below. Complete the algorithm by filling in the blanks. Assume all variables are suitably declared.

 while _____do
     while _____do
     j:=(j+K) mod n;
 if j
Theory Explanation.
Question 14

(a) Use the patterns given to prove that

(You are not permitted to employ induction)

(b) Use the result obtained in (a) to prove that

Theory Explanation.
Question 15

Consider the following recursive function:

 function fib (1:integer);integer;
 if (n=0) or (n=1) then fib:=1
 else fib:=fib(n-1) + fib(n-2)

The above function is run on a computer with a stack of 64 bytes. Assuming that only return address and parameter and passed on the stack, and that an integer value and an address takes 2 bytes each, estimate the maximum value of n for which the stack will not overflow. Give reasons for your answer.

Theory Explanation.
Question 16

An independent set in a graph is a subset of vertices such that no two vertices in the subset are connected by an edge. An incomplete scheme for a greedy algorithm to find a maximum independent set in a tree is given below:

                      V: Set of all vertices in the tree;        I:=φ;
    While             V ≠ φdo
                      select a vertex u; ∈ V such that
                      V:= V – {u};
                      if u is such that
                      then 1:= I ∪ {u}

(a) Complete the algorithm by specifying the property of vertex u in each case
(b) What is the time complexity of the algorithm.

Theory Explanation.
Question 17

Merge sort uses

Divide and conquer strategy
Backtracking approach
Heuristic search
Greedy approach
Question 17 Explanation: 
Merge sort uses the divide and conquer strategy.
Question 18

The postfix expression for the infix expression
A + B*(C + D)/F + D*E is:

AB + CD + *F/D + E*
ABCD + *F/DE*++
A *B + CD/F *DE++
A + *BCD/F* DE++
None of the above
Question 18 Explanation: 
The postfix expression will be,
A B C D + * F / + D E * +
Question 19

Which of the following statements is true?

    I. As the number of entries in a hash table increases, the number of collisions increases.
    II. Recursive programs are efficient
    III. The worst case complexity for Quicksort is O(n2)
I and II
II and III
I and IV
I and III
Question 19 Explanation: 
Binary search using linked list is not efficient as it will not give O(log n), because we will not be able to find the mid in constant time. Finding mid in linked list takes O(n) time.
Recursive program requires stack for storing the function state. Any recursive program can be rewritten as an iterative one. Advantage of recursion is "less programming effort", while it always lags behind iterative one in terms of performance.
Question 20

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.

Theory Explanation.
Question 21

How many minimum spanning trees does the following graph have? Draw them. (Weights are assigned to the edge).

Theory Explanation.
Question 22

For parameters a and b, both of which are ω(1), T(n) = T(n1/a)+1, and T(b)=1.

Then T(n) is

θ(loga logb n)
θ(logb loga n)
θ(log2 log2 n)
θ(logab n)
Question 22 Explanation: 
T(n) = T(n1/a+1, T(b) = 1
T(n) = [T(n1/a2)+1] + 1
= [T(n1/a3)+1] + 2
= [T(n1/a3)] + 3
= [T(n1/ak)] + b
= logb n = ak
= log logb n = k log a
= k= loga logb n
T(n)=1+loga logb n
T(n)=O(loga logb n)
Question 23

Let G = (V,E) be a weighted undirected graph and let T be a Minimum Spanning Tree (MST) of G maintained using adjacency lists. Suppose a new weighted edge (u,v) ∈ V×V is added to G. The worst case time complexity of determining if T is still an MST of the resultant graph is

θ(|E| log|V|)
Question 23 Explanation: 
• As T is a minimum spanning tree and we need to add a new edge to existing spanning tree.
• Later we need to check still T is a minimum spanning tree or not, So we need to check all vertices whether there is any cycle present after adding a new edge.
• All vertices need to traverse to confirm minimum spanning tree after adding new edge then time complexity is O(V).
Time Complexity:
Total vertices: V, Total Edges : E
• O(logV) – to extract each vertex from the queue. So for V vertices – O(VlogV)
• O(logV) – each time a new pair object with a new key value of a vertex and will be done at most once for each edge. So for total E edge – O(ElogV)
• So overall complexity: O(VlogV) + O(ElogV) = O((E+V)logV) = O(ElogV)
Note: Method-1 is the most appropriate answer for giving a question.
Question 24

Consider a graph G = (V, E), where V = {v1, v2, …, v100}, E = {(vi, vj) | 1 ≤ i < j ≤ 100}, and weight of the edge (vi, vj) is |i - j|. The weight of the minimum spanning tree of G is ______.

Question 24 Explanation: 
• If there are n vertices in the graph, then each spanning tree has n − 1 edges.
• N =100
• Edge weight is |i-j| for Edge (vi,vj) {1<=i<=100}
• The weight of edge(v1,v2) is 1 , edge(v5,v6) is 1.
• So, 99 edges of weight is 99.
Question 25

Assuming P ≠ NP, which of the following is TRUE?

