Problem statement

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.
    Link to problem

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.

Solutions

Approach 1: Hotshot

Using the count_ones built-in in Rust.

Implementation in Rust

fn hamming_weight(n: u32) -> i32 {
    n.count_ones() as i32
}

Time Complexity: $O(n)$
Runtime on LeetCode: $0$ms

Approach 2: Modulo and Division

The normal way is to check every ending digit and if it is a one, we add it to our count.
Dividing the number by 10 gets rid of the last digit, and you can use shift operator in this case num >> 1

Pseudocode

while num > 0
    count += num % 2
    num /= 10

Implementation in Rust

fn hamming_weight(mut n: u32) -> i32 {
    let mut count = 0;
    while n > 0 {
        count += n % 2;
        n = n >> 1;
    }
    count as i32
}

Time Complexity: $O(n)$
Runtime on LeetCode: $0$ms

Approach 3: Bit Manipulation

Quite a clever approach where you utilize number and number - 1.
Basically, anding those two together will yield you a number or a zero.

Consider the following example: 1011

Subtracting it with previous number we get 1010
Anding those two together we get 1010, add 1 to our count.

Now and 1010 with 1001, which is 1000, add 1 to our count.

Now 1000 and 0111, which is 0000, add 1 to our count.

The number is now 0 so stop the loop.

Implementation in Rust

fn hamming_weight(mut n: u32) -> i32 {
    let mut count = 0;
    while n > 0 {
        n = n & (n-1);
        count += 1;
    }
    count
}

Time Complexity: $O(n)$
Runtime on LeetCode: $0$ms