Cracking the Coding Interview, 4 Edition - 150 Programming Interview Questions and Solutions

Solutions to Chapter 5 | Bit Manipulation
5.4 Explain what the following code does: ((n & (n-1)) == 0).
SOLUTION
pg 58
We can work backwards to solve this question.
What does it mean if A & B == 0?
It means that A and B never have a 1 bit in the same place. So if n & (n-1) == 0, then n and
n-1 never share a 1.
What does n-1 look like (as compared with n)?
Try doing subtraction by hand (in base 2 or 10). What happens?
1101011000 [base 2]
- 1
= 1101010111 [base 2]
593100 [base 10]
- 1
= 593099 [base 10]
When you subtract 1 from a number, you look at the least significant bit. If it’s a 1 you change
it to zero and you are done. If it’s a zero, you must “borrow” from a larger bit. So, you go to
increasingly larger bits, changing each bit from a 0 to a 1, until you find a 1. You flip that one
to a 0 and you are done.
Thus, n-1 will look like n, except that n’s initial 0s will be 1’s in n-1, and n’s least significant 1
will be a 0 in (n-1). That is:
if n = abcde1000
then n-1 = abcde0111
So what does n & (n-1) == 0 indicate?
n and (n-1) must have no 1s in common. Given that they look like this:
if n = abcde1000
then n-1 = abcde0111
abcde must be all 0s, which means that n must look like this: 00001000. n is therefore a
power of two.
So, we have our answer: ((n & (n-1)) == 0) checks if n is a power of 2 (or 0).
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