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Swap two integer variables

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How to swap two integer variables without using the third variable ?
See Solution: Swap two integer variables

I know three solutions, comment if you know more 🙂

Solution 1:

x = x+y;
y  = x-y;
x  = x-y;

Solution 2:

y = 2*((x+y)/2) – (x =y);

Solution 3:

x = x^y;
y = x^y;
x = x^y;

Note: However solution 1 & 2 has problem of overflow doing addition of two integers might overflow. But solution 3 will not have any such problem

How to measure 4 gallon of water

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How to measure exactly 4 gallon of water from 3 gallon and 5 gallon jars, Given, you have unlimited water supply from a running tap.

It was also solved by Bruce Willis in one of his Die Hard movie to stop a bomb.

See Solution: How to measure 4 gallon water from 3 gallon and 5 gallon jar
There are 2 possible solutions i can think of

Solution 1:

  1. Fill 3 gallon jar with water.
  2. Pour all its water into 5 gallon jar.
  3. Fill 3 gallon jar again.
  4. Pour its water into 5 gallon jar untill it is full. Now you will have exactly 1 gallon water remaining in 3 gallon jar.
  5. Empty 5 gallon jar, pour 1 gallon water from 3 gallon jar into it. Now 5 gallon jar has exactly 1 gallon of water.
  6. Fill 3 gallon jar again and pour all its water into 5 gallon jar, thus 5 gallon jar will have exactly 4 gallon of water.

Solution 2:

  1. Fill 5 gallon jar.
  2. Pour all its water to 3 gallon jar, thus leaving exactly 2 gallon water in 5 gallon jar.
  3. Empty 3 gallon jar and pour all of 5 gallon jar water into it, thus now 3 gallon jar will have exactly 2 gallon of water.
  4. Fill 5 gallon jar again and pour its water to 3 gallon jar until that is full. So exactly 4 gallon water will remain in 5 gallon jar.

Objective Questions set C/C++ – Part 5

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4 Objective type interview questions on C/C++ with answers

Before you start:-

  1. You can check your answer by filling a,b,c,d or e in the text box.
  2. You can get the answer by clicking Give Up Button.

Try yourself before giving up 🙂

Q1)

#include
int main(){
    char a[5]="hello";
    printf("%s",a);
}

Options: a) will print “hello” only b) will print some garbage values c) will print “hello” atleast d) None

Tried enough already?
[Read more…]

Objective Questions set C/C++ – Part 4

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10 Objective type interview questions with answers on C/C++

Before you start:-

  1. You can check your answer by filling a,b,c,d or e in the text box.
  2. You can get the answer by clicking Give Up Button.

Try yourself before giving up 🙂

Q1)

#include
main()
{
    unsigned char i;
    int sum=0;
    for(i=0;i<300;i++)
        sum=sum+i;
    printf("%d",sum);
}

Options: a) 44850 b) 45150 c) 300 d) Infinite Loop

Tried enough already?

Q2)

#include
fun(int i)
{
    int j=0;
    while (i & i-1) {
        j++;   
    }
    printf("%d",j);
}

main(){
    f(100);
}

Options: a. 100 b. 99 c. 2 d. 3

Tried enough already?

Q3)

Code 1 :
        for(i=0; i<1000; i++)
                for(j=0; j<100; j++)
                        x = y;

Code 2 :
        for(i=0; i<100; i++)
                for(j=0; j<1000; j++)
                        x = y;

Which code will execute faster ?
[Read more…]

Objective Questions set C/C++ – Part 3

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10 Objective type interview questions with answers on C/C++

Before you start:-

  1. You can check your answer by filling a,b,c,d or e in the text box.
  2. You can get the answer by clicking Give Up Button.

Try yourself before giving up 🙂

Q1) What will the below program prints ?

