## Technical Aptitude Questions- Data Structure

1. What is data structure?

A data structure is a way of organizing data that considers not only the items stored, but also their relationship to each other. Advance knowledge about the relationship between data items allows designing of efficient algorithms for the manipulation of data.

1. List out the areas in which data structures are applied extensively?

• Compiler Design,

• Operating System,

• Database Management System,

• Statistical analysis package,

• Numerical Analysis,

• Graphics,

• Artificial Intelligence,

• Simulation

1. What are the major data structures used in the following areas : RDBMS, Network data model & Hierarchical data model.

• RDBMS – Array (i.e. Array of structures)

• Network data model – Graph

• Hierarchical data model – Trees

1. If you are using C language to implement the heterogeneous linked list, what pointer type will you use?

The heterogeneous linked list contains different data types in its nodes and we need a link, pointer to connect them. It is not possible to use ordinary pointers for this. So we go for void pointer. Void pointer is capable of storing pointer to any type as it is a generic pointer type.

1. Minimum number of queues needed to implement the priority queue?

Two. One queue is used for actual storing of data and another for storing priorities.

1. What is the data structures used to perform recursion?

Stack. Because of its LIFO (Last In First Out) property it remembers its ‘caller’ so knows whom to return when the function has to return. Recursion makes use of system stack for storing the return addresses of the function calls.

Every recursive function has its equivalent iterative (non-recursive) function. Even when such equivalent iterative procedures are written, explicit stack is to be used.

1. What are the notations used in Evaluation of Arithmetic Expressions using prefix and postfix forms?

Polish and Reverse Polish notations.

1. Convert the expression ((A + B) * C – (D – E) ^ (F + G)) to equivalent Prefix and Postfix notations.

Prefix Notation:

^ – * +ABC – DE + FG

Postfix Notation:

AB + C * DE – – FG + ^

1. Sorting is not possible by using which of the following methods?

(a) Insertion

(b) Selection

(c) Exchange

(d) Deletion

(d) Deletion.

Using insertion we can perform insertion sort, using selection we can perform selection sort, using exchange we can perform the bubble sort (and other similar sorting methods). But no sorting method can be done just using deletion.

1. A binary tree with 20 nodes has null branches? 21

Let us take a tree with 5 nodes (n=5)

It will have only 6 (ie,5+1) null branches. In general,

A binary tree with n nodes has exactly n+1 null nodes.

1. What are the methods available in storing sequential files ?

• Straight merging,

• Natural merging,

• Polyphase sort,

• Distribution of Initial runs.

1. How many different trees are possible with 10 nodes ?

1014

For example, consider a tree with 3 nodes(n=3), it will have the maximum combination of 5 different (ie, 23 – 3 = 5) trees. In general:

If there are n nodes, there exist 2n-n different trees.

1. List out few of the Application of tree data-structure?

• The manipulation of Arithmetic expression,

• Symbol Table construction,

• Syntax analysis.

1. List out few of the applications that make use of Multilinked Structures?

• Sparse matrix,

• Index generation.

1. In tree construction which is the suitable efficient data structure?

(a) Array (b) Linked list (c) Stack (d) Queue (e) none

1. What is the type of the algorithm used in solving the 8 Queens problem?

Backtracking

1. In an AVL tree, at what condition the balancing is to be done?

If the ‘pivotal value’ (or the ‘Height factor’) is greater than 1 or less than –1.

1. What is the bucket size, when the overlapping and collision occur at same time?

One. If there is only one entry possible in the bucket, when the collision occurs, there is no way to accommodate the colliding value. This results in the overlapping of values.

1. Traverse the given tree using Inorder, Preorder and Postorder traversals. • Inorder : D H B E A F C I G J

• Preorder: A B D H E C F G I J

• Postorder: H D E B F I J G C A

1. There are 8, 15, 13, 14 nodes were there in 4 different trees. Which of them could have formed a full binary tree?

15.

In general:

There are 2n-1 nodes in a full binary tree.

By the method of elimination:

Full binary trees contain odd number of nodes. So there cannot be full binary trees with 8 or 14 nodes, so rejected. With 13 nodes you can form a complete binary tree but not a full binary tree. So the correct answer is 15.

Note:

Full and Complete binary trees are different. All full binary trees are complete binary trees but not vice versa.

1. In the given binary tree, using array you can store the node 4 at which location? At location 6

 1 2 3 – – 4 – – 5
 Root LC1 RC1 LC2 RC2 LC3 RC3 LC4 RC4

where LCn means Left Child of node n and RCn means Right Child of node n

1. Sort the given values using Quick Sort?

 65 70 75 80 85 60 55 50 45

Sorting takes place from the pivot value, which is the first value of the given elements, this is marked bold. The values at the left pointer and right pointer are indicated using L and R respectively.

