Algorithms
Question 1 
An array of integers of size n can be converted into a heap by adjusting the heaps rooted at each internal node of the complete binary tree starting at the node ⌊(n – 1) /2⌋, and doing this adjustment up to the root node (root node is at index 0) in the order ⌊(n – 1)/2⌋, ⌊(n – 3)/2⌋, ….., 0. The time required to construct a heap in this manner is
A  O(log n) 
B  O(n) 
C  O(n log log n) 
D  O(n log n) 
And we know that time complexity for building the heap is O(n).
Question 2 
Which one of the following binary trees has its inorder and preorder traversals as BCAD and ABCD, respectively?
A  
B  
C  
D 
Left root right.
Preorder traversal is,
Root left right.
Question 3 
Let f(n), g(n) and h(n) be functions defined for positive inter such that f(n) = O(g(n)), g(n) ≠ O(f(n)), g(n) = O(h(n)), and h(n) = O(g(n)). Which one of the following statements is FALSE?
A  f(n) + g(n) = O(h(n)) + h(n)) 
B  f(n) = O(h(n)) 
C  fh(n) ≠ O(f(n)) 
D  f(n)h(n) ≠ O(g(n)h(n)) 
So, f(n)g(n) = O(g(n)h(n)) is True.
Hence, (D) is false.
Question 4 
Consider the undirected graph below:
Using Prim’s algorithm to construct a minimum spanning tree starting with node A, which one of the following sequences of edges represents a possible order in which the edges would be added to construct the minimum spanning tree?
A  (E, G), (C, F), (F, G), (A, D), (A, B), (A, C) 
B  (A, D), (A, B), (A, C), (C, F), (G, E), (F, G) 
C  (A, B), (A, D), (D, F), (F, G), (G, E), (F, C) 
D  (A, D), (A, B), (D, F), (F, C), (F, G), (G, E)

(C) produces connected component every instead a new edge is added but when first vertex is chosen (first vertex is chosen randomly) first edge must be minimum weight edge that is chosen. Therefore, (A, D) must be chosen before (A, B). Therefore (C) is false.
Question 5 
Consider a list of recursive algorithms and a list of recurrence relations as shown below. Each recurrence relation corresponds to exactly one algorithm and is used to derive the time complexity of the algorithm.
RECURSIVE ALGORITHM RECURRENCE RELATION P. Binary search I. T(n) = T(nk) + T(k) + cn Q. Merge sort II. T(n) = 2T(n1) + 1 R. Quick sort III. T(n) = 2T(n/2) + cn S. Tower of Hanoi IV. T(n) = T(n/2) + 1
A  PII, QIII, RIV, SI 
B  PIV, QIII, RI, SII 
C  PIII, QII, RIV, SI 
D  PIV, QII, RI, SIII 
Binary search – T(n/2) + 1
Merge sort – T(n) = 2T(n/2)+ cn
Tower of hanoi – T(n) = 2T(n1) +1
Question 6 
For parameters a and b, both of which are ω(1), T(n) = T(n^{1/a})+1, and T(b)=1.
Then T(n) is
A  θ(log_{a} log_{b} n) 
B  θ(log_{b} log_{a} n)

C  θ(log_{2} log_{2} n)

D  θ(log_{ab} n)

T(n) = [T(n^{1/a2})+1] + 1
= [T(n^{1/a3})+1] + 2
= [T(n^{1/a3})] + 3
= [T(n^{1/ak})] + b
= log_{b} n = a^{k}
= log log_{b} n = k log a
= k= log_{a} log_{b} n
T(n)=1+log_{a} log_{b} n
T(n)=O(log_{a} log_{b} n)
Question 7 
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
A  θ(E+V) 
B  θ(E logV) 
C  θ(EV) 
D  θ(V) 
• 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).
Method2:
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: Method1 is the most appropriate answer for giving a question.
Question 8 
Consider a graph G = (V, E), where V = {v_{1}, v_{2}, …, v_{100}}, E = {(v_{i}, v_{j})  1 ≤ i < j ≤ 100}, and weight of the edge (v_{i}, v_{j}) is i – j. The weight of the minimum spanning tree of G is ______.
A  99 
• N =100
• Edge weight is ij 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 9 
A  f ( n ^2 ) ( f ( n ) ^2 ), when f ( n ) is a polynomial 
B  f ( n ^2 ) o ( f ( n ) ^2 ) 
C  f ( n ^2 ) O ( f ( n ) ^2 ), when f ( n ) is an exponential function 
D 
f(n)=n^c where c is a constant
f(n^2) = (n^2)^c = n^2c
(f(n))^2 = (n^c)^2 = n^2c
f(n^2) = (f(n))^2 is TRUE asymptotically.