#include
void e( int );
main( )
{
    int a;
    a=3;
    e(a);
}

void e(int n)
{
    if (n>0)
    {
        e(--n);
        printf("%d",n);
        e(--n);
    }
}

Options: a) 0120 b) 0121 c) 0201 d) 0211

Tried enough already?
Q2)

#include
main( )
{
    int a[5]= {1,2,3,4,5};
    int*ptr = (int*)(&a+1);
    printf("%d %d", *(a+1), *(ptr-1));
}

Options: a)2 2 b) 2 1 c) 2 5 d) None

Tried enough already?
Q3)

#include
void foo(int [][3]);
main( )
{
    int a[3][3]={ {1,2,3}, {4,5,6}, {7,8,9} };
    foo(a);
    printf("%d", a[2][1]);
}

void foo(int b[][3])
{
    ++b;
    b[1][1]=35;
}

Options: a)8, b) 9, c) 7, d)35

Tried enough already?
Q4)
If performance is a critical requirement, then storage class should be used is ?
a)auto b) register c) extern d) static
Tried enough already?

Q5)
In a program
1-> char str1[ ] = “abc”;
2-> char *str2 = “xyz”;
3-> str2 = str1;
4-> str1 = str2;

Options:
a)compile error at line 3
b)compile error at line 4
c)no problem in the code
d)run time error

Tried enough already?

Q6)

int A = sizeof(char*)
int B = sizeof(int*)

a)A=B b)A<B c)A>B d) depends on compiler

Tried enough already?

Q7)

void func (int y)
{
static int x;
.
.
.
}

x and y will be stored on respectively in ?

Options:
a) stack and heap b) stack and data section c) data section and stack d) heap and data section

Tried enough already?

Q8)

#include

main(){
    int i=5;
    ++i++;
    printf("%d", i);
}

Options:
a)5 b) 6 c) 7 d) compilation error

Tried enough already?

Q9)
Statement 1-> const char *p1=”hello world”;
Statement 2-> char *const p2=”hello world”;
Statement 3-> p1=”welcome”;
Statement 4-> p2=”good bye”;
Statement 5-> *(p1+3) = ‘A’;
Statement 6­-> *(p2+3) = ‘A’;

Which statement cause problem ?

Options: a)4 & 5 b) 1 & 3 c) 3 & 5 d) 2 & 3

Tried enough already?

Q10)

#include
int function()
{
    static int a=0;
    a=a+1;
    return a;
}

main()
{
    function();
    function();
    printf("%d", function());
}

Options: a) 1 b) 2 c) 3 d) None

Tried enough already?

Objective Questions set C/C++ – Part 2

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10 Objective type interview questions with answers on C/C++

Before you start:-

  1. You can check your answer by filling a,b,c,d or e in the text box.
  2. You can get the answer by clicking Give Up Button.

Try yourself before giving up 🙂

Q1)

#include
struct XX{
    int a:10;
    int b:6;
    char c;
} structure;
int main(){
    int i = sizeof(structure);
    printf("%d", i);
}

Options: a) 4 b) 8 c) 9 d) 3 e) Compilation error

Tried enough already?

Q2)

#include
#include
main()
{
    char *s;
    s="hot java";
    strcpy(s,"solaris java");
    printf("%s", s);
}

Options:a) hot java b) solaris java c) compilation error d) Segmentation fault

Tried enough already?
[Read more…]

Objective Questions set C/C++ – Part 1

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10 Objective type interview questions with answers on C/C++

Before you start:-

  1. You can check your answer by filling a,b,c,d or e in the text box.
  2. You can get the answer by clicking Give Up Button.

Try yourself before giving up 🙂

Q1) What will be the output of the program below

#include
main(){
    int num[] = {1,4,8,12,16};
    int *p,*q;
    int i;
    p =  num;
    q = num+2;
    i = *p++;
    printf("%d, %d, %d\n",i, *p, *q);
}

Options a) 4, 4, 8 b) 1, 4, 8 c) 2, 1, 8 d) 2, 4, 8

Tried enough already?
Q2) What will be the output of the program below

#include

main(){
char *a[4]={"jaya","mahesh","chandra","swapant"};
int i = sizeof(a)/sizeof(char *);
printf("i = %d\n", i);
}

Options: a) 4 b) 7 c) 8 d) 1

Tried enough already?
[Read more…]

Red and blue balls

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You have three bags and three labels. One bag has only red balls, one has only blue balls and one has both red and blue balls. Three labels are R, B and RB. R label was meant for the bag with only red balls, B label was meant for the bag with only blue balls and RB for the bag with both red and blue balls. Ram by mistake labelled the bags wrongly such that all the labels are wrong, how many minimum number of balls he should pick and from what bags to correct the labels? It is given that each bag has unlimited number of balls to be picked.