 65 70L 75 80 85 60 55 50 45R

Since pivot is not yet changed the same process is continued after interchanging the values at L and R positions

 65 45 75 L 80 85 60 55 50 R 70
 65 45 50 80 L 85 60 55 R 75 70
 65 45 50 55 85 L 60 R 80 75 70
 65 45 50 55 60 R 85 L 80 75 70

When the L and R pointers cross each other the pivot value is interchanged with the value at right pointer. If the pivot is changed it means that the pivot has occupied its original position in the sorted order (shown in bold italics) and hence two different arrays are formed, one from start of the original array to the pivot position-1 and the other from pivot position+1 to end.

 60 L 45 50 55 R 65 85 L 80 75 70 R
 55 L 45 50 R 60 65 70 R 80 L 75 85
 50 L 45 R 55 60 65 70 80 L 75 R 85

In the next pass we get the sorted form of the array.

 45 50 55 60 65 70 75 80 85
1. For the given graph, draw the DFS and BFS? • BFS: A X G H P E M Y J

• DFS: A X H P E Y M J G

1. Classify the Hashing Functions based on the various methods by which the key value is found.

• Direct method,

• Subtraction method,

• Modulo-Division method,

• Digit-Extraction method,

• Mid-Square method,

• Folding method,

• Pseudo-random method.

1. What are the types of Collision Resolution Techniques and the methods used in each of the type?

• Open addressing (closed hashing),

The methods used include:

Overflow block,

• Closed addressing (open hashing)

The methods used include:

Binary tree…

1. In RDBMS, what is the efficient data structure used in the internal storage representation?

B+ tree. Because in B+ tree, all the data is stored only in leaf nodes, that makes searching easier. This corresponds to the records that shall be stored in leaf nodes.

1. Draw the B-tree of order 3 created by inserting the following data arriving in sequence – 92 24 6 7 11 8 22 4 5 16 19 20 78 1. Of the following tree structure, which is, efficient considering space and time complexities?
1. Incomplete Binary Tree

2. Complete Binary Tree

3. Full Binary Tree

(b) Complete Binary Tree.

By the method of elimination:

Full binary tree loses its nature when operations of insertions and deletions are done. For incomplete binary trees, extra storage is required and overhead of NULL node checking takes place. So complete binary tree is the better one since the property of complete binary tree is maintained even after operations like additions and deletions are done on it.

1. What is a spanning Tree?

A spanning tree is a tree associated with a network. All the nodes of the graph appear on the tree once. A minimum spanning tree is a spanning tree organized so that the total edge weight between nodes is minimized.

1. Does the minimum spanning tree of a graph give the shortest distance between any 2 specified nodes?

No.

Minimal spanning tree assures that the total weight of the tree is kept at its minimum. But it doesn’t mean that the distance between any two nodes involved in the minimum-spanning tree is minimum.

1. Convert the given graph with weighted edges to minimal spanning tree. the equivalent minimal spanning tree is: 1. Which is the simplest file structure?

1. Sequential

2. Indexed

3. Random

(a) Sequential

1. Whether Linked List is linear or Non-linear data structure?

According to Access strategies Linked list is a linear one.

According to Storage Linked List is a Non-linear one.

1. Draw a binary Tree for the expression :

A * B – (C + D) * (P / Q) 1. For the following COBOL code, draw the Binary tree?

01 STUDENT_REC.

02 NAME.

03 FIRST_NAME PIC X(10).

03 LAST_NAME PIC X(10).

02 YEAR_OF_STUDY.

03 FIRST_SEM PIC XX.

03 SECOND_SEM PIC XX. ## Technical Aptitude Question – C Language – Part-1

Note : All the programs are tested under Turbo C/C++ compilers. It is assumed that,

 Programs run under DOS environment,
 The underlying machine is an x86 system,
 Program is compiled using Turbo C/C++ compiler.

The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed).

Predict the output or error(s) for the following:

1. void main()

{

int const * p=5;

printf(“%d”,++(*p));

}

Compiler error: Cannot modify a constant value.

Explanation:

p is a pointer to a “constant integer”. But we tried to change the value of the “constant integer”.

2. main()

{

char s[ ]=”man”;
int i;
for(i=0;s[ i ];i++)
printf(“\n%c%c%c%c”,s[ i ],*(s+i),*(i+s),i[s]);

}

mmmm

aaaa

nnnn

Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea.

Generally array name is the base address for that array. Here s is the base address. i is the

index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].

3. main()

{

float me = 1.1; double you = 1.1;

if(me==you)

printf(“I love U”);

else

printf(“I hate U”);

}

I hate U

Explanation:

For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.

Rule of Thumb:

Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .

4. main()

{

static int var = 5;

printf(“%d “,var – -);

if(var)

main();

}

5432 1

Explanation:

When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.

