unsigned int reverseBits(unsigned int num)
{
unsigned int NO_OF_BITS = sizeof(num) * 8;
unsigned int reverse_num = 0;
int i;
for (i = 0; i

Question

unsigned int reverseBits(unsigned int num)
{
    unsigned int  NO_OF_BITS = sizeof(num) * 8;
    unsigned int reverse_num = 0;
    int i;
    for (i = 0; i < NO_OF_BITS; i++)
    {
        if((num & (1 << i)))
           reverse_num |= 1 << ((NO_OF_BITS - 1) - i);
   }
    return reverse_num;
}

what is the complexity of the above program and how??
 is complexity is O(log n)  how??

anonymous · Answer

seems like it is at least linear - dependent on the size of num - if not more.
doesn't look like it is less than linear

anonymous · Answer

no i got the answer. it is O(log n) only ...... since the no of bits  is 2^n. that is if the number is 8 then we need only 3 bits to represent. so it is not linear time . it is O(log n)

anonymous · Answer

ok - well i was takin a leap there since i didn't even know what language that was.  this course is using python - you might get better answers from us if you post python code