Number of 1 Bits

---

Number of 1 Bits: Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight).

• 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.

---

Try this Problem on your own or check similar problems:

1. Reverse Bits
2. Power of Two
3. Counting Bits

Solution:

Java

// you need to treat n as an unsigned value
public int hammingWeight(int n) {
    int totalCount = 0;
    while(n != 0){
        totalCount += n & 1;
        n >>>= 1;
    }
    return totalCount;
}

JavaScript

/**
 * @param {number} n - a positive integer
 * @return {number}
 */
var hammingWeight = function (n) {
  let totalCount = 0;
  while (n !== 0) {
    totalCount += n & 1;
    n >>>= 1;
  }
  return totalCount;
};

Python

class Solution:
    def hammingWeight(self, n: int) -> int:
        totalCount = 0
        while n != 0:
            totalCount += n & 1
            n >>= 1
        return totalCount

C++

class Solution {
public:
    int hammingWeight(uint32_t n) {
         int totalCount = 0;
        while (n != 0) {
            totalCount += n & 1;
            n >>= 1;
        }
        return totalCount;
    }
};

---

Time/Space Complexity:

• Time Complexity: O(1)
• Space Complexity: O(1)

---

Explanation:

Since we have limited length of binary representation we achieve constant space and time complexity. There are two pieces to the code above:

• First we need to do unsigned right-shift operator >>> since the >> carries over the value of most significant bits (which will then not work for negative numbers), but you could however do >> 32 times controlling the loop in other way. Therefore it's important to remember that >>>`` will always put a 0 in the left most bit, while >>` will put a 1 or a 0 depending on if the number is negative or positive.
• So we do >>> until we reach 0 (nothing left to shift), each time checking if the last significant bit is 1 by doing n & 1 and incrementing totalCount if it is.

Finally we return totalCount as our final result.