Answers withexplanations

1) The number of distinct minimum spanning trees for the weighted graph below is ____
(A) 6
(B) 7
(C) 8
(D) 9
Answer: (A)6
Highlighted (in green) are the edges picked to make a MST. In the right side of MST, we could either pick
edge ‘a’ or ‘b’. In the left side, we could either pick ‘c’ or ‘d’ or ‘e’ in MST.
There are 2 options for one edge to be picked and 3 options for another edge to be picked. Therefore,
total 2*3 possible MSTs.

2) Consider the following rooted tree with the vertex P labeled as root
The order in which the nodes are visited during in-order traversal is
(A) SQPTRWUV
(B) SQPTURWV
(C) SQPTWUVR
(D) SQPTRUWV
Answer: (A)
The only confusion in this question is, there are 3 children of R. So when should R appear – after U or
after R? There are two possibilities: SQPTRWUV and SQPTWURV. Only 1st possibility is present as an
option A, the IInd possibility is not there. Therefore option A is the right answer.
3) The minimum number of arithmetic operations required to evaluate the polynomial P(X) = X5
+
4X3
+ 6X + 5 for a given value of X using only one temporary variable.
(A) 6
(B) 7
(C) 8
(D) 9
Answer: B
we can parenthesize the polynomial to minimize the number of operations (See Horner’s Method). We
get X(X2
(X2
+ 4) + 6) + 5 after parenthesization.
Following is sequence of operations to be used.
Note that we are allowed to use only one variable.
res = X*X
res = res + 4
res = X*res
res = X*res
res = res + 6
res = X*res
res = res + 5
4) 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(nLogn)
(C) Θ(nLogn)
(D) O(n3
)

Answer: (A)
The middle element may always be an extreme element (minimum or maximum) in sorted order,
therefore time complexity in worst case becomes O(n2
)
5) Consider the pseudocode given below. The function DoSomething() takes as argument a
pointer to the root of an arbitrary tree represented by the leftMostChild-rightSibling
representation. Each node of the tree is of type treeNode.
typedef struct treeNode* treeptr;
struct treeNode {
treeptr leftMostChild, rightSibling;
};
int DoSomething (treeptr tree) {
int value = 0;
if (tree != NULL) {
if (tree->leftMostChild == NULL)
value = 1;
else
value = DoSomething(tree->leftMostChild);
value = value + DoSomething(tree->rightSibling);
}
return(value);
When the pointer to the root of a tree is passed as the argument to DoSomething, the value
returned by the function corresponds to the
(A) number of internal nodes in the tree.
(B) height of the tree.
(C) number of nodes without a right sibling in the tree.
(D) number of leaf nodes in the tree.
Answer: (D)
The important thing to note in this question is tree’s representation. Tree is represented as leftmost
child and right sibling form. So if the leftmost child is NULL for a node, then there is no child of this node.
If we take a look at the function, we can notice that the function increments the “value” by 1 only for a
leaf node.
6) Suppose a polynomial time algorithm is discovered that correctly computes the largest clique
in a given graph. In this scenario, which one of the following represents the correct Venn diagram
of the complexity classes P, NP and NP Complete (NPC)?

(A) A
(B) B
(C) C
(D) D
Answer: (D)
Largest Clique is an NP complete problem. If one NP complete problem can be solved in polynomial time,
then all of them can be. So NPC set becomes equals to P.
7) Consider two strings A = “qpqrr” and B = “pqprqrp”. Let x be the length of the longest common
subsequence (not necessarily contiguous) between A and B and let y be the number of such longest
common subsequences between A and B. Then x + 10y = ___.
(A) 43
(B) 34
(C) 57
(D) 75
Answer: (A)34
The longest length is 4. There are 3 LCS of length 4 “qprr”, “pqrr” and “qpqr”.
8) Suppose P, Q, R, S, T are sorted sequences having lengths 20, 24, 30, 35, 50 respectively. They are to
be merged into a single sequence by merging together two sequences at a time. The number of
comparisons that will be needed in the worst case by the optimal algorithm for doing this is ____.
Answer: (B)358
To merge two lists of size m and n, we need to do m+n-1 comparisons in worst case. Since we need to
merge 2 at a time, the optimal strategy would be to take smallest size lists first. The reason for picking
smallest two items is to carry minimum items for repetition in merging.
We first merge 20 and 24 and get a list of 44 using 43 worst case comparisons. Then we merge 30 and
35 into a list of 65 using 64 worst case comparisons. Then we merge 50 and 44 into a list of 94 using 93
comparisons. Finally we merge 94 and 65 using 158 comparisons. So total number of comparisons is 43
+ 64 + 93 + 158 which is 358.
9) Consider the tree arcs of a BFS traversal from a source node W in an unweighted, connected,
undirected graph. The tree T formed by the tree arcs is a data structure for computing.
(A) the shortest path between every pair of vertices.
(B) the shortest path from W to every vertex in the graph.

