191-number-of-1-bits

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Description

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?

Solution(Python)

class Solution:
    def hammingWeight(self, n: int) -> int:
        return self.bitmanipulation(n)

    # Time Complexity: O(2^logn)
    # Space Complexity: O(logn)
    def bruteforce(self, n):
        n = bin(n)

        def recur(n):
            if not n:
                return 0
            return recur(n[1:]) + 1 if n[0] == "1" else recur(n[1:])

        return recur(n)

    # Time Complexity: O(log(n))
    # Space Complexity: O(logn)
    def naive(self, n):
        return bin(n).count("1")

    # Time Complexity: O(log(n))
    # Space Complexity: O(1)
    def bitmanipulation(self, n):
        cnt = 0
        while n:
            n &= n - 1
            cnt += 1
        return cnt