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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