(C) the shortest paths from W to only those nodes that are leaves of T.
(D) the longest path in the graph
Answer: (B)
BFS always produces shortest paths from source to all other vertices in an unweighted graph. The
reason is simple, in BFS, we first explore all vertices which are 1 edge away from source, then we explore
all vertices which are 2 edges away from the source and so on. This property of BFS makes it useful in
many algorithms like Edmonds–Karp algorithm.
10) Consider the following pseudo code. What is the total number of multiplications to be performed?
D = 2
for i = 1 to n do
for j = i to n do
for k = j + 1 to n do
D = D * 3
(A) Half of the product of the 3 consecutive integers.
(B) One-third of the product of the 3 consecutive integers.
(C) One-sixth of the product of the 3 consecutive integers.
(D) None of the above.
Answer (D)
The statement “D = D * 3″ is executed n*(n+1)*(n-1)/6 times. Let us see how.
For i = 1, the multiplication statement is executed (n-1) + (n-2) + .. 2 + 1 times.
For i = 2, the statement is executed (n-2) + (n-3) + .. 2 + 1 times
………………………..
……………………….
For i = n-1, the statement is executed once.
For i = n, the statement is not executed at all
So overall the statement is executed following times
[(n-1) + (n-2) + .. 2 + 1] + [(n-2) + (n-3) + .. 2 + 1] + … + 1 + 0
The above series can be written as
S = [n*(n-1)/2 + (n-1)*(n-2)/2 + ….. + 1]
The sum of above series can be obtained by trick of subtraction the series from standard Series S1 = n2
+
(n-1)2
+ .. 12
. The sum of this standard series is n*(n+1)*(2n+1)/6
S1 – 2S = n + (n-1) + … 1 = n*(n+1)/2
2S = n*(n+1)*(2n+1)/6 – n*(n+1)/2
S = n*(n+1)*(n-1)/6
11) Consider a hash table with 9 slots. The hash function is h(k) = k mod 9. The collisions are
resolved by chaining. The following 9 keys are inserted in the order: 5, 28, 19, 15, 20, 33, 12, 17, 10.
The maximum, minimum, and average chain lengths in the hash table, respectively, are
(A) 3, 0, and 1
(B) 3, 3, and 3
(C) 4, 0, and 1
(D) 3, 0, and 2
Answer: (A)
Following are values of hash function for all keys

5 --> 5
28 --> 1
19 --> 1 [Chained with 28]
15 --> 6
20 --> 2
33 --> 6 [Chained with 15]
12 --> 3
17 --> 8
10 --> 1 [Chained with 28 and 19]
The maximum chain length is 3. The keys 28, 19 and 10 go to same slot 1, and form a chain of length 3.
The minimum chain length 0, there are empty slots (0, 4 and 7).
Average chain length is (0 + 3 + 1 + 1 + 0 + 1 + 2 + 0 + 1)/9 = 1
12) A priority queue is implemented as a Max-Heap. Initially, it has 5 elements. The level-order
traversal of the heap is: 10, 8, 5, 3, 2. Two new elements 1 and 7 are inserted into the heap in that
order. The level-order traversal of the heap after the insertion of the elements is:
(A) 10, 8, 7, 3, 2, 1, 5
(B) 10, 8, 7, 2, 3, 1, 5
(C) 10, 8, 7, 1, 2, 3, 5
(D) 10, 8, 7, 5, 3, 2, 1
Answer: (A)
Initially heap has 10, 8, 5, 3, 2
10
/
8 5
/
3 2
After insertion of 1
10
/
8 5
/ /
3 2 1
No need to heapify as 5 is greater than 1.
After insertion of 7
10
/
8 5
/ /
3 2 1 7
Heapify 5 as 7 is greater than 5
10
/
8 7
/ /
3 2 1 5
No need to heapify any further as 10 is
greater than 7

13) Which one of the following correctly determines the solution of the recurrence relation with T(1) =
1?
T(n) = 2T(n/2) + Logn
(A) Θ(n)
(B) Θ(nLogn)
(C) Θ(n*n)
(D) Θ(log n)
Answer: (A)
This can be solved using Master Method. It falls in case 1.
14) Suppose implementation supports an instruction REVERSE, which reverses the order of elements
on the stack, in addition to the PUSH and POP instructions. Which one of the following statements is
TRUE with respect to this modified stack?
(A) A queue cannot be implemented using this stack.
(B) A queue can be implemented where ENQUEUE takes a single instruction and DEQUEUE takes a
sequence of two instructions.
(C) A queue can be implemented where ENQUEUE takes a sequence of three instructions and
DEQUEUE takes a single instruction.
(D) A queue can be implemented where both ENQUEUE and DEQUEUE take a single instruction each.
Answer: (C)
To DEQUEUE an item, simply POP.
To ENQUEUE an item, we can do following 3 operations
1) REVERSE
2) PUSH
3) REVERSE
15) Let G be a graph with n vertices and m edges. What is the tightest upper bound on the running
time on Depth First Search of G? Assume that the graph is represented using adjacency matrix.
(A) O(n)
(B) O(m+n)
(C) O(n2
)
(D) O(mn)
Answer: (C)
Explanation: Depth First Search of a graph takes O(m+n) time when the graph is represented using
adjacency list.
In adjacency matrix representation, graph is represented as an “n x n” matrix. To do DFS, for every
vertex, we traverse the row corresponding to that vertex to find all adjacent vertices (In adjacency list
representation we traverse only the adjacent vertices of the vertex). Therefore time complexity becomes
O(n2
)
16) Consider the directed graph given below. Which one of the following is TRUE?

(A) The graph doesn’t have any topological ordering
(B) Both PQRS and SRPQ are topological ordering
(C) Both PSRQ and SPRQ are topological ordering
(D) PSRQ is the only topological ordering
Answer: (C)
Explanation: The graph doesn’t contain any cycle, so there exist topological ordering.
P and S must appear before R and Q because there are edges from P to R and Q, and from S to R and
Q. See Topological Ordering for more details.
17) Let P be a QuickSort Program to sort numbers in ascending order using the first element as
pivot. Let t1 and t2 be the number of comparisons made by P for the inputs {1, 2, 3, 4, 5} and {4, 1,
5, 3, 2} respectively. Which one of the following holds?
(A) t1 = 5
(B) t1 < t2
(C) t1 > t2
(D) t1 = t2
Answer: (C)
Explanation: When first element or last element is chosen as pivot, Quick Sort‘s worst case occurs for
the sorted arrays.
In every step of quick sort, numbers are divided as per the following recurrence.
T(n) = T(n-1) + O(n)
18) Consider the following C function in which size is the number of elements in the array E:
The value returned by the function MyX is the
int MyX(int *E, unsigned int size)
{
int Y = 0;
int Z;
int i, j, k;
for (i = 0; i < size; i++)
Y = Y + E[i];
for (i = 0; i < size; i++)
for (j = i; j < size; j++)
{
Z = 0;
for (k = i; k <= j; k++)
Z = Z + E[k];
if (Z > Y)
Y = Z;
}

