01. Number of 1 Bits

The problem can be found at the following link: Question Link

My approaches

Since this question is relatively simple and can be solved in multiple ways, I would like to present two commonly used approaches.

Approach 1