hi all..
I am getting a lot of enquiries about the Kanbay
pattern.... I will tell you whatever I know...
I wrote this paper long back - March 12th to be exact.
It was a walkin. There were no Technical questions
till the last interview... so guys.. put your heads
down and work on your speed and accuracy... Most of
the ques were of the type you see in Agarwal(quants).
Analytical is of medium toughnes.. That too tests your
speed and accuracy (not like the stupid stuff infy
throws at you ). There is negative marking(1 mark for
a correct answer and -1 for wrong). So concentrate on
accuracy... They shortlisted around 160 from the more
than 3000 who wrote the test. Then we had gds in
batches of 10. From every batch an average of 2 ppl
were shortlisted. Then we had an interview chaired by
both hr and tech ppl (30 - 45 mins). Well I got
chucked out at that point (At that time I did not know
even C properly). Thats all I know....
ALL THE BEST Guyzzz!!!!
KANBAY
The GD on aug 2nd was:
you are each given a role to play..u have to justify why yours is
best suited to the question and final reach a general consensus..tips:
be cool..
listen carefully..
n talk..tlak what u want but do say something..
don't be too aggressive nor too submissive n yes..don't address any
one member only.
out of 800 students, 63 of us got thru written and 17 of us from
these 63 got thru GD..yeah!i cracked the GD.
then comes the interview.
prepare tech guys..they ask a lot of branch ques..n if you have
mentioned any languages in your resume be well prepared.
be preapred to answer ques from A to Z of whatever u have written in
your CV.
they also gave us puzzles like:
3 switches in the bottom floor, 3 bulbs on top floor..u can go
upstairs only once.find out which switch is for which bulb.
then , out of a group of 10 matchboxes..each with 10 matchsticks of
10 gms. if one weighs 9 gms..how will u find it in minimum no. of
weighings..
be confidant and preapre ques like
why sbhould we take u?
what are ur assets,hobbies,the last book that read , the last film
that u saw ..name of characters in that film etc..
good luck,nikky.
KANBAY
I had attende Kanbay test long back.
They had three sets of papers..to ensure people dont copy.
But generally the problems are te same.
There are two sections ...Aptitude and Maths.
Aptitude is easy but time is a major constraint.In maths all I remember is there were
problems on
1)Compound Interest (3 or 4)
2)Two taps are filling a tank and the third one is emptying it .. sort.
3)Problems on trains
Maybe RS AGarwal will help a lot.In aptitude there were problems on RELATIONS
and on triangle area..Like
One equilateral triangle is divide into say into 3 equilateral traingle.What is the area
proportion of the outermost to the innermost?
It was easy but one FALLS SHORT ON time.Practise a lot and I am sure you'll get through.
C Puzzles : Questions and Answers - Level 1
----------------------------------------------------------------------------
[Image] Questions
L1.Q1 : Write the output of the following program
#include
#define ABC 20
#define XYZ 10
#define XXX ABC - XYZ
void main()
{
int a;
a = XXX * 10;
printf("%d\n", a);
}
Solution for L1.Q1
----------------------------------------------------------------------------
L1.Q2 : Write the output of this program
#include
#define calc(a, b) (a * b) / (a - b)
void main()
{
int a = 20, b = 10;
printf("%d\n", calc(a + 4, b -2));
}
Solution for L1.Q2
----------------------------------------------------------------------------
L1.Q3 : What will be output of the following program ?
#include
void main()
{
int cnt = 5, a;
do {
a /= cnt;
} while (cnt --);
printf ("%d\n", a);
}
Solution for L1.Q3
----------------------------------------------------------------------------
L1.Q4 : Print the output of this program
#include
void main()
{
int a, b, c, abc = 0;
a = b = c = 40;
if (c) {
int abc;
abc = a*b+c;
}
printf ("c = %d, abc = %d\n", c, abc);
}
Solution for L1.Q4
----------------------------------------------------------------------------
L1.Q5 : Print the output of this program
#include
main()
{
int k = 5;
if (++k < 5 && k++/5 || ++k <= 8);
printf("%d\n", k);
}
Solution for L1.Q5
----------------------------------------------------------------------------
L1.Q6 : What is the output of this program ?