return Y;
}
(A) maximum possible sum of elements in any sub-array of array E.
(B) maximum element in any sub-array of array E.
(C) sum of the maximum elements in all possible sub-arrays of array E
(D) the sum of all the elements in the array E.
Answer: (A)
Explanation: The function does following
Y is used to store maximum sum seen so far and Z is used to store current sum
1) Initialize Y as sum of all elements
2) For every element, calculate sum of all subarrays starting with arr[i]. Store the current sum in Z. If Z is
greater than Y, then update Y.
19) The preorder traversal sequence of a binary search tree is 30, 20, 10, 15, 25, 23, 39, 35, 42.
Which one of the following is the postorder traversal sequence of the same tree?
(A) 10, 20, 15, 23, 25, 35, 42, 39, 30
(B) 15, 10, 25, 23, 20, 42, 35, 39, 30
(C) 15, 20, 10, 23, 25, 42, 35, 39, 30
(D) 15, 10, 23, 25, 20, 35, 42, 39, 30
Ans (D)
The following is the constructed tree
30
/
20 39
/ /
10 25 35 42
/
15 23
20) Consider the following function
int unknown(int n) {
int i, j, k = 0;
for (i = n/2; i <= n; i++)
for (j = 2; j <= n; j = j * 2)
k = k + n/2;
return k;
}
What is the returned value of the above function?
(A) Θ(n^2)
(B) Θ(n^2Logn)
(C) Θ(n^3)
(D) Θ(n^3Logn)
Answer (B)
The outer loop runs n/2 or Θ(n) times. The inner loop runs Θ(Logn) times (Note that j is divide by 2 in
every iteration). So the statement "k = k + n/2;" runs Θ(nLogn) times. The statement increases value of k
by n/2. So the value of k becomes n/2*Θ(nLogn) which is Θ(n^2Logn)
21) Which of the following statements is/are TRUE for an undirected graph?
P: Number of odd degree vertices is even
Q: Sum of degrees of all vertices is even

A) P Only
B) Q Only
C) Both P and Q
D) Neither P nor Q
Answer (C)
Q is true: Since the graph is undirected, every edge increases the sum of degrees by 2.
P is true: If we consider sum of degrees and subtract all even degrees, we get an even number (because
Q is true). So total number of odd degree vertices must be even.
22) Consider an undirected random graph of eight vertices. The probability that there is an edge
between a pair of vertices is 1/2. What is the expected number of unordered cycles of length
three?
(A) 1/8
(B) 1
(C) 7
(D) 8
Answer (C)
A cycle of length 3 can be formed with 3 vertices. There can be total 8C3 ways to pick 3 vertices from 8.
The probability that there is an edge between two vertices is 1/2. So expected number of unordered
cycles of length 3 = (8C3)*(1/2)^3 = 7
23) Which of the following statements are TRUE?
(1) The problem of determining whether there exists a cycle in an undirected graph is in P.
(2) The problem of determining whether there exists a cycle in an undirected graph is in NP.
(3) If a problem A is NP-Complete, there exists a non-deterministic polynomial time algorithm to
solve A.
(A) 1,2 and 3
(B) 1 and 2 only
(C) 2 and 3 only
(D) 1 and 3 only
Answer (A)
1 is true because cycle detection can be done in polynomial time using DFS
2 is true because P is a subset of NP.
3 is true because NP complete is also a subset of NP and NP means Non-deterministic Polynomial time
solution exists.
24) Consider the directed graph shown in the figure below. There are multiple shortest paths
between vertices S and T. Which one will be reported by Dijstra?s shortest path algorithm?
Assume that, in any iteration, the shortest path to a vertex v is updated only when a strictly
shorter path to v is discovered.

(A) SDT
(B) SBDT
(C) SACDT
(D) SACET
Answer (D)
25) Suppose a circular queue of capacity (n – 1) elements is implemented with an array of n elements.
Assume that the insertion and deletion operation are carried out using REAR and FRONT as array
index variables, respectively. Initially, REAR = FRONT = 0. The conditions to detect queue full and
queue empty are
(A) Full: (REAR+1) mod n == FRONT, empty: REAR == FRONT
(B) Full: (REAR+1) mod n == FRONT, empty: (FRONT+1) mod n == REAR
(C) Full: REAR == FRONT, empty: (REAR+1) mod n == FRONT
(D) Full: (FRONT+1) mod n == REAR, empty: REAR == FRONT
Answer (A)
26) Assuming P != NP, which of the following is true ?
(A) NP-complete = NP
(B) NP-complete ∩ P = Φ
(C) NP-hard = NP
(D) P = NP-complete
Answer (B)
The answer is B (no NP-Complete problem can be solved in polynomial time). Because, if one NP-
Complete problem can be solved in polynomial time, then all NP problems can solved in polynomial time.
If that is the case, then NP and P set become same which contradicts the given condition.
27) An undirected graph G(V, E) contains n ( n > 2 ) nodes named v1 , v2 ,….vn. Two nodes vi , vj
are connected if and only if 0 < |i – j| <= 2. Each edge (vi, vj ) is assigned a weight i + j. A sample
graph with n = 4 is shown below.

What will be the cost of the minimum spanning tree (MST) of such a graph with n nodes?
(A) 1/12(11n^2 – 5n)
(B) n^2 – n + 1
(C) 6n – 11
(D) 2n + 1
Answer: (B)
Minimum spanning tree for 2 nodes would be
(v1) _ (v2)
Total weight 3
(v1) _ (v2)
|
(v3)
Total weight= 3 + 4 = 7
(v1) _ (v2) _ (v4)
|
(v3)
Total weight= 3 + 4 + 6 = 13
(v1) _ (v2) _ (v4)
|
(v3)
|
(v5)
Total weight= 3 + 4 + 6 + 8 = 21
(v1) _ (v2) _ (v4) _ (v6)
|
(v3)
|
(v5)
Total weight= 3 + 4 + 6 + 8 + 10 = 31
We can observe from above examples that when we add kth node, the weight of spanning tree
increases by 2k-2. Let T(n) be the weight of minimum spanning tree. T(n) can be written as
T(n) = T(n-1) + (2n-2) for n > 2
T(1) = 0, T(2) = 0 and T(2) = 3
The recurrence can be written as sum of series (2n – 2) + (2n-4) + (2n-6) + (2n-8) + …. 3 and solution of
this recurrence is n^2 – n + 1.

