Following questions have been asked in GATE CS 2014 exam.

1) The number of distinct minimum spanning trees for the weighted graph below is ____

Answer: 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) Let A be a square matrix of size n x n. Consider the following program. What is the expected output?

(A) The matrix A itself
(B) Transpose of matrix A
(C) Adding 100 to the upper diagonal elements and subtracting 100 from diagonal elements of A
(D) None of the above

Answer: A
If we take look at the inner statements of first loops, we can notice that the statements swap A[i][j] and A[j][i] for all i and j.
Since the loop runs for all elements, every element A[l][m] would be swapped twice, once for i = l and j = m and then for i = m and j = l. Swapping twice means the matrix doesn’t change.

4) 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

5) 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).

6) 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.

(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.

7) 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.

8) The minimum number of comparisons required to find the minimum and the maximum of 100 numbers is _________________.

Answer: 147
The minimum number of comparisons required is 3n/2 – 3 for n numbers. Please refer method 3 of Maximum and minimum of an array using minimum number of comparisons for more details.

9) 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 = ___.

Answer: 34
The longest length is 4. There are 3 LCS of length 4 “qprr”, “pqrr” and “qpqr”.

10) 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: 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.

11) 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.

12) 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 (C)
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

13) 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

4) 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

15) 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.

17) 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

18) 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)

19) Consider a rooted Binary tree represented using pointers. The best upper bound on the time required to determine the number of subtrees having having exactly 4 nodes O(n^a Logn ^b). Then the value of a + 10b is ________

Answer: 1
Explanation: We can find the subtree with 4 nodes in O(n) time. Following can be a simple approach.
1) Traverse the tree in bottom up manner and find size of subtree rooted with current node
2) If size becomes 4, then print the current node.

20) 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.