NP-complete = NP
NP-complete ∩ P = ∅
NP-hard = NP
P = NP-complete
Question 25 Explanation: 
Note: Out of syllabus.
The definition of NP-complete is,
A decision problem p is NP-complete if:
1. p is in NP, and
2. Every problem in NP is reducible to p in polynomial time.
It is given that assume P ≠ NP , hence NP-complete ∩ P = ∅ .
This is due to the fact that, if NP-complete ∩ P ≠ ∅ i.e. there are some problem (lets say problem P1) which is in P and in NP-complete also, then it means that P1 (NP-complete problem) can be solved in polynomial time also (as it is also in P class) and this implies that every NP problem can be solve in polynomial time, as we can convert every NP problem into NP-complete problem in polynomial time.
Which means that we can convert every NP problem to P1 and solve in polynomial time and hence P = NP, which is contradiction to the given assumption that P ≠ NP.
Question 26

Let W(n) and A(n) denote respectively, the worst case and average case running time of an algorithm executed on an input of size n. Which of the following is ALWAYS TRUE?

A(n) = Ω (W(n))
A(n) = Θ (W(n))
A(n) = O (W(n))
A(n) = o (W(n))
Question 26 Explanation: 
The average case time can be lesser than or even equal to the worst case.
So, A(n) would be upper bounded by W(n) and it will not be strict upper bound as it can even be same (e.g. Bubble Sort and merge sort).
A(n) = O(W(n))
Note: Option A is wrong because A(n) is not equal to Ω(w(n)) .
Question 27

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;
(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]:= .............
Theory Explanation.
Question 28

Insert the characters of the string K R P C S N Y T J M into a hash table of size 10.
Use the hash function

 h(x) = (ord(x) – ord("a") + 1) mod10 

and linear probing to resolve collisions.
(a) Which insertions cause collisions?
(b) Display the final hash table.

Theory Explanation.
Question 29

A complete, undirected, weighted graph G is given on the vertex {0, 1,...., n−1} for any fixed ‘n’. Draw the minimum spanning tree of G if
(a) the weight of the edge (u,v) is ∣u − v∣
(b) the weight of the edge (u,v) is u + v

Theory Explanation.
Question 30

Let G be the directed, weighted graph shown in below figure.

We are interested in the shortest paths from A.
(a) Output the sequence of vertices identified by the Dijkstra’s algorithm for single source shortest path when the algorithm is started at node A.
(b) Write down sequence of vertices in the shortest path from A to E.
(c) What is the cost of the shortest path from A to E?

Theory Explanation.
Question 31

Consider the following program that attempts to locate an element x in a sorted array a[] using binary search. Assume N>1. The program is erroneous. Under what conditions does the program fail?

      var          i,j,k: integer; x: integer;
                   a:= array; [1...N] of integer;
      begin	   i:= 1; j:= N;
      repeat	   k:(i+j) div 2;
                   if a[k] < x then i:= k 
                   else j:= k 
      until (a[k] = x) or (i >= j);
      if (a[k] = x) then
      writeln ('x is in the array')
      writeln ('x is not in the array')
Theory Explanation.
Question 32

(a) Solve the following recurrence relation:

  xn = 2xn-1 - 1, n>1
  x1 = 2 

(b) Consider the grammar

  S →  Aa | b
  A → Ac | Sd | ε 

Construct an equivalent grammar with no left recursion and with minimum number of production rules.

Theory Explanation.
Question 33

Let A be an n×n matrix such that the elements in each row and each column are arranged in ascending order. Draw a decision tree which finds 1st, 2nd and 3rd smallest elements in minimum number of comparisons.

Theory Explanation.
Question 34

(a) Consider the following algorithm. Assume procedure A and procedure B take O(1) and O(1/n) unit of time respectively. Derive the time complexity of the algorithm in O-notation.

         algorithm what (n)      
                  if n = 1 then call A 
             else begin
                   what (n-1);
                   call B(n)

(b) Write a constant time algorithm to insert a node with data D just before the node with address p of a singly linked list.

Theory Explanation.
Question 35

The root directory of a disk should be placed

at a fixed address in main memory
at a fixed location on the disk
anywhere on the disk
at a fixed location on the system disk
anywhere on the system disk
Question 35 Explanation: 
Root directory can points to the various user directories. Then they will be stored in a way that user can't be easily modify them. Then they should be at fixed location on the disk.
Question 36

Consider a simple connected graph G with n vertices and n-edges (n>2). Then, which of the following statements are true?