28) Consider two binary operators ‘↑ ‘ and ‘↓’ with the precedence of operator ↓ being lower than
that of the ↑ operator. Operator ↑ is right associative while operator ↓ is left associative. Which
one of the following represents the parse tree for expression (7 ↓ 3 ↑ 4 ↑ 3 ↓ 2)?
Answer: (B)
Let us consider the given expression (7 ↓ 3 ↑ 4 ↑ 3 ↓ 2).
Since the precedence of ↑ is higher, the sub-expression ([3 ↑ 4 ↑ 3) will be evaluated first. In this sub-
expression, 4 ↑ 3 would be evaluated first because ↑ is right to left associative. So the expression is
evaluated as ((7 ↓ (3 ↑ (4 ↑ 3))) ↓ 2). Also, note that among the two ↓ operators, first one is evaluated
before the second one because the associativity of ↓ is left to right.
29) A max-heap is a heap where the value of each parent is greater than or equal to the values of
its children. Which of the following is a max-heap?

Answer: (B)
A binary tree is max-heap if it is a complete binary tree (A complete binary tree is a binary tree in which
every level, except possibly the last, is completely filled, and all nodes are as far left as possible) and it
follows the max-heap property (value of each parent is greater than or equal to the values of its children).
A) is not a max-heap because it is not a complete binary tree
B) is a max-heap because it is complete binary tree and follows max-heap property.
C) is not a max-heap because 8 is a chile of 5 in this tree, so violates the max-heap property.
D) is not a max-heap because 8 is a chile of 5 in this tree, so violates the max-heap property. There are
many other nodes in this tree which violate max-heap property in this tree.
30) Four matrices M1, M2, M3 and M4 of dimensions pxq, qxr, rxs and sxt respectively can be
multiplied is several ways with different number of total scalar multiplications. For example, when
multiplied as ((M1 X M2) X (M3 X M4)), the total number of multiplications is pqr + rst + prt. When
multiplied as (((M1 X M2) X M3) X M4), the total number of scalar multiplications is pqr + prs + pst.
If p = 10, q = 100, r = 20, s = 5 and t = 80, then the number of scalar multiplications needed is
A) 248000
B) 44000
C) 19000
D) 25000
Answer (C)
We get minimum number of multiplications using ((M1 X (M2 X M3)) X M4).
Total number of multiplications = 100x20x5 (for M2 x M3) + 10x100x5 + 10x5x80 = 19000.
31) Which of the given options provides the increasing order of asymptotic complexity of
functions f1, f2, f3 and f4?
f1(n) = 2^n
f2(n) = n^(3/2)
f3(n) = nLogn
f4(n) = n^(Logn)
A) f3, f2, f4, f1
B) f3, f2, f1, f4
C) f2, f3, f1, f4
D) f2, f3, f4, f1
Answer (A)
f1(n) = 2^n
f2(n) = n^(3/2)
f3(n) = nLogn
f4(n) = n^(Logn)
Except f3, all other are exponential. So f3 is definitely first in output. Among remaining, n^(3/2) is
next.One way to compare f1 and f4 is to take Log of both functions. Order of growth of Log(f1(n)) is Θ(n)
and order of growth of Log(f4(n)) is Θ(Logn * Logn). Since Θ(n) has higher growth than Θ(Logn * Logn),
f1(n) grows faster than f4(n).
32) We are given a set of n distinct elements and an unlabeled binary tree with n nodes. In how
many ways can we populate the tree with the given set so that it becomes a binary search tree?
A) 0
B) 1
C) n!
D) (1/(n+1)).2nCn
Answer (B)

33) An algorithm to find the length of the longest monotonically increasing sequence of numbers
in an array A[0 :n-1] is given below.
Let Li denote the length of the longest monotonically increasing sequence starting at index i in
the array
Which of the following statements is TRUE?
(A) The algorithm uses dynamic programming paradigm
(B) The algorithm has a linear complexity and uses branch and bound paradigm
(C) The algorithm has a non-linear polynomial complexity and uses branch and bound paradigm
(D) The algorithm uses divide and conquer paradigm.
Answer: (A)
34) consider a complete undirected graph with vertex set {0, 1, 2, 3, 4}. Entry Wij in the matrix W
below is the weight of the edge {i, j}.
What is the minimum possible weight of a spanning tree T in this graph such that vertex 0 is a leaf
node in the tree T?
(A) 7
(B) 8
(C) 9
(D) 10
Answer (D)
To get the minimum spanning tree with vertex 0 as leaf, first remove 0th row and 0th column and then get
the minimum spanning tree (MST) of the remaining graph. Once we have MST of the remaining graph,
connect the MST to vertex 0 with the edge with minimum weight (we have two options as there are two 1s
in 0th row).
35) In the graph given in question 1, what is the minimum possible weight of a path P from vertex
1 to vertex 2 in this graph such that P contains at most 3 edges?
(A) 7
(B) 8
(C) 9
(D) 10
Answer (B)
Path: 1 -> 0 -> 4 -> 2
Weight: 1 + 4 + 3

