CareerCup

To swap two numbers w/o using a temp variable, use XOR (^):

a = a ^ b
b = a ^ b
a = a ^ b

Try it with this and see for yourself:

a = 0011
b = 0101

Q2.)
let two numbers are a and b, they can be swapped like below
a = a + b
b = a - b
a = a - b

Q3.)
Maintain two pointers slow and fast.
Iterate thru the list by moving slow pointer once and fast pointer twice. When fast pointer will point to null, slow pointer will be pointing to middle element.

to get number of bits:
divide by 2 until u get a 0 using a while loop. increment a counter in each iteration.

Count number of bits set

unsigned int countOnes(unsigned int i)
{
   unsigned int c;
   for (c=0; i; c++)
       i &= i-1;
   return c;
}

Q1)

int count_bits(int n)
{

    int x = 1;
    int res = 0;

    while(x < n)
    {
        if( n & x != 0 )
            res++;

        x = x << 1;
    }

    return res;
}

counting number of bits program is wrong........this program counts only number of 1's in the bits. (what about zero's bit)

/* find the mid of the list */
for (i=0,j=0;i<n;i++)
{
if (i/2 > j) j++;
printf("i=%d, j=%d\n",i,j);
}

int count_bits(int n)
{
int x = 1;
int res = 0;
while(x < n)
{
if( n & x != 0 )
res++;
x = x << 1;
}

return res;
}

if you think this gives only 1 the ans is

int count_bits(int n)
{

return (sizeof(int)*8)
}

return res;
}

{
int n=1,c=0;
while(n!=0)
{
n=n<<1;
c++;
}
return c;
}

i = 0;
do{
num = num/2;
remain[i] = num%2;
i++;
}while(num!=0)
print(i); // i = count of bits

Take 2 pointers x1 and x2.
Keep x1 at the starting point of link list and x2 to start->next.
Increament x1 by 1 and x2 by 2 till reach end of list.
x1->num is the answer.