#include
void fn(int, int);
main()
{
int a = 5;
printf("Main : %d %d\n", a++, ++a);
fn(a, a++);
}
void fn(int a, int b)
{
printf("Fn : a = %d \t b = %d\n", a, b);
}
Solution for L1.Q6
----------------------------------------------------------------------------
[Image] Answers
L1.A1
Solution for L1.Q1
a = xxx * 10
which is => a = ABC - XYZ * 10
=> a = 20 - 10 * 10
=> a = 20 - 100
=> a = -80
----------------------------------------------------------------------------
L1.A2
Solution for L1.Q2
Actual substitution is like this :
calc(20+4, 10 -2) is calculated as follows
(20+4 * 10-2) / (20+4 - 10-2)
(20+40-2) / 12
58 / 12 = 4.8
since it is printed in %d the ans is 4
----------------------------------------------------------------------------
L1.A3
Solution for L1.Q3
This problem will compile properly, but it will give run time
error. It will give divide-by-zero error. Look in to the do
loop portion
do {
a /= cnt;
} while (cnt --);
when the 'cnt' value is 1, it is decremented in
'while (cnt --)' and on next reference of 'cnt' it
becomes zero.
a /= cnt; /* ie. a /= 0 */
which leads to divide-by-zero error.
----------------------------------------------------------------------------
L1.A4
Solution for L1.Q4
the result will be
c = 40 and abc = 0;
because the scope of the variable 'abc' inside if(c) {.. } is
not valid out side that if (.) { .. }.
----------------------------------------------------------------------------
L1.A5
Solution for L1.Q5
The answer is 7. The first 2 conditions (++k < 5 and k++/5) are
checked and it is found that it is true. So, there is no need to
check the 3rd condition (++k <= 8). At this point k value is
incremented by twice, hence the value of k becomes 7.
----------------------------------------------------------------------------
L1.A6
Solution for L1.Q6
The solution depends on the implementation of stack. (Depends on OS)
In some machines the arguments are passed from left to right to the
stack. In this case the result will be
Main : 5 7
Fn : 7 7
Other machines the arguments may be passed from right to left to the
stack. In that case the result will be
Main : 6 6
Fn : 8 7
----------------------------------------------------------------------------
Goto C Puzzle
* level 2
* level 3
[Image] For any further clarifications/ informations and to contribute some
more puzzles for this page please feel free to contact Achutha Raman
----------------------------------------------------------------------------
[ About me ] [ Publications ] [ C puzzles ] [ Home ] [ Form ]
[
Music library ] [ Picture library ]
C Puzzles : Questions and Answers - Level 2
----------------------------------------------------------------------------
[Image] Questions
L2.Q1 : Write the output of this program
#include
main()
{
int *a, *s, i;
s = a = (int *) malloc( 4 * sizeof(int));
for (i=0; i<4; i++) *(a+i) = i * 10;
printf("%d\n", *s++);
printf("%d\n", (*s)++);
printf("%d\n", *s);
printf("%d\n", *++s);
printf("%d\n", ++*s);
}
Solution for L2.Q1
----------------------------------------------------------------------------
L2.Q2 : Checkout this program result
#include
void fn(int);
static int val = 5;
main()
{
while (val --) fn(val);
printf("%d\n", val);
}
void fn(int val)
{
static int val = 0;
for (; val < 5; val ++) printf("%d\n", val);
}
Solution for L2.Q2
----------------------------------------------------------------------------
L2.Q3 : Can you predict the output of this program ?
