Algorithms
Question 1 |
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 2 |
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 3 |
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 4 |
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?
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:
Call-by-reference = 8
Call-by-value = 1
Question 5 |
Which one of the following statements is false?
A | Optimal binary search tree construction can be performed efficiently using dynamic programming. |
B | Breadth-first 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 | Depth-first search can be used to find connected components of a graph. |
Question 6 |
Consider the following two functions:
Which of the following is true?
A | g1(n) is O(g2(n)) |
B | g1 (n) is O(3) |
C | g2 (n) is O(g1 (n)) |
D | g2 (n) is O(n) |
E | Both A and B |
Growth rate of g1 is less than that of g2 i.e., g1(n) = O(g2(n)) = O(n).
Question 7 |
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.
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 8 |
(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 9 |
Consider the following recursive function:
function fib (1:integer);integer; begin if (n=0) or (n=1) then fib:=1 else fib:=fib(n-1) + fib(n-2) 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 10 |
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 11 |
Merge sort uses
A | Divide and conquer strategy |
B | Backtracking approach |
C | Heuristic search |
D | Greedy approach |
Question 12 |
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 13 |
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)
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 14 |
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 15 |
How many minimum spanning trees does the following graph have? Draw them. (Weights are assigned to the edge).
A | Theory Explanation. |
Question 16 |
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
A | θ(loga logb n) |
B | θ(logb loga n)
|
C | θ(log2 log2 n)
|
D | θ(logab n)
|
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 17 |
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| log|V|) |
C | θ(|E||V|) |
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).
Method-2:
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 18 |
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 ______.
A | 99 |
• 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 19 |
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 20 |
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 21 |
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 22 |
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 23 |
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 24 |
(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.
A | Theory Explanation. |
Question 25 |
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.
A | Theory Explanation. |
Question 26 |
(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) begin if n = 1 then call A else begin what (n-1); 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 27 |
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 28 |
Consider a simple connected graph G with n vertices and n-edges (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 29 |
where O(n) stands for order n is:
A | O(n) |
B | O(n2) |
C | O(n3) |
D | O(3n2) |
E | O(1.5n2) |
F | B, C, D and E |
⇒ 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 30 |
Consider the recursive algorithm given below:
procedure bubblersort (n); var i,j: index; temp : item; begin for i:=1 to n-1 do if A[i] > A [i+1] then begin temp : A[i]; A[i]:=A[i+1]; A[i+1]:=temp end; bubblesort (n-1) end
Let an be the number of times the ‘if…then….’ Statement gets executed when the algorithm is run with value n. Set up the recurrence relation by defining an in terms of an-1. Solve for an.
A | Theory Explanation. |
Question 31 |
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 __________
A | O(m log n) |
Question 32 |
Following algorithm(s) can be used to sort n integers in the range [1...n3] in O(n) time
A | Heapsort |
B | Quicksort |
C | Mergesort |
D | Radixsort |
As Radix sort is not comparison based sort (it is counting sort), so can be done in linear time.
Question 33 |
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.
A | n |
→ The array is already sorted in ascending order.
→ The array is already sorted in descending order.
Question 34 |
Consider the function F(n) for which the pseudo code is given below:
Function F(n) begin F1 ← 1 if(n=1) then F ← 3 else For i = 1 to n do begin C ← 0 For j = 1 to F(n – 1) do begin C ← C + 1 end F1 = F1 * C end F = F1 end [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).
A | Theory Explanation. |
Question 35 |
The minimum number of comparisons required to sort 5 elements is _____
A | 7 |
= ⌈log(n!)⌉
= ⌈log(5!)⌉
= ⌈log(120)⌉
= 7
Question 36 |
The weighted external path length of the binary tree in figure is _____
A | 144 |
Question 37 |
If the binary tree in figure is traversed in inorder, then the order in which the nodes will be visited is ______
A | 4, 1, 6, 7, 3, 2, 5, 8 |
(Left, Root, Right)
So, the order will be
4, 1, 6, 7, 3, 2, 5, 8
Question 38 |
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 | A-S, B-R, C-P, D-Q |
Question 39 |
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:
A | O(n2) |
B | O(mn) |
C | O(m+n) |
D | O(m log n) |
E | O(m2 |
F | B, D and E |
→ Where m is no. of edges and n is number of vertices then n = O(m2)
→ O(m logn) < O(mn)
Question 40 |
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), POPThe sequence of values popped out is:
A | 20, 10, 20, 10, 20 |
B | 20, 20, 10, 10, 20 |
C | 10, 20, 20, 10, 20 |
D | 20, 20, 10, 20, 10 |
→ 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 41 |
Which one of the following options arranges the functions in the increasing order of asymptotic growth rate?
A | f2, f3, f1 |
B | f3, f2, f1 |
C | f2, f1, f3 |
D | f1, f2, f3 |
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.
Method-1:
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
Method-2:
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 42 |
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?
A | R7is achieved by three different solutions.
|
B | R7=18 |
C | R7=19 |
D | R7cannot be achieved by a solution consisting of three pieces. |
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 43 |
The number of minimum-weight spanning trees of the graph is _______
A | 3 |
To find the number of spanning trees using 2 methods:
- If graph is complete, use n^n-2 formula
- 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 44 |
Which one of the following options is correct?
A | |
B | |
C | |
D |
Question 45 |
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?
A | Quicksort using the last element as pivot |
B | Insertion sort |
C | Selection sort |
D | Mergesort |
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 46 |
A | |
B | |
C | |
D |
Let assume n=512
Method-1:
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
T(n)=2T(n/2)+2
=2(2T(n/4)+2)+2
=4T(n/4)+4+2
፧
=2k-1T(2)+1ik-12i
=2k-1+2k-2
=3n/2-2
→ The given example n=512
Apply into 3n/2 -2
= (3*512)/2 -2
= 768-2
= 766
Method-2:
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
Method-3:
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.
- After the first round of Tournament , there will be exactly n/2 numbers =256 that will lose the round.
- 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 47 |
A | |
B | |
C | |
D |
In this question they given three main constraints
- The input array is in sorted order
- Use recursive binary search
- 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 48 |
Which one of the following options is correct about the recurrence T(n)?
A | |
B | |
C | |
D |
Question 49 |
S1: There exists a minimum weighted edge in G which is present in every minimum spanning tree of G.
S2:If every edge in G has distinct weight, then G has a unique minimum spanning tree.
Which of the following options is correct?
A | S1is false and S2is true.
|
B | S1is true and S2is false. |
C | Both S1 and S2are false. |
D | Both S1 and S2are true. |
Statement-1: FALSE: The given statement is not valid for all the cases because they are not mentioned, edge weights are in distinct or duplicate. So, we can take any random edge weights for the given statement.
Example:
Statement-2: TRUE: Using the kruskal’s (or) prim's algorithm we get a unique MST when there is a unique edge weight.
Example:
Based on the above graph, we get the MST is
Question 50 |
The sum of the quality-scores of all the vertices in the graph shown above is _________.
A | 929 |