OptionB: FALSE: The small omega function indicates the tightest upper bound.
f(n)^2 < o(f(n)^2) is FALSE asymptotically.
OptionC: FALSE: Consider f(n)=logn
f(n^2)=logn^2 = 2*logn
(f(n))^2 = (logn)^2 = logn * logn
f(n^2) <=Ω(f(n))^2 is FALSE asymptotically.
OptionD: FALSE:
Exponential values(Eg: 2^n, 3^n,..,k^n where k is a constant ”
f(n)=3^n
f(n^2)=f(3^(n^2)
(f(n))^2 = (3^n)^2 = 3^2n
f(n^2) >= O(f(n))^2 is FALSE asymptotically.
Question 10 
A  m/n 
B  n/m 
C  2n/m 
D  n/2m 
How?
For 1st element the probability of key1 ends up in slot 1 is 1/m.
For 2nd element the probability of key2 ends up in slot 2 is 1/m
..
..
For nth element the probability of keyn ends up in slot n is 1/m
Hence expected number of elements in a slot is:
1/m+1/m+… (n times)= n/m
In the given question h1 is for elements at even sequence position and h2 is for odd number of sequence positions. This will not affect the overall probability.
Hence here also it is n/m
Question 11 
A  509 
Now, the 3^rd largest number will be at location of max3. We can find what can be the probable index values of max1 at different levels. They will be 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023 . . .
∴ 511 is index value of our max1
510 is index value of our max2
509 is index value of our max3
& Hence, Answer is 509
Question 12 
A  The edge with the second smallest weight is always part of any minimum spanning tree of G . 
B  One or both of the edges with the third smallest and the fourth smallest weights are part of any minimum spanning tree of G . 
C  
D  G can have multiple minimum spanning trees. 
OptionA: TRUE: As per the above graph, the second minimum edge weight is also part of the MST.
The second smallest weight is always in MST because it will not form a cycle.
OptionB: TRUE: Graph G is having more than 4 vertices. Suppose 3rd smallest element is forming a cycle then it takes 4th smallest element. So, the given statement is TRUE.
OptionC: TRUE: As per the example graph, it is always correct.
OptionD: FALSE: We will get a unique minimum spanning tree if edge weights are distinct.
Question 13 
Then, which of the following statements is/are TRUE?
A  f (2^ n 1) 2^ n 1 
B  f (2 ^n ) 1 
C  f (5 . 2 ^n ) 2^ n + 1 1 
D  f (2^ n + 1) 2 ^n + 1 
Based on the “n” value we can get OptionA, B and C are correct.
Question 14 
A  24 
Graph have 3 elements –> We get 2 MSTs
Graph have 4 elements –> We get 6 MSTs (3*2*1)
Graph have 5 elements –> We get 24 MSTs(4*3*2*1)
Question 15 
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 16 
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.
A  Theory Explanation. 
Question 17 
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
A  Theory Explanation. 
Question 18 
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?
A  Theory Explanation. 
Question 19 
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') else writeln ('x is not in the array') end;
A  Theory Explanation. 
Question 20 
(a) Solve the following recurrence relation:
x_{n} = 2x_{n1}  1, n>1 x_{1} = 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.
A  Theory Explanation. 
Question 21 
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 1^{st}, 2^{nd} and 3^{rd} smallest elements in minimum number of comparisons.
A  Theory Explanation. 
Question 22 
(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 Onotation.
algorithm what (n) begin if n = 1 then call A else begin what (n1); call B(n) end end.
(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.
A  Theory Explanation. 
Question 23 
Merge sort uses
A  Divide and conquer strategy 
B  Backtracking approach 
C  Heuristic search 
D  Greedy approach 
Question 24 
The postfix expression for the infix expression
A + B*(C + D)/F + D*E is:
A  AB + CD + *F/D + E* 
B  ABCD + *F/DE*++ 
C  A *B + CD/F *DE++ 
D  A + *BCD/F* DE++ 
E  None of the above 
A B C D + * F / + D E * +
Question 25 
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(n^{2})
A  I and II 
B  II and III 
C  I and IV 
D  I and III 
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 26 
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 27 
How many minimum spanning trees does the following graph have? Draw them. (Weights are assigned to the edge).
A  Theory Explanation. 
