191. Number of 1 Bits

Description of the Problem

Write a function that takes the binary representation of 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 1

Tags: Bit Manipulation

Code (Rust)

impl Solution {
    pub fn hammingWeight (n: u32) -> i32 {
        let mut n = n;
        let mut count = 0;
        while n != 0 {
            count += n & 1;
            n  = n >> 1; // Unsigned Right Shift
        }
        return count as i32;
    }
}

Complexity

  • d is the number of digits of the input number

Time Complexity:

  • \(T(d) = O(d)\)

Auxiliary Space:

  • \(T(d) = O(1)\)

Solution 2 (Answer to the follow-up)

Tags: Precomputation Bit Manipulation

Explanation

Since the function is called many times, we must reduce the executeion time. Space can buy time; we can pre-compute the number of '1' bit in a 4-bit numbers.

Code (Rust)

impl Solution {
    pub fn hammingWeight (n: u32) -> i32 {

        let precomputed = [
            0,1,1,2, // 0000 0001 0010 0011
            1,2,2,3, // 0100 0101 0110 0111
            1,2,2,3, // 1000 1001 1010 1011
            2,3,3,4  // 1100 1101 1110 1111
        ];

        let mut n = n;
        let mut count = 0;
        while n != 0 {
            count += precomputed[(n & 0b1111) as usize];
            n  = n >> 4; // Unsigned Right Shift
        }
        return count;
    }
}

Complexity

  • d is the number of digits of the input number

Time Complexity:

  • \(T(d) = \Theta(\lceil d/4 \rceil)\)

Auxiliary Space:

  • \(T(d) = O(1)\)