## Problem – Number of 1 Bits

Write a function that takes 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`

.

**Follow up:** If this function is called many times, how would you optimize it?

### Number of 1 Bits LeetCode Solution in Java

```
public static int hammingWeight(int n) {
int ones = 0;
while(n!=0) {
ones = ones + (n & 1);
n = n>>>1;
}
return ones;
}
```

### Number of 1 Bits LeetCode Solution in C++

```
int hammingWeight(uint32_t n) {
int count = 0;
while (n) {
n &= (n - 1);
count++;
}
return count;
}
```

### Number of 1 Bits LeetCode Solution in Python

```
def hammingWeight(self, n):
"""
:type n: int
:rtype: int
"""
return bin(n).count('1')
```

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