36) The degree sequence of a simple graph is the sequence of the degrees of the nodes in the
graph in decreasing order. Which of the following sequences cannot be the degree sequence of
any graph?
I. 7, 6, 5, 4, 4, 3, 2, 1
II. 6, 6, 6, 6, 3, 3, 2, 2
III. 7, 6, 6, 4, 4, 3, 2, 2
IV. 8, 7, 7, 6, 4, 2, 1, 1
(A) I and II
(B) III and IV
(C) IV only
(D) II and IV
Answer (D)
In sequence IV, we have a vertex with degree 8 which is not possible in a simple graph (no self-loops and
no multiple edges) with total vertex count as 8. Maximum possible degree in such a graph is 7.
In sequence II, four vertices are connected to 6 other vertices, but remaining 4 vertices have degrees as
3, 3, 2 and 2 which are not possible in a simple graph (no self-loops and no multiple edges).
37) Consider a B+-tree in which the maximum number of keys in a node is 5. What is the minimum
number of keys in any non-root node?
(A) 1
(B) 2
(C) 3
(D) 4
Answer (B)
Since the maximum number of keys is 5, maximum number of children a node can have is 6. By definition
of B Tree, minimum children that a node can have would be 6/2 = 3. Therefore, minimum number of keys
that a node can have becomes 2 (3-1).
38) A hash table of length 10 uses open addressing with hash function h(k)=k mod 10, and linear
probing. After inserting 6 values into an empty hash table, the table is as shown below.
Which one of the following choices gives a possible order in which the key values could have
been inserted in the table?
(A) 46, 42, 34, 52, 23, 33
(B) 34, 42, 23, 52, 33, 46
(C) 46, 34, 42, 23, 52, 33
(D) 42, 46, 33, 23, 34, 52
Answer (C)
The sequence (A) doesn’t create the hash table as the element 52 appears before 23 in this sequence.
The sequence (B) doesn’t create the hash table as the element 33 appears before 46 in this sequence.

The sequence (C) creates the hash table as 42, 23 and 34 appear before 52 and 33, and 46 appears
before 33.
The sequence (D) doesn’t create the hash table as the element 33 appears before 23 in this sequence.
39) How many different insertion sequences of the key values using the same hash function and
linear probing will result in the hash table shown above?
(A) 10
(B) 20
(C) 30
(D) 40
Answer (C)
In a valid insertion sequence, the elements 42, 23 and 34 must appear before 52 and 33, and 46 must
appear before 33.
Total number of different sequences = 3! x 5 = 30
In the above expression, 3! is for elements 42, 23 and 34 as they can appear in any order, and 5 is for
element 46 as it can appear at 5 different places.
40) Which one of the following is a key factor for preferring B-trees to binary search trees for
indexing database relations?
(a) Database relations have a large number of records
(b) Database relations are sorted on the primary key
(c) B-trees require less memory than binary search trees
(d) Data transfer form disks is in blocks.
Answer (d)
A disk block contains fairly large number of keys. Unlike BST where each node contains only one key, B-
Tree is designed to contain large number of keys so that tree height is small.
41) How many distinct binary search trees can be created out of 4 distinct keys?
(a) 5
(b) 14
(c) 24
(d) 42
Answer (b)
Here is a systematic way to enumerate these BSTs. Consider all possible binary search trees with each
element at the root. If there are n nodes, then for each choice of root node, there are n – 1 non-root
nodes and these non-root nodes must be partitioned into those that are less than a chosen root and those
that are greater than the chosen root.
Let’s say node i is chosen to be the root. Then there are i – 1 nodes smaller than i and n – i nodes bigger
than i. For each of these two sets of nodes, there is a certain number of possible subtrees.
Let t(n) be the total number of BSTs with n nodes. The total number of BSTs with i at the root is t(i – 1) t(n
– i). The two terms are multiplied together because the arrangements in the left and right subtrees are
independent. That is, for each arrangement in the left tree and for each arrangement in the right tree, you
get one BST with i at the root.
Summing over i gives the total number of binary search trees with n nodes.
The base case is t(0) = 1 and t(1) = 1, i.e. there is one empty BST and there is one BST with one node.

42) In a complete k-ary tree, every internal node has exactly k children. The number of leaves in
such a tree with n internal nodes is:
(a) nk
(b) (n – 1) k+ 1
(c) n( k – 1) + 1
(d) n(k – 1)
Answer (c)
43) A program P reads in 500 integers in the range [0..100] representing the scores of 500
students. It then prints the frequency of each score above 50. What would be the best way for P to
store the frequencies?
(a) An array of 50 numbers
(b) An array of 100 numbers
(c) An array of 500 numbers
(d) A dynamically allocated array of 550 numbers
Answer (a)
An array of size 50 looks the best option to store number of students for each score. We need to store
frequencies of scores above 50. We can ignore scores below 50 and to index the scores above 50, we
can subtract 50 from the score value/
44) An undirected graph G has n nodes. Its adjacency matrix is given by an n × n square matrix
whose (i) diagonal elements are 0‘s and (ii) non-diagonal elements are 1‘s. which one of the
following is TRUE?
(a) Graph G has no minimum spanning tree (MST)
(b) Graph G has a unique MST of cost n-1
(c) Graph G has multiple distinct MSTs, each of cost n-1
(d) Graph G has multiple spanning trees of different costs
Answer (c)
If all non-diagonal elements are 1, then every vertex is connected to every other vertex in the graph with
an edge of weight 1. Such a graph has multiple distinct MSTs with cost n-1. See the below example.
The connected graph:
Below are three Minimum Spanning trees each of cost 2.0.
Minimum Spanning Tree 1

