# Number of 1 Bits LeetCode Solution | Easy Approach Share:

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`.

Time: O(32)=O(1)
Space: O(1)

### C++

``````class Solution {
public:
int hammingWeight(uint32_t n) {
int ans = 0;

for (int i = 0; i < 32; ++i)
if ((n >> i) & 1)
++ans;

return ans;
}
};
``````

### Java

`````` public class Solution {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {
int ans = 0;

for (int i = 0; i < 32; ++i)
if (((n >> i) & 1) == 1)
++ans;

return ans;
}
}

``````

### Python

``````class Solution:
def hammingWeight(self, n: int) -> int:
ans = 0

for i in range(32):
if (n >> i) & 1:
ans += 1

return ans

``````

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