## Instructions

## Approach

There are several ways to solve this problem. Let's look at a few.

#### Using count()

We convert the decimal number to binary, replace "0b" with no value and count the number of 1's

### Implementation

```
def hammingWeight(self, n: int) -> int:
bits = bin(n).replace("0b", "")
return bits.count('1')
```

## Using the bit_count() method

The method returns the number of ones in the binary representation of the absolute value of the integer.

```
def hammingWeight(self, n: int) -> int:
return n.bit_count()
```

You can review this resource for a better understanding of the method.

### Bit Manipulation

We can loop through the number and count the number of 1's

Using the modulus operator(%) we can get the number of 1's and shift n to the right until all 32bits are zeros.

- n%2 will return 0 or 1 so the count only changes when it is a 1.

#### Implementation

```
def hammingWeight(self, n: int) -> int:
count = 0
while n:
count += n%2
n = n >> 1
return count
```

This algorithm has a constant time complexity O(1) because we iterate through 32 bits.

The space complexity is O(1) because we do not need extra memory.

I hope you found this helpful. Let me know of other solution approaches.

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