number_of_1_bits 

 1# @leet start
 2class Solution:
 3    def hammingWeight(self, n: int) -> int:
 4        """
 5        The count of one bits in a number can be done by bitwise ANDing the
 6        number `n` with itself minus 1 `n - 1`.
 7        The reason why this works is that if a bit is set in `n`, in `n - 1`,
 8        its least significant bit will become 0.
 9        Thus, if we AND the numbers together, the least significant bit is removed.
10        We count how many times we do this until the number hits 0 to remove
11        all of the one bits.
12        """
13        count = 0
14        while n:
15            count += 1
16            n &= n - 1
17        return count
18
19
20# @leet end
21sol = Solution()
22
23
24def test():
25    assert sol.hammingWeight(11) == 3
26    assert sol.hammingWeight(128) == 1
class Solution: 
 3class Solution:
 4    def hammingWeight(self, n: int) -> int:
 5        """
 6        The count of one bits in a number can be done by bitwise ANDing the
 7        number `n` with itself minus 1 `n - 1`.
 8        The reason why this works is that if a bit is set in `n`, in `n - 1`,
 9        its least significant bit will become 0.
10        Thus, if we AND the numbers together, the least significant bit is removed.
11        We count how many times we do this until the number hits 0 to remove
12        all of the one bits.
13        """
14        count = 0
15        while n:
16            count += 1
17            n &= n - 1
18        return count
def hammingWeight(self, n: int) -> int: 
 4    def hammingWeight(self, n: int) -> int:
 5        """
 6        The count of one bits in a number can be done by bitwise ANDing the
 7        number `n` with itself minus 1 `n - 1`.
 8        The reason why this works is that if a bit is set in `n`, in `n - 1`,
 9        its least significant bit will become 0.
10        Thus, if we AND the numbers together, the least significant bit is removed.
11        We count how many times we do this until the number hits 0 to remove
12        all of the one bits.
13        """
14        count = 0
15        while n:
16            count += 1
17            n &= n - 1
18        return count

The count of one bits in a number can be done by bitwise ANDing the number n with itself minus 1 n - 1. The reason why this works is that if a bit is set in n, in n - 1, its least significant bit will become 0. Thus, if we AND the numbers together, the least significant bit is removed. We count how many times we do this until the number hits 0 to remove all of the one bits.

sol = <Solution object>
def test(): 
25def test():
26    assert sol.hammingWeight(11) == 3
27    assert sol.hammingWeight(128) == 1