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?

Approach 1: Built-in Function

class Solution {
public:
    int hammingWeight(uint32_t n) {
        // or return bitset<32>(n).count();
        return __builtin_popcount(n);
    }
};

class Solution:
    def hammingWeight(self, n: int) -> int:
        return bin(n).count('1')

func hammingWeight(num uint32) int {
    return bits.OnesCount32(num);
}

impl Solution {
    pub fn hammingWeight (n: u32) -> i32 {
        n.count_ones() as i32
    }
}

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        return Integer.bitCount(n);
    }
}

/**
 * @param {number} n - a positive integer
 * @return {number}
 */
var hammingWeight = function (n) {
  return n.toString(2).split('1').length - 1;
};

Approach 2: Bit Manipulation

We check each parity of teach bit. Increase $ans$ by 1 if the bit is set.

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int ans = 0;
        while (n) {
            if (n & 1) ans++;
            n >>= 1; // or n /= 2;
        }
        return ans;
    }
};

public class Solution {
    // you need to treat n as an unsigned value
    //
    public int hammingWeight(int n) {
        int ones = 0;
        // n > 0, fails to return the correct the answer because of Integer MAX_VALUE.
        // Integer.MAX_VALUE + 1 is -2147483648, so it's not greater than 0, so n will not enter into loop
        while (n != 0) {
            ones += (n & 1);
            // Why can't we use n >>= 1?
            // Since n is 32 bit binary number, >> operator does shift by keeping signed bit position same as before
            // Take a look at SO reference - https://stackoverflow.com/questions/2811319/difference-between-and
            n >>>= 1;
        }
        return ones;
    }
}

class Solution:
    def hammingWeight(self, n: int) -> int:
        res = 0
        while n:
            if (n & 1): res += 1
            # same as --> n //= 2
            n >>= 1
        return res

/**
 * @param {number} n - a positive integer
 * @return {number}
 */
var hammingWeight = function (n) {
  let res = 0;
  while (n != 0) {
    if (n & 1) res++;
    n /= 2;
  }
  return res;
};

Approach 3: n & (n - 1)

We can optimise approach 2. Instead of checking all digits, we can use n & (n - 1) to remove the rightmost set bit.

n     n     n - 1  n & (n - 1)
--   ----   ----   -------
 0   0000   0111    0000
 1   0001   0000    0000
 2   0010   0001    0000
 3   0011   0010    0010
 4   0100   0011    0000
 5   0101   0100    0100
 6   0110   0101    0100
 7   0111   0110    0110
 8   1000   0111    0000
 9   1001   1000    1000
10   1010   1001    1000
11   1011   1010    1010
12   1100   1011    1000
13   1101   1100    1100
14   1110   1101    1100
15   1111   1110    1110

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int ans = 0;
        for (; n; n = n & (n - 1)) ans++;
        return ans;
    }
};

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int ones = 0;
        // Since n is 32 bit binary number, count 1's till that range
        for (int i = 0; i < 32; i++) {
            ones += (n & 1);
            n >>= 1;
        }
        return ones;
    }
}

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

0191 - Number of 1 Bits (Easy)

Problem Link

Problem Statement

Approach 1: Built-in Function

Approach 2: Bit Manipulation

Approach 3: n & (n - 1)

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