G has no cycles.
The graph obtained by removing any edge from G is not connected.
G has at least one cycle.
The graph obtained by removing any two edges from G is not connected.
Both C and D.
Question 36 Explanation: 
If a graph have n vertices and n edges (n>2) then it is to be cyclic graph. Then it have atleast one cycle and if we remove two edges then it is not connected.
For example let us consider, n=3.
Question 37

where O(n) stands for order n is:

B, C, D and E
Question 37 Explanation: 

⇒ In this 'n' is constant. So, n is added to n times itself which is O(n2).
Hence, (a) is wrong. And rest (B), (C), (D), (E) are correct.
Question 38

Consider the recursive algorithm given below:

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

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.

Theory Explanation.
Question 39

Complexity of Kruskal’s algorithm for finding the minimum spanning tree of an undirected graph containing n vertices and m edges if the edges are sorted is __________

O(m log n)
Question 39 Explanation: 
Though the edges are to be sorted still due to union find operation complexity is O(m log n).
Question 40

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

Question 40 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 41

Assume that the last element of the set is used as partition element in Quicksort. If n distinct elements from the set [1…..n] are to be sorted, give an input for which Quicksort takes maximum time.

Question 41 Explanation: 
For n distinct elements the algorithm will take maximum time when:
→ The array is already sorted in ascending order.
→ The array is already sorted in descending order.
Question 42

Consider the function F(n) for which the pseudo code is given below:

     Function F(n)
     F1 ← 1
     if(n=1) then F ← 3
     else For i = 1 to n do
                       C ← 0
     For               j = 1 to F(n – 1) do
                       begin C ← C + 1 end
                       F1 = F1 * C
                       F = F1
     [n is a positive integer greater than zero] 

(a) Derive a recurrence relation for F(n)
(b) Solve the recurrence relation for a closed form solutions of F(n).

Theory Explanation.
Question 43

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

Question 43 Explanation: 
Minimum no. of comparisons
= ⌈log(n!)⌉
= ⌈log(5!)⌉
= ⌈log(120)⌉
= 7
Question 44

The weighted external path length of the binary tree in figure is _____

Question 44 Explanation: 
Question 45

If the binary tree in figure is traversed in inorder, then the order in which the nodes will be visited is ______

4, 1, 6, 7, 3, 2, 5, 8
Question 45 Explanation: 
Inorder traversal is
(Left, Root, Right)
So, the order will be
4, 1, 6, 7, 3, 2, 5, 8
Question 46

Match the pairs in the following questions by writing the corresponding letters only.

(A) The number distinct binary trees with n nodes       (P) n!/2        
(B) The number of binary strings of length of 2n        (Q) (3n/n)
    with an equal number of 0’s and 1’s                                     
(C) The number of even permutations of n objects        (R) (2n/n) 
(D) The number of binary strings of length 6m which     (S) 1/n+1(2n/n) 
    are palindromes with 2n 0’s.
A-S, B-R, C-P, D-Q
Question 47

Choose the correct alternatives (more than one may be correct) and write the corresponding letters only: Kruskal’s algorithm for finding a minimum spanning tree of a weighted graph G with vertices and m edges has the time complexity of:

O(m log n)
B, D and E
Question 47 Explanation: 
Though the edges are sorted still due to finding union operation complexity is O(m log n).
→ Where m is no. of edges and n is number of vertices then n = O(m2)
→ O(m logn) < O(mn)
Question 48

Choose the correct alternatives (more than one may be correct) and write the corresponding letters only: The following sequence of operations is performed on a stack:

 PUSH (10), PUSH (20), POP, PUSH (10), PUSH (20), POP, POP, POP, PUSH (20), POP
The sequence of values popped out is:
20, 10, 20, 10, 20
20, 20, 10, 10, 20
10, 20, 20, 10, 20
20, 20, 10, 20, 10
Question 48 Explanation: 
PUSH(10), PUSH(20), POP = 20 (i)
→ PUSH(10), PUSH(10), PUSH(20), POP = 20 (ii)
→ PUSH(10), PUSH(10), POP = 10 (iii)
→ PUSH(10), POP = 10 (iv)
→ PUSH(20)
→ PUSH(20), POP = 20 (v)
20, 20, 10, 10, 20
Question 49
What is the worst-case number of arithmetic operations performed by recursive binary search on a sorted array of size n?
Question 49 Explanation: 

In this question they given three main constraints

  1. The input array is in sorted order
  2. Use recursive binary search
  3. Worst case number of operations

If the array is in sorted order then the worst case time complexity is O(logn)

Ex: 10, 20, 30

The binary search approach is using either recursive or iterative method is

Step-1: element = middle, the element is found and return the index.

Step-2: element > middle, search for the element in the sub-array starting from middle+1 index to n.

Step-3: element < middle, search for element in the sub-array starting from 0 index to middle -1.

Note: The worst case happens when we want to find the smallest/largest farthest node. So, it will not take more than O(logn) time. 

Question 50
For constants a ≥ 1 and b > 1, consider the following recurrence defined on the non-negative integers:

Which one of the following options is correct about the recurrence T(n)?
Question 50 Explanation: 
There are 50 questions to complete.

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