#include
main()
{
typedef union {
int a;
char b[10];
float c;
} Union;
Union x, y = { 100 };
x.a = 50;
strcpy (x.b, "hello");
x.c = 21.50;
printf ("Union 2 : %d %s %f\n", x.a, x.b, x.c);
printf ("Union Y : %d %s %f\n", y.a, y.b, y.c);
}
Solution for L2.Q3
----------------------------------------------------------------------------
L2.Q4 : Print the output of the program
#include
main()
{
struct Data {
int a;
int b;
} y[4] = { 1, 10, 3, 30, 2, 20, 4, 40};
struct Data *x = y;
int i;
for(i=0; i<4; i++) {
x->a = x->b, ++x++->b;
printf("%d %d\t", y[i].a, y[i].b);
}
}
Solution for L2.Q4
----------------------------------------------------------------------------
L2.Q5 : Write the output of this program
#include
main()
{
typedef struct {
int a;
int b;
int c;
char ch;
int d;
}xyz;
typedef union {
xyz X;
char y[100];
}abc;
printf("sizeof xyz = %d sizeof abc = %d\n",
sizeof(xyz), sizeof(abc));
}
Solution for L2.Q5
----------------------------------------------------------------------------
L2.Q6 : Find out the error in this code
#include
#include
#define Error(str) printf("Error : %s\n", str); exit(1);
main()
{
int fd;
char str[20] = "Hello! Test me";
if ((fd = open("xx", O_CREAT | O_RDWR)) < 0))
Error("open failed");
if (write(fd, str, strlen(str)) < 0)
Error("Write failed");
if (read(fd, str, strlen(str)) < 0)
Error("read failed");
printf("File read : %s\n", str);
close(fd);
}
Solution for L2.Q6
----------------------------------------------------------------------------
L2.Q7 : What will be the output of this program ?
#include
main()
{
int *a, i;
a = (int *) malloc(10*sizeof(int));
for (i=0; i<10; i++)
*(a + i) = i * i;
for (i=0; i<10; i++)
printf("%d\t", *a++);
free(a);
}
Solution for L2.Q7
----------------------------------------------------------------------------
L2.Q8 :
Write a program to calculate number of 1's (bit) in a given integer number
i.e) Number of 1's in the given integer's equivalent binary representation.
Solution for L2.Q8
----------------------------------------------------------------------------
[Image] Answers
L2.A1
Solution for L2.Q1
The output will be : 0 10 11 20 21
*s++ => *(s++)
*++s => *(++s)
++*s => ++(*s)
----------------------------------------------------------------------------
L2.A2
Solution for L2.Q2
Some compiler (ansi) may give warning message, but it will
compile without errors.
The output will be : 1 2 3 4 5 and -1
----------------------------------------------------------------------------
L2.A3
Solution for L2.Q3
This is the problem about Unions. Unions are similar to structures
but it differs in some ways. Unions can be assigned only with one
field at any time. In this case, unions x and y can be assigned
with any of the one field a or b or c at one time. During
initialisation of unions it takes the value (whatever assigned )
only for the first field. So, The statement y = {100} intialises
the union y with field a = 100.
In this example, all fields of union x are assigned with some values.
But at any time only one of the union field can be assigned. So,
for the union x the field c is assigned as 21.50.
Thus, The output will be
Union 2 : 22 22 21.50
Union Y : 100 22 22
( 22 refers unpredictable results )
----------------------------------------------------------------------------
L2.A4
Solution for L2.Q4
The pointer x points to the same location where y is stored.
So, The changes in y reflects in x.
The output will be :
10 11 30 31 20 21 40 41
----------------------------------------------------------------------------
L2.A5
Solution for L2.Q5
The output of this program is purely depends on the processor
architecuture. If the sizeof integer is 4 bytes and the size of
character is 1 byte (In some computers), the output will be
sizeof xyz = 20 sizeof abc = 100
The output can be generalized to some extent as follows,
sizeof xyz = 4 * sizeof(int) + 1 * sizeof(char) + padding bytes
sizeof abc = 100 * sizeof(char) + padding bytes
To keep the structures/unions byte aligned, some padding bytes are
added in between the sturcture fields. In this example 3 bytes are
padded between ' char ch' and 'int d' fields. The unused bytes are
called holes. To understand more about padding bytes (holes) try
varing the field types of the structures and see the output.