5. main()

{

int c[ ]={2.8,3.4,4,6.7,5};

int j,*p=c,*q=c;

for(j=0;j<5;j++)

{

printf(” %d “,*c);

++q;

}

for(j=0;j<5;j++)

{

printf(” %d “,*p); ++p;

}

}

2 222223465

Explanation:

Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

6. main()

{

extern int i;

i=20;

printf(“%d”,i);

}

Linker Error : Undefined symbol ‘_i’

Explanation:

extern storage class in the following declaration,

extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

7. main()

{

int i=-1,j=-1,k=0,l=2,m;

m=i++&&j++&&k++||l++;

printf(“%d %d %d %d %d”,i,j,k,l,m);

}

0013 1

Explanation :

Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

8. main()

{

char *p;
printf(“%d %d “,sizeof(*p),sizeof(p));

}

1   2

Explanation:

The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

9. main()

{

int i=3;

switch(i)

{
default:printf(“zero”);

case 1: printf(“one”);

break;
case 2:printf(“two”);

break;

case 3: printf(“three”);

break;

}

}

three

Explanation :

The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn’t match.

10. main()

{

printf(“%x”,-1<<4);

}

fff0

Explanation :

-1 is internally represented as all 1’s. When left shifted four times the least significant 4 bits are filled with 0’s.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

11. main()

{

char string[]=”Hello World”;

display(string);

}

void display(char *string)

{

printf(“%s”,string);

}

Compiler Error : Type mismatch in redeclaration of function display

Explanation :

In third line, when the function display is encountered, the compiler doesn’t know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

12. main()

{

int c=- -2;

printf(“c=%d”,c);

}

c=2;

Explanation:

Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.

Note:

However you cannot give like –2. Because — operator can only be applied to variables as a decrement operator (eg., i–). 2 is a constant and not a variable.

13.

#define int char

main()

{
int i=65;

printf(“sizeof(i)=%d”,sizeof(i));

}

sizeof(i)=1

Explanation:

Since the #define replaces the string int by the macro char

14. main()

{

int i=10;

i=!i>14;

printf(“i=%d”,i);

}

i=0

Explanation:

In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

15.

#include<stdio.h>

main()

{
char s[]={‘a’,’b’,’c’,’\n’,’c’,”}; char *p,*str,*str1;
p=&s;
str=p;
str1=s;
printf(“%d”,++*p + ++*str1-32);

}

77

Explanation:

p is pointing to character ‘\n’. str1 is pointing to character ‘a’ ++*p. “p is pointing to ‘\n’ and that is incremented by one.” the ASCII value of ‘\n’ is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to ‘a’ that is incremented by 1 and it becomes ‘b’. ASCII value of ‘b’ is 98.

Now performing (11 + 98 – 32), we get 77(“M”); So we get the output 77 :: “M” (Ascii is 77).

16.

#include<stdio.h>

main()

{
int a = { {10,2,3,4}, {5,6,7,8} }; int *p,*q;
p=&a;
*q=***a;
printf(“%d—-%d”,*p,*q);

}

SomeGarbageValue—1

Explanation:

p=&a you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.

17.

#include<stdio.h>

main()

{

struct xx

{
int x=3;

char name[]=”hello”;

};

struct xx *s;

printf(“%d”,s->x);

printf(“%s”,s->name);

}

Compiler Error

Explanation:

You should not initialize variables in declaration

18.

#include<stdio.h>

main()

{
struct xx

{
int x;

struct yy

{

char s;

struct xx *p;

};

struct yy *q;

};

}

Compiler Error

Explanation:

The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

19. main()

{

printf(“\nab”);

printf(“\bsi”);

printf(“\rha”);

}

hai

Explanation:

\n – newline

\b – backspace

\r – linefeed

20. main()

{

int i=5;

printf(“%d%d%d%d%d%d”,i++,i–,++i,–i,i);

}

45545

Explanation:

The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the result.

21.

#define square(x) x*x

main()

{
int i;

i = 64/square(4);

printf(“%d”,i);

}

64

Explanation:

the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

22. main()

{

char *p=”hai friends”,*p1; p1=p;
while(*p!=”) ++*p++;

printf(“%s %s”,p,p1);

}

ibj!gsjfoet

Explanation:

++*p++ will be parse in the given order

•   *p that is value at the location currently pointed by p will be taken
•   ++*p the retrieved value will be incremented
•   when ; is encountered the location will be incremented that is p++ will be executed

Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj! gsjfoet” and since p reaches ‘’ and p1 points to p thus p1doesnot print anything.

23.

#include <stdio.h>

#define a 10

main()

{

#define a 50

printf(“%d”,a);

}

50

Explanation:

The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

24.

#define clrscr() 100

main()

{
clrscr();

printf(“%d\n”,clrscr());

}

100

Explanation:

Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :

main()

{

100;

printf(“%d\n”,100);

}

Note100; is an executable statement but with no action. So it doesn’t give any problem

25.

main()

{

printf(“%p”,main);

}