45) The subset-sum problem is defined as follows. Given a set of n positive integers, S = {a1 ,a2
,a3 ,…,an} and positive integer W, is there a subset of S whose elements sum to W? A dynamic
program for solving this problem uses a 2-dimensional Boolean array X, with n rows and W+1
columns. X[i, j],1 <= i <= n, 0 <= j <= W, is TRUE if and only if there is a subset of {a1 ,a2 ,...,ai}
whose elements sum to j. Which of the following is valid for 2 <= i <= n and ai <= j <= W?
(A) X[i, j] = X[i – 1, j] V X[i, j -ai]
(B) X[i, j] = X[i – 1, j] V X[i – 1, j – ai]
(C) X[i, j] = X[i – 1, j] V X[i, j – ai]
(D) X[i, j] = X[i – 1, j] V X[i -1, j – ai]
Answer (B)
X[I, j] (2 <= i <= n and ai <= j <= W), is true if any of the following is true 1) Sum of weights excluding ai is
equal to j, i.e., if X[i-1, j] is true. 2) Sum of weights including ai is equal to j, i.e., if X[i-1, j-ai] is true so that
we get (j – ai) + ai as j.
46) A set X can be represented by an array x[n] as follows:
Consider the following algorithm in which x,y and z are Boolean arrays of size n:
algorithm zzz(x[] , y[], z [])
{
int i;
for (i=O; i<n; ++i)
z[i] = (x[i] ^ ~y[i]) V (~x[i] ^ y[i])
}
The set Z computed by the algorithm is:
(A) (X Intersection Y)
(B) (X Union Y)

(C) (X-Y) Intersection (Y-X)
(D) (X-Y) Union (Y-X)
Answer (D)
The expression x[i] ^ ~y[i]) results the only 1s in x where corresponding entry in y is 0. An array with these
set bits represents set X – Y
The expression ~x[i] ^ y[i]) results the only 1s in y where corresponding entry in x is 0. An array with these
set bits represents set Y – X.
The operator “V” results in Union of the above two sets.
47) Let X be a problem that belongs to the class NP. Then which one of the following is TRUE?
(A) There is no polynomial time algorithm for X.
(B) If X can be solved deterministically in polynomial time, then P = NP.
(C) If X is NP-hard, then it is NP-complete.
(D) X may be undecidable.
Answer (C)
(A) is incorrect because set NP includes both P(Polynomial time solvable) and NP-Complete .
(B) is incorrect because X may belong to P (same reason as (A))
(C) is correct because NP-Complete set is intersection of NP and NP-Hard sets.
(D) is incorrect because all NP problems are decidable in finite set of operations.
48) Consider the polynomial p(x) = a0 + a1x + a2x^2 +a3x^3, where ai != 0, for all i. The minimum
number of multiplications needed to evaluate p on an input x is:
(A) 3
(B) 4
(C) 6
(D) 9
Answer (A)
Multiplications can be minimized using following order for evaluation of the given expression.
p(x) = a0 + x(a1 + x(a2 + a3x))
49) To implement Dijkstra’s shortest path algorithm on unweighted graphs so that it runs in linear
time, the data structure to be used is:
(A) Queue
(B) Stack
(C) Heap
(D) B-Tree
Answer(A)
The shortest path in an un-weighted graph means the smallest number of edges that must be traversed in
order to reach the destination in the graph. This is the same problem as solving the weighted version
where all the weights happen to be 1. If we use Queue (FIFO) instead of Priority Queue (Min Heap), we
get the shortest path in linear time O(|V| + |E|). Basically we do BFS traversal of the graph to get the
shortest paths.
50) A 3-ary max heap is like a binary max heap, but instead of 2 children, nodes have 3 children. A
3-ary heap can be represented by an array as follows: The root is stored in the first location, a[0],
nodes in the next level, from left to right, is stored from a[1] to a[3]. The nodes from the second
level of the tree from left to right are stored from a[4] location onward. An item x can be inserted
into a 3-ary heap containing n items by placing x in the location a[n] and pushing it up the tree to
satisfy the heap property.
Which one of the following is a valid sequence of elements in an array representing 3-ary max
heap?
(A) 1, 3, 5, 6, 8, 9
(B) 9, 6, 3, 1, 8, 5
(C) 9, 3, 6, 8, 5, 1
(D) 9, 5, 6, 8, 3, 1

Answer (D)
9
/ |
/ |
5 6 8
/ |
/ |
3 1
51) An implementation of a queue Q, using two stacks S1 and S2, is given below:
void insert(Q, x) {
push (S1, x);
}
void delete(Q){
if(stack-empty(S2)) then
if(stack-empty(S1)) then {
print(“Q is empty”);
return;
}
else while (!(stack-empty(S1))){
x=pop(S1);
push(S2,x);
}
x=pop(S2);
}
Let n insert and m (<=n) delete operations be performed in an arbitrary order on an empty queue
Q. Let x and y be the number of push and pop operations performed respectively in the process.
Which one of the following is true for all m and n?
(A) n+m <= x < 2n and 2m <= y <= n+m
(B) n+m <= x < 2n and 2m<= y <= 2n
(C) 2m <= x < 2n and 2m <= y <= n+m
(D) 2m <= x <2n and 2m <= y <= 2n
Answer(A)
The order in which insert and delete operations are performed matters here. The best case: Insert and
delete operations are performed alternatively. In every delete operation, 2 pop and 1 push operations are
performed. So, total m+ n push (n push for insert() and m push for delete()) operations and 2m pop
operations are performed.
The worst case: First n elements are inserted and then m elements are deleted. In first delete operation, n
+ 1 pop operations and n push operation are performed. Other than first, in all delete operations, 1 pop
operation is performed. So, total m + n pop operations and 2n push operations are performed (n push for
insert() and m push for delete())

52) Consider the following graph:
Which one of the following cannot be the sequence of edges added, in that order, to a minimum
spanning tree using Kruskal’s algorithm?
(A) (a—b),(d—f),(b—f),(d—c),(d—e)
(B) (a—b),(d—f),(d—c),(b—f),(d—e)
(C) (d—f),(a—b),(d—c),(b—f),(d—e)
(D) (d—f),(a—b),(b—f),(d—e),(d—c)
Answer (D)
The edge (d-e) cannot be considered before (d-c) in Kruskal’s minimum spanning tree algorithm because
Kruskal’s algorithm picks the edge with minimum weight from the current set of edges at each step.
53) The median of n elements can be found in O(n)time. Which one of the following is correct
about the complexity of quick sort, in which median is selected as pivot?
(A) Θ(n)
(B) Θ(nlogn)
(C) Θ(n^2)
(D) Θ(n^3)
Answer (B)
If median is always used as pivot, then recursion remains T(n) = 2T(n/2) + cn for all the cases where cn is
combined time for median finding and partition. So, worst case time complexity of this quick sort becomes
Θ(nlogn). In practical implementations, however, this variant is considerably slower on average
54) Consider a binary max-heap implemented using an array. Which one of the following array
represents a binary max-heap?
(A) 25,12,16,13,10,8,14
(B) 25,14,13,16,10,8,12
(C) 25,14,16,13,10,8,12
(D) 25,14,12,13,10,8,16
Answer (C)
A tree is max-heap if data at every node in the tree is greater than or equal to it’s children’ s data.
In array representation of heap tree, a node at index i has its left child at index 2i + 1 and right child at
index 2i + 2.
25
/