----------------------------------------------------------------------------
L2.A6
Solution for L2.Q6
Just try to execute this file as such. You can find out that it will
exit immediately. Do you know why?
With this hint, we can trace out the error. If you look into
the macro 'Error', you can easily identify that there are two
separete statements without brases '{ ..}'. That is the problem.
So, it exits after the calling open(). The macro should be
put inside the brases like this.
#define Error(str) { printf("Error : %s\n", str); exit(1); }
----------------------------------------------------------------------------
L2.A7
Solution for L2.Q7
This program will fault (Memory fault/segmentation fault). Can you
predict Why?
Remove the statment 'free(a);' from the program, then execute the
program. It will run. It gives the results correctly.
What causes 'free(a)' to generate fault?
Just trace the address location of pointer variable 'a'. The variable
'a' is incremented inside the 'for loop'. Out side the 'for loop' the
variable 'a' will point to 'null'. When the free() call is made, it will
free the data area from the base_address (which is passed as the
argument of the free call) upto the length of the data allocated previously.
In this case, free() tries to free the length of 10 *sizeof(int) from the
base pointer location passed as the argument to the free call, which is
'null' in this case. Thus, it generates memory fault.
----------------------------------------------------------------------------
L2.A8
Solution for L2.Q8
#include
main(argc, argv)
int argc;
char *argv[];
{
int count = 0, i;
int v = atoi(argv[1]);
for(i=0; i<8*sizeof(int); i++)
if(v &(1<
}
----------------------------------------------------------------------------
Goto C Puzzle
* level 1
* level 3
[Image] For any further clarifications/ informations and to contribute some
more puzzles for this page please feel free to contact Achutha Raman
----------------------------------------------------------------------------
[ About me ] [ Publications ] [ C puzzles ] [ Home ] [ Form ]
[ Music library ] [ Picture library ]
C Puzzles : Questions and Answers - Level 3
----------------------------------------------------------------------------
[Image] Questions
L3.Q1 :
Write a function revstr() - which reverses the given string
in the same string buffer using pointers.
(ie) Should not use extra buffers for copying the
reverse string.
Solution for L3.Q1
----------------------------------------------------------------------------
L3.Q2 :
Write a program to print the series 2 power x, where x >= 0
( 1, 2, 4, 8, 16, .... ) without using C math library and
arithmatic operators ( ie. *, /, +, - and math.h are not allowed)
Solution for L3.Q2
----------------------------------------------------------------------------
L3.Q3 :
Write a program to swap two integers without using 3rd integer
(ie. Without using any temporary variable)
Solution for L3.Q3
----------------------------------------------------------------------------
L3.Q4 :
Write a general swap macro in C :
- A macro which can swap any type of data (ie. int, char,
float, struct, etc..)
Solution for L3.Q4
----------------------------------------------------------------------------
L3.Q5 :
Write a program to delete the entry from the doubly linked list
without saving any of the entries of the list to the temporary variable.
Solution for L3.Q5
----------------------------------------------------------------------------
L3.Q6 : What will be the output of this program ?