Question 28 
FORTRAN implementation do not permit recursion because
A  they use static allocation for variables 
B  they use dynamic allocation for variables 
C  stacks are not available on all machines 
D  it is not possible to implement recursion on all machines 
→ Recursion requires dynamic allocation of data.
Question 29 
The recurrence relation that arises in relation with the complexity of binary search is:
A  T(n) = T(n/2) + k, k a constant 
B  T(n) = 2T(n/2) + k, k a constant 
C  T(n) = T(n/2) + log n 
D  T(n) = T(n/2) + n 
∴ T(n) = 2T(n/2) + k, k a constant
Question 30 
Which of the following algorithm design techniques is used in the quicksort algorithm?
A  Dynamic programming 
B  Backtracking 
C  Divide and conquer 
D  Greedy method 
Question 31 
In which one of the following cases is it possible to obtain different results for callby reference and callbyname parameter passing methods?
A  Passing a constant value as a parameter 
B  Passing the address of an array as a parameter 
C  Passing an array element as a parameter 
D  Passing an array following statements is true 
{ ........
a[ ] = {1, 2, 3, 4}
i = 0
fun(a[i]);
print a[0];
}
fun(int x)
{
int i = 1;
x = 8;
}
O/p:
Callbyreference = 8
Callbyvalue = 1
Question 32 
Which one of the following statements is false?
A  Optimal binary search tree construction can be performed efficiently using dynamic programming. 
B  Breadthfirst search cannot be used to find connected components of a graph. 
C  Given the prefix and postfix walks over a binary tree, the binary tree cannot be uniquely constructed. 
D  Depthfirst search can be used to find connected components of a graph. 
Question 33 
Consider the following two functions:
Which of the following is true?
A  g_{1}(n) is O(g_{2}(n)) 
B  g_{1} (n) is O(^{3}) 
C  g_{2} (n) is O(g_{1} (n)) 
D  g_{2} (n) is O(n) 
E  Both A and B 
Growth rate of g_{1} is less than that of g_{2} i.e., g_{1}(n) = O(g_{2}(n)) = O(n).
Question 34 
An array A contains n integers in locations A[0],A[1], …………… A[n1]. It is required to shift the elements of the array cyclically to the left by K places, where 1≤K≤n1. 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.
min:=n; i=0; while _____do begin temp:=A[i]; j:=i; while _____do begin A[j]:=_____; j:=(j+K) mod n; if j
A  Theory Explanation. 
Question 35 
(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
A  Theory Explanation. 
Question 36 
Consider the following recursive function:
function fib (1:integer);integer; begin if (n=0) or (n=1) then fib:=1 else fib:=fib(n1) + fib(n2) end;
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.
A  Theory Explanation. 
Question 37 
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 begin select a vertex u; ∈ V such that V:= V – {u}; if u is such that then 1:= I ∪ {u} end; output(I);
(a) Complete the algorithm by specifying the property of vertex u in each case
(b) What is the time complexity of the algorithm.
A  Theory Explanation. 
Question 38 
The root directory of a disk should be placed
A  at a fixed address in main memory 
B  at a fixed location on the disk 
C  anywhere on the disk 
D  at a fixed location on the system disk 
E  anywhere on the system disk 
Question 39 
Consider a simple connected graph G with n vertices and nedges (n>2). Then, which of the following statements are true?
A  G has no cycles. 
B  The graph obtained by removing any edge from G is not connected. 
C  G has at least one cycle. 
D  The graph obtained by removing any two edges from G is not connected. 
E  Both C and D. 
For example let us consider, n=3.
Question 40 
where O(n) stands for order n is:
A  O(n) 
B  O(n^{2}) 
C  O(n^{3}) 
D  O(3n^{2}) 
E  O(1.5n^{2}) 
F  B, C, D and E 
⇒ In this 'n' is constant. So, n is added to n times itself which is O(n^{2}).
Hence, (a) is wrong. And rest (B), (C), (D), (E) are correct.
Question 41 
Consider the recursive algorithm given below:
procedure bubblersort (n); var i,j: index; temp : item; begin for i:=1 to n1 do if A[i] > A [i+1] then begin temp : A[i]; A[i]:=A[i+1]; A[i+1]:=temp end; bubblesort (n1) end
Let a_{n} 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 a_{n} in terms of a_{n1}. Solve for a_{n}.
A  Theory Explanation. 
Question 42 
The numbers 1, 2, .... n are inserted in a binary search tree in some order. In the resulting tree, the right subtree of the root contains p nodes. The first number to be inserted in the tree must be
A  p 
B  p + 1 
C  n  p 
D  n  p + 1 
RST contains elements = p
Root contains = 1 element
1^{st} contains = n  (p + 1) element
Root contains the value is n  p.