See Solution : Red and blue balls
The minimum number of balls he should pick is 1 from the bag with label RB.

How ?

As the label of all the bags are wrong, bag with label RB can either have all red balls or all blue balls.

Now ram will either get red ball or the blue ball.

Case 1: Ram gets the red ball
Bag with label RB is the bag with all the red balls(R).
Bag with label B can either be RB or R(as all the labels are wrong) but as we already know R bag so it can only be RB.
We know 2 bags correctly so third bag with label R should be B.

Case 2: Ram gets the blue ball
Bag with label RB is the bag with all the blue balls(B).
Bag with label R can either be B or RB(as all the labels are wrong) but as we already know B bag so it can only be RB.
we know 2 bags correctly so third bag with label B should be R.

River crossing – the harder one

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A dysfunctional family has to cross the river. On one side of the river are a mom and 2 daughters, dad and 2 sons, the maid and the dog. There is a boat only big enough to hold 2 people (counting the dog as 1 person). Only the adults are capable of operating the boat. Everyone has to get to the other side, without anything bad happening.
Now the difficulties are : if the dog is left with anyone and the maid isn’t there to control him, he’ll bite. The dad can’t be left with any of the daughters when the mom isn’t there. Likewise, the mom can’t be trusted alone with either of the sons when the dad isn’t there.
Remember! only an adult can operate the boat, and the boat can’t drive itself.See Solution : River crossing - the harder one

Say
West shore is {W} and East shore is {E}
Lets give the nick names to each family member and dog 😉
Mother – {m}, Father – {F}, Daughters – {d1, d2}, Sons – {s1, s2}
House maid – {h}, Dog – {d}

Initially,
W = {m, d1, d2, f, s1, s2, h, d}
E = {…}

let’s move everyone, over…

housemaid and dog go east, and the housemaid comes back:

W = {m, d1, d2, f, s1, s2, h}
E = {d}

housemaid and s1 go east, h and d come back:

W = {m, d1, d2, f, s2, h, d}

E = {s1}

father and s2 go east, father comes back:

W = {m, d1, d2, f, h, d}

E = {s1, s2}

mother and father go east, mother comes back:

W = {m, d1, d2, h, d}

E = {f, s1, s2}

h and d go east, father comes back:

W = {m, d1, d2, f}

E = {s1, s2, h, d}

father and mother go east, mother comes back:

W = {m, d1, d2}

E = {f, s1, s2, h, d}

mother and d1 go east, housemaid and d come back:

W = {d2, h, d}

E = {m, d1, f, s1, s2}

h and d2 go east, h comes back

W = {h, d}

E = {m, d1, d2, f, s1, s2}

h and d go east

W = {}

E = {m, d1, d2, f, s1, s2, h, d}

And we are done.

Defective coins

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We have 10 bags of 1 Rupee coins. One bag contains all the defective coins, the weight of each coin in that bag is 1 gram lesser than the weight of a normal 1 Rupee coin. You have a spring balance, which tells the exact weight. After how many minimum no. of weighs you can separate the bag with defective coins.

Other variant of the same problem:
There are 10 machines in a factory. Each produces coins weighing 10 grams each. One day the factory owner cones to know that one of the machine is not functioning properly and produces coins of weight 9 grams. You have to find out the faulted machine. You ONLY have a weighing machine and you can use it only ONCE.

See Solution : Defective coins
The minimum no. of weighs required is one.

Want to try now ?

solution :

Let the weight of a normal 1 Rupee coin is W.
Now number the bags from 1 to 10 and take out the no. of coins equal to bag’s number from each bag, for example from 1st bag, take one coin, from 5th bag take 5 coins and so on.
Let P is the no. of the bag which contains the defective coins.
Now put all the coins on spring balance and weigh them. Read the spring balance reading, say it is S grams.
Then S = (W grams)*(1+2+3….+10) – P*(1 gram)
P will come out to be a number between 1 to 10 and thats the required bag.

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