#include
main()
{
int *a, *savea, i;
savea = a = (int *) malloc(4 * sizeof(int));
for (i=0; i<4; i++) *a++ = 10 * i;
for (i=0; i<4; i++) {
printf("%d\n", *savea);
savea += sizeof(int);
}
}
Solution for L3.Q6
----------------------------------------------------------------------------
LX.Q7 : Trace the program and print the output
#include
typedef int abc(int a, char *b);
int func2(int a, char *b)
{
a *= 2;
strcat(b, "func2 ");
return a;
}
int func1(int a, char *b)
{
abc *fn = func2;
a *= a;
strcat(b, "func1 ");
return (fn(a, b));
}
main()
{
abc *f1, *f2;
int res;
static char str[50] = "hello! ";
f1 = func1;
res = f1(10, str);
f1 = func2;
res = f1(res, str);
printf("res : %d str : %s\n", res, str);
}
Solution for LX.Q7
----------------------------------------------------------------------------
LX.Q8 :
Write a program to reverse a Linked list within the same list
Solution for LX.Q8
----------------------------------------------------------------------------
LX.Q9 : What will be the output of this program
#include
main()
{
int a=3, b = 5;
printf(&a["Ya!Hello! how is this? %s\n"], &b["junk/super"]);
printf(&a["WHAT%c%c%c %c%c %c !\n"], 1["this"],
2["beauty"],0["tool"],0["is"],3["sensitive"],4["CCCCCC"]);
}
Solution for LX.Q9
----------------------------------------------------------------------------
LX.Q10 :
Solution for LX.Q10
----------------------------------------------------------------------------
[Image] Answers
L3.A1
Solution for L3.Q1
#include
char *rev_str(char *str)
{
char *s = str, *e = s + strlen(s) -1;
char *t = "junk";
while (s < e) { *t = *e; *e-- = *s; *s++ = *t; }
return(str);
}
main (int argc, char **argv)
{
printf("%s\n", rev_str(argv[1]));
}
----------------------------------------------------------------------------
L3.A2
Solution for L3.Q2
#include
void main()
{
int i;
for(i=0; i< 10; i++)
printf("%d\t", 2 << i);
}
----------------------------------------------------------------------------
L3.A3
Solution for L3.Q3
#include
main()
{
int a, b;
printf("Enter two numbers A, B : ");
scanf("%d %d", &a, &b);
a^=b^=a^=b; /* swap A and B */
printf("\nA = %d, B= %d\n", a, b);
}
----------------------------------------------------------------------------
L3.A4
Solution for L3.Q4
#include
/* Generic Swap macro*/
#define swap(a, b, type) { type t = a; a = b; b = t; }
/* Verification routines */
main()
{
int a=10, b =20;
float e=10.0, f = 20.0;
char *x = "string1", *y = "string2";
typedef struct { int a; char s[20]; } st;
st s1 = {50, "struct1"}, s2 = {100, "struct2"};
swap(a, b, int);
printf("%d %d\n", a, b);
swap(e, f, float );
printf("%f %f\n", e, f);
swap(x, y, char *);
printf("%s %s\n", x, y);
swap(s1, s2, st);
printf("S1: %d %s \tS2: %d %s\n", s1.a, s1.s, s2.a, s2.s);
ptr_swap();
}
ptr_swap()
{
int *a, *b;
float *c, *d;
a = (int *) malloc(sizeof(int));
b = (int *) malloc(sizeof(int));
*a = 10; *b = 20;
swap(a, b, int *);
printf("%d %d\n", *a, *b);
c = (float *) malloc(sizeof(float));
d = (float *) malloc(sizeof(float));
*c = 10.01; *d = 20.02;
swap(c, d, float *);
printf("%f %f\n", *c, *d);
}
----------------------------------------------------------------------------
L3.A5
Solution for L3.Q5
#include
/* Solution */
typedef struct Link{
int val;
struct Link *next;
struct Link *prev;
} Link;
void DL_delete(Link **, int);
void DL_delete(Link **head, int val)
{
Link **tail;
while ((*head)) {
if ((*head)->next == NULL) tail = head;
if ((*head)->val == val) {
*head = (*head)->next;
}
else head = &(*head)->next;
}
while((*tail)) {
if ((*tail)->val == val) {
*tail = (*tail)->prev;
}
else tail= &(*tail)->prev;
}
}
/* Supporting (Verification) routine */
Link *DL_build();
void DL_print(Link *);
main()
{
int val;
Link *head;
head = DL_build();
DL_print(head);
printf("Enter the value to be deleted from the list : ");
scanf("%d", &val);
DL_delete(&head, val);
DL_print(head);
}
Link *DL_build()
{
int val;
Link *head, *prev, *next;
head = prev = next = NULL;
while(1) {
Link *new;
printf("Enter the value for the list element (0 for end) : ");
scanf("%d", &val);
if (val == 0) break;
new = (Link *) malloc(sizeof(Link));
new->val = val;
new->prev = prev;
new->next = next;
if (prev) prev->next = new;
else head = new;
prev = new;
}
return (head);
}
void DL_print(Link *head)
{
Link *shead = head, *rhead;
printf("\n****** Link List values ********\n\n");
while(head) {
printf("%d\t", head->val);
if (head->next == NULL) rhead = head;
head = head->next;
}
printf("\n Reverse list \n");
while(rhead)
{
printf("%d\t", rhead->val);
rhead = rhead->prev;
}
printf("\n\n");
}
----------------------------------------------------------------------------
L3.A6
Solution for L3.Q6
The first value will be 0, the rest of the three values will not be
predictable. Actually it prints the values of the following location
in each step
* savea
* (savea + sizeof(int) * sizeof(int))
etc...
