[link](https://leetcode.com/problems/number-of-1-bits/)

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Write a function that takes the binary representation of an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight).

Note:

- Note that in some languages, such as Java, there is no unsigned integer type. In this case, the input will be given as a signed integer type. It should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.
- In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 3, the input represents the signed integer. -3.

#### Example 1:

Input: n = 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.

#### Example 2:

Input: n = 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.

#### Example 3:

Input: n = 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.

#### Constraints:

- The input must be a binary string of length 32.

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The variable count is initialized to 0, which will keep track of the count of '1' bits. The code enters a while loop that continues as long as n is greater than 0.

Inside the loop, the code checks if the least significant bit of n is equal to 1 by performing a bitwise AND operation between n and 1 (n & 1). If the result is 1, it means the least significant bit is 1, so the count is incremented by 1. After checking the least significant bit, the code shifts n to the right by 1 bit (n = n >> 1). This effectively discards the least significant bit and brings the next bit to the rightmost position.

The loop continues until all bits in n have been checked. Once the loop ends, the final count of '1' bits is returned as the result.

#### Solution 1

class Solution:
    def hammingWeight(self, n: int) -> int:
        count = 0
        while n > 0:
            if n & 1 == 1:
                count += 1
            n = n >> 1
        return count
        

O(T): O(log(n))
O(S): O(1)