# C++ Solution: n & (n - 1)

• ``````int hammingWeight(uint32_t n) {
int count = 0;

while (n) {
n &= (n - 1);
count++;
}

return count;
}
``````

n & (n - 1) drops the lowest set bit. It's a neat little bit trick.

Let's use n = 00101100 as an example. This binary representation has three 1s.

If n = 00101100, then n - 1 = 00101011, so n & (n - 1) = 00101100 & 00101011 = 00101000. Count = 1.

If n = 00101000, then n - 1 = 00100111, so n & (n - 1) = 00101000 & 00100111 = 00100000. Count = 2.

If n = 00100000, then n - 1 = 00011111, so n & (n - 1) = 00100000 & 00011111 = 00000000. Count = 3.

n is now zero, so the while loop ends, and the final count (the numbers of set bits) is returned.

• Worth mentioning that this solution will only loop as many times as the number of 1 bits whereas the solutions shifting the bits of n will loop also for every 0 bit in between.

• Yes, thank you. This should have been explained, as this is the crux of the algorithm.

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• This post is deleted!

• Thanks for the explanation!

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