/
14 16
/ /
/ /
13 10 8 12
55) What is the content of the array after two delete operations on the correct answer to the
previous question?
(A) 14,13,12,10,8
(B) 14,12,13,8,10
(C) 14,13,8,12,10
(D) 14,13,12,8,10
Answer(D)
For Heap trees, deletion of a node includes following two operations.
1) Replace the root with last element on the last level.
2) Starting from root, heapify the complete tree from top to bottom.
56) 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(n^2)
(D) Θ(n2
log n)
Answer(B)
The recursion expression becomes:
T(n) = T(n/4) + T(3n/4) + cn
After solving the above recursion, we get Θ(nLogn).
57) 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
Answer (B) 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.
58) Dijkstra’s single source shortest path algorithm when run from vertex a in the below graph,
computes the correct shortest path distance to

(A) only vertex a
(B) only vertices a, e, f, g, h
(C) only vertices a, b, c, d
(D) all the vertices
Answer (D)
Dijkstra’s single source shortest path is not guaranteed to work for graphs with negative weight edges,
but it works for the given graph.
Let us see…
Let us run the 1st pass
b 1
b is minimum, so shortest distance to b is 1.
After 1st pass, distances are
c 3, e -2.
e is minimum, so shortest distance to e is -2
After 2nd pass, distances are
c 3, f 0.
f is minimum, so shortest distance to f is 0
After 3rd pass, distances are
c 3, g 3.
Both are same, let us take g. so shortest distance to g is 3.
After 4th pass, distances are
c 3, h 5
c is minimum, so shortest distance to c is 3
After 5th pass, distances are
h -2
h is minimum, so shortest distance to h is -2
59) The following C function takes a single-linked list of integers as a parameter and rearranges
the elements of the list. The function is called with the list containing the integers 1, 2, 3, 4, 5, 6, 7
in the given order. What will be the contents of the list after the function completes execution?
struct node
{
int value;
struct node *next;
};
void rearrange(struct node *list)
{
struct node *p, * q;
int temp;
if ((!list) || !list->next)
return;
p = list;
q = list->next;
while(q)
{
temp = p->value;

p->value = q->value;
q->value = temp;
p = q->next;
q = p?p->next:0;
}
}
(A) 1,2,3,4,5,6,7
(B) 2,1,4,3,6,5,7
(C) 1,3,2,5,4,7,6
(D) 2,3,4,5,6,7,1
Answer (B)
The function rearrange() exchanges data of every node with its next node. It starts exchanging data from
the first node itself.
60) The Breadth First Search algorithm has been implemented using the queue data structure.
One possible order of visiting the nodes of the following graph is
(A) MNOPQR
(B) NQMPOR
(C) QMNPRO
(D) QMNPOR
Answer (C)
61) Consider the following functions:
f(n) = 2^n
g(n) = n!
h(n) = n^logn
Which of the following statements about the asymptotic behaviour of f(n), g(n), and h(n) is true?
(A) f(n) = O(g(n)); g(n) = O(h(n))
(B) f(n) = Ω(g(n)); g(n) = O(h(n))
(C) g(n) = O(f(n)); h(n) = O(f(n))
(D) h(n) = O(f(n)); g(n) = Ω(f(n))
Answer (D)
According to order of growth: h(n) < f(n) < g(n) (g(n) is asymptotically greater than f(n) and f(n) is
asymptotically greater than h(n) ) We can easily see above order by taking logs of the given 3 functions
lognlogn < n < log(n!) (logs of the given f(n), g(n) and h(n)).
Note that log(n!) = Θ(nlogn)
62) The number of leaf nodes in a rooted tree of n nodes, with each node having 0 or 3 children
is:
a) n/2
b) (n-1)/3
c) (n-1)/2
d) (2n+1)/3
Answer(d)

Let L be the number of leaf nodes and I be the number of internal nodes, then following relation holds for
above given tree
L = (3-1)I + 1 = 2I + 1
Total number of nodes(n) is sum of leaf nodes and internal nodes
n = L + I
After solving above two, we get L = (2n+1)/3
63) A weight-balanced tree is a binary tree in which for each node. The number of nodes in the left
sub tree is at least half and at most twice the number of nodes in the right sub tree. The maximum
possible height (number of nodes on the path from the root to the farthest leaf) of such a tree on n
nodes is best described by which of the following?
a) log2 n
b) log4/3 n
c) log3 n
d) log3/2 n
Answer(d)
Let the maximum possible height of a tree with n nodes is represented by H(n).
The maximum possible value of H(n) can be approximately written using following recursion
H(n) = H(2n/3) + 1
The solution of above recurrence is log3/2 n. We can simply get it by drawing a recursion tree.
64) Let w be the minimum weight among all edge weights in an undirected connected graph. Let e
be a specific edge of weight w . Which of the following is FALSE?
(A) There is a minimum spanning tree containing e.
(B) If e is not in a minimum spanning tree T, then in the cycle formed by adding e to T, all edges
have the same weight.
(C) Every minimum spanning tree has an edge of weight w .
(D) e is present in every minimum spanning tree.
Answer (D)
(A), (B) and (C) are correct.
(D) is incorrect as there may be many edges of wight w in the graph and e may not be picked up in some
of the minimum spanning trees.
65) An array of n numbers is given, where n is an even number. The maximum as well as the
minimum of these n numbers needs to be determined. Which of the following is TRUE about the
number of comparisons needed?
(A) At least 2n – c comparisons, for some constant c, are needed.
(B) At most 1.5n – 2 comparisons are needed.
(C) At least nLog2n comparisons are needed.
(D) None of the above.
Answer (B)
66) In an unweighted, undirected connected graph, the shortest path from a node S to every other
node is computed most efficiently, in terms of time complexity by
(A) Dijkstra’s algorithm starting from S.
(B) Warshall’s algorithm
(C) Performing a DFS starting from S.
(D) Performing a BFS starting from S.