Question 43 
In a depthfirst traversal of a graph G with n vertices, k edges are marked as tree edges. The number of connected components in G is
A  k 
B  k + 1 
C  n  k  1 
D  n  k 
In this question, we are going to applying the depth first search traversal on each component of graph where G is connected (or) disconnected which gives minimum spanning tree
i.e., k = np
p = n  k
Question 44 
In the following table, the left column contains the names of standard graph algorithms and the right column contains the time complexities of the algorithms. Match each algorithm with its time complexity.
1. BellmanFord algorithm A: O ( m log n) 2. Kruskal’s algorithm B: O (n^{3}) 3. FloydWarshall algorithm C: O (nm) 4. Topological sorting D: O (n + m)
A  1→ C, 2 → A, 3 → B, 4 → D 
B  1→ B, 2 → D, 3 → C, 4 → A 
C  1→ C, 2 → D, 3 → A, 4 → B 
D  1→ B, 2 → A, 3 → C, 4 → D 
Krushkal's algorithm → O(m log n)
FloydWarshall algorithm → O(n^{3})
Topological sorting → O(n+m)
Question 45 
A hash table contains 10 buckets and uses linear probing to resolve collisions. The key values are integers and the hash function used is key % 10. If the values 43, 165, 62, 123, 142 are inserted in the table, in what location would the key value 142 be inserted?
A  2 
B  3 
C  4 
D  6 
165%10 = 5 [occupy 5]
62%10 = 2 [occupy 2]
123%10 = 3 [3 already occupied, so occupies 4]
142%10 = 2 [2, 3, 4, 5 are occupied, so it occupies 6]
Question 46 
In a binary tree, for every node the difference between the number of nodes in the left and right subtrees is at most 2. If the height of the tree is h > 0, then the minimum number of nodes in the tree is:
A  2^{h1} 
B  2^{h1} + 1 
C  2^{h}  1 
D  2^{h} 
Above tree satisfies the property given in the question. Hence, only option (B) satisfies it.
Question 47 
Let T(n) be a function defined by the recurrence T(n) = 2T(n/2) + √n for n ≥ 2 and T(1) = 1 Which of the following statements is TRUE?
A  T(n) = θ(log n) 
B  T(n) = θ(√n) 
C  T(n) = θ(n) 
D  T(n) = θ(n log n) 
Question 48 
Let G be a weighted undirected graph and e be an edge with maximum weight in G. Suppose there is a minimum weight spanning tree in G containing the edge e. Which of the following statements is always TRUE?
A  There exists a cutset in G having all edges of maximum weight 
B  There exists a cycle in G having all edges of maximum weight 
C  Edge e cannot be contained in a cycle 
D  All edges in G have the same weight 
(B) False, because the cutset of heaviest edge may contain only one edge.
(C) False. The cutset may form cycle with other edge.
(D) False. Not always true.
Question 49 
A binary search tree contains the numbers 1, 2, 3, 4, 5, 6, 7, 8. When the tree is traversed in preorder and the values in each node printed out, the sequence of values obtained is 5, 3, 1, 2, 4, 6, 8, 7. If the tree is traversed in postorder, the sequence obtained would be
A  8, 7, 6, 5, 4, 3, 2, 1 
B  1, 2, 3, 4, 8, 7, 6, 5 
C  2, 1, 4, 3, 6, 7, 8, 5 
D  2, 1, 4, 3, 7, 8, 6, 5 
5, 3, 1, 2, 4, 6, 8, 7
Inorder is the sorted sequence, i.e.,
1, 2, 3, 4, 5, 6, 7, 8
And we know that with given inorder and preorder, we can draw a unique BST.
So, the BST will be,
Hence, postorder traversasl sequence is,
2, 1, 4, 3, 7, 8, 6, 5
Question 50 
Let G be a directed graph whose vertex set is the set of numbers from 1 to 100. There is an edge from a vertex i to a vertex j iff either j = i + 1 or j = 3i. The minimum number of edges in a path in G from vertex 1 to vertex 100 is
A  4 
B  7 
C  23 
D  99 
j = i +1
(or)
j = 3i
Second option will help us reach from 1 to 100 rapidly. The trick to solve this question is to think in reverse way. Instead of finding a path from 1 to 100, try to find a path from 100 to 1.
So, the edge sequence with minimum number of edges is
1 → 3 → 9 → 10 → 11 → 33 → 99 → 100
which consists of 7 edges.