Number of 1 Bits LeetCode Solution | Easy Approach

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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:

  • 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.

Number of 1 Bits Solutions

Time: O(32)=O(1)
Space: O(1)

C++

class Solution {
 public:
  int hammingWeight(uint32_t n) {
    int ans = 0;

    for (int i = 0; i < 32; ++i)
      if ((n >> i) & 1)
        ++ans;

    return ans;
  }
};

Java

 public class Solution {
  // you need to treat n as an unsigned value
  public int hammingWeight(int n) {
    int ans = 0;

    for (int i = 0; i < 32; ++i)
      if (((n >> i) & 1) == 1)
        ++ans;

    return ans;
  }
}

 

Python

class Solution:
  def hammingWeight(self, n: int) -> int:
    ans = 0

    for i in range(32):
      if (n >> i) & 1:
        ans += 1

    return ans

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