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??

anonymous · Answer

n=NO_OF_BITS,hoping that u r stating of time complexity the enitre for loop iterates n times hence its order of n or in notations : O(n)

anonymous · Answer

no actually the complexity of the program in O(log n)
how??