ie. savea += sizeof(int) => savea = savea + sizeof(savea_type) * sizeof(int)
( by pointer arithmatic)
=> save = savea + sizeof(int) * sizeof(int)
Note: You can verify the above by varing the type of 'savea' variable to
char, double, struct, etc.
Instead of statement 'savea += sizeof(int)' use savea++
then the values 0, 10, 20 and 30 will be printed.
This behaviour is because of pointer arithmatic.
----------------------------------------------------------------------------
LX.A7
Solution for LX.Q7
Two function pointers f1 and f2 are declared of the type abc.
whereas abc is a pointer to a function returns int.
func1() which is assigned to f1 is called first. It modifies the values of
the parameter 'a' and 'b' to 100, "hello! func1 ". In func1(), func2() is
called which further modifies the value of 'a' and 'b' to 200, "hello! func1
func2 " and returns the value of 'a' which is 200 to the main. Main calls
f1() again after assigning func2() to f1. So, func2() is called and it
returns the following value which will be the output of this program.
res : 400 str : hello! func1 func2 func2
The output string shows the trace of the functions called : func1() and func2()
then again func2().
----------------------------------------------------------------------------
LX.A8
Solution for LX.Q8
#include
typedef struct Link {
int val;
struct Link *next;
} Link;
/* Reverse List function */
Link *SL_reverse(Link *head)
{
Link *revlist = (Link *)0;
while(head) {
Link *tmp;
tmp = head;
head = head->next;
tmp->next = revlist;
revlist = tmp;
}
return revlist;
}
/* Supporting (Verification) routines */
Link *SL_build();
main()
{
Link *head;
head = SL_build();
head = SL_reverse(head);
printf("\nReversed List\n\n");
while(head) {
printf("%d\t", head->val);
head = head->next;
}
}
Link *SL_build()
{
Link *head, *prev;
head = prev = (Link *)0;
while(1) {
Link *new;
int val;
printf("Enter List element [ 0 for end ] : ");
scanf("%d", &val);
if (val == 0) break;
new = (Link *) malloc(sizeof(Link));
new->val = val;
if (prev) prev->next = new;
else head = new;
prev = new;
}
prev->next = (Link *)0;
return head;
}
----------------------------------------------------------------------------
LX.A9
Solution for LX.Q9
In C we can index an array in two ways.
For example look in to the following lines
int a[3] = {10, 20, 30, 40};
In this example index=3 of array 'a' can be represented in 2 ways.
1) a[3] and
2) 3[a]
i.e) a[3] = 3[a] = 40
Extend the same logic to this problem. You will get the output as follows
Hello! how is this? super
That is C
----------------------------------------------------------------------------
LX.A10
Solution for LX.Q10
----------------------------------------------------------------------------
Goto C Puzzle
* level 2
* level 1
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