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# Number of 1 Bits LeetCode Solution

## 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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