CareerCup

Here is thought. Let the number is x.Compute k=log(x+1)on base 2. If((1<<k)==(x+1)) then true otherwise false

if (n & (n +1)) == 0)
printf("Number is 1 less than power of 2\n") ;
else
printf("Number is NOT 1 less than power of 2\n") ;

int powof2less1(int n)
{
// for eg with input n = 7
// n + 1 = 8
// (8 & 7) == 0 so return 1
// and so on
return (((n+1) & (n)) == 0);
}

int pwr2ls1(int number)
{
int n= number +1;

if( (n) & (n-1) == 0)
return 1 ;
else return 0;
}

Would be your lucky day, if amazon asks this question. :)

number of bits == 2 && first bit == 1
optimization: when number of bits count exceeds 2 break, the number is not the answer