Answer(D)
* Time Comlexity of the Dijkstra’s algorithm is O(|V|^2 + E)
* Time Comlexity of the Warshall’s algorithm is O(|V|^3)
* DFS cannot be used for finding shortest paths
* BFS can be used for unweighted graphs. Time Complexity for BFS is O(|E| + |V|)
67) A complete n-ary tree is a tree in which each node has n children or no children. Let I be the
number of internal nodes and L be the number of leaves in a complete n-ary tree. If L = 41, and I =
10, what is the value of n?
(A) 3
(B) 4
(C) 5
(D) 6
Answer (C)
For an n-ary tree where each node has n children or no children, following relation holds
L = (n-1)*I + 1
Where L is the number of leaf nodes and I is the number of internal nodes.
Let us find out the value of n for the given data.
L = 41 , I = 10
41 = 10*(n-1) + 1
(n-1) = 4
n = 5
68) The height of a binary tree is the maximum number of edges in any root to leaf path. The
maximum number of nodes in a binary tree of height h is:
(A) 2^h -1
(B) 2^(h-1) – 1
(C) 2^(h+1) -1
(D) 2*(h+1)
Answer (C)
Maximum number of nodes will be there for a complete tree.
Number of nodes in a complete tree of height h = 1 + 2 + 2^2 + 2*3 + …. 2^h = 2^(h+1) – 1
69) The maximum number of binary trees that can be formed with three unlabeled nodes is:
(A) 1
(B) 5
(C) 4
(D) 3
Answer (B)
O
/
O O
(i)
O
/
O
/
O
(ii)
O
/
O

O
(iii)
O

O

O
(iv)
O

O
/
O
(v)
Note that nodes are unlabeled. If the nodes are labeled, we get more number of trees.
70) Which of the following sorting algorithms has the lowest worst-case complexity?
(A) Merge sort
(B) Bubble sort
(C) Quick sort
(D) Selection sort
Answer (A)
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
71) Suppose the numbers 7, 5, 1, 8, 3, 6, 0, 9, 4, 2 are inserted in that order into an initially empty
binary search tree. The binary search tree uses the usual ordering on natural numbers. What is
the in-order traversal sequence of the resultant tree?
(A) 7 5 1 0 3 2 4 6 8 9
(B) 0 2 4 3 1 6 5 9 8 7
(C) 0 1 2 3 4 5 6 7 8 9
(D) 9 8 6 4 2 3 0 1 5 7
Answer (c)
In-order traversal of a BST gives elements in increasing order. So answer c is correct without any doubt.
72) Let S be a stack of size n >= 1. Starting with the empty stack, suppose we push the first n
natural numbers in sequence, and then perform n pop operations. Assume that Push and Pop
operation take X seconds each, and Y seconds elapse between the end of one such stack
operation and the start of the next operation. For m >= 1, define the stack-life of m as the time
elapsed from the end of Push(m) to the start of the pop operation that removes m from S. The
average stack-life of an element of this stack is
(A) n(X+ Y)
(B) 3Y + 2X
(C) n(X + Y)-X
(D) Y + 2X
Answer(c)
We can easily arrive at the result by taking few examples.

73) Let G be an undirected connected graph with distinct edge weight. Let emax be the edge with
maximum weight and emin the edge with minimum weight. Which of the following statements is
false?
(a) Every minimum spanning tree of G must contain emin
(b) If emax is in a minimum spanning tree, then its removal must disconnect G
(c) No minimum spanning tree contains emax
(d) G has a unique minimum spanning tree
Answer (c)
(a) and (b) are always true.
(c) is false because (b) is true.
(d) is true because all edge weights are distinct for G.
74) Let G be an undirected graph. Consider a depth-first traversal of G, and let T be the resulting
depth-first search tree. Let u be a vertex in G and let v be the first new (unvisited) vertex visited
after visiting u in the traversal. Which of the following statements is always true?
(a) {u,v} must be an edge in G, and u is a descendant of v in T
(b) {u,v} must be an edge in G, and v is a descendant of u in T
(c) If {u,v} is not an edge in G then u is a leaf in T
(d) If {u,v} is not an edge in G then u and v must have the same parent in T
Answer (c)
75) Consider an undirected unweighted graph G. Let a breadth-first traversal of G be done starting
from a node r. Let d(r, u) and d(r, v) be the lengths of the shortest paths from r to u and v
respectively, in G. lf u is visited before v during the breadth-first traversal, which of the following
statements is correct?
a) d(r, u) < d (r, v)
b) d(r, u) > d(r, v)
c) d(r, u) <= d (r, v)
d) None of the above
Answer (c)
d(r, u) and d(r, v) will be equal when u and v are at same level, otherwise d(r, u) will be less than d(r, v)
76) How many undirected graphs (not necessarily connected) can be constructed out of a given
set V= {V 1, V 2,…V n} of n vertices ?
a) n(n-l)/2
b) 2^n
c) n!
d) 2^(n(n-1)/2)
Answer (d)
In an undirected graph, there can be maximum n(n-1)/2 edges. We can choose to have (or not have) any
of the n(n-1)/2 edges. So, total number of undirected graphs with n vertices is 2^(n(n-1)/2).

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