Binary Bits Count

Write a function that takes the binary representation of an unsigned integer and returns the number of '1' bits it has.

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.

Solution 1

public int CountBits(uint n) {
    int count = 0;
    while (n != 0) {
        count += (int)(n & 1);
        n >>= 1;
    }
    return count;
}

This function works by iterating through the bits of n from right to left. In each iteration, it adds the least significant bit of n to count, and then shifts n to the right to process the next bit. This continues until all bits of n have been processed.

The time complexity of the algorithm is O(log(n)), where n is the input to the function. This is because in each iteration of the while loop, we’re dividing n by 2, effectively reducing the problem size by a factor of 2. Hence, the number of iterations is proportional to the number of bits in n, which is log(n) in base 2.

The space complexity of the algorithm is O(1), which means it uses a constant amount of space. This is because we’re only using a fixed number of variables and not any data structures that grow with the size of the input. The integer count and the input n are the only variables being used, and their size does not change with different inputs.

Solution 2

public int CountBits(uint n) {
    int count = 0;
    while (n != 0) {
        n &= (n - 1);
        count++;
    }
    return count;
}

In this algorithm, the expression n & (n - 1) removes the least significant ‘1’ bit from n. This is repeated until n becomes 0, and the number of iterations is equal to the number of ‘1’ bits in n.

The time complexity of this algorithm is O(k), where k is the number of ‘1’ bits in n. This is because each iteration of the loop eliminates one ‘1’ bit. In the worst case, when n is a power of 2, the time complexity is O(log(n)), which is the same as your first algorithm. However, on average, this algorithm will be faster if n has relatively few ‘1’ bits.

The space complexity remains O(1), as we’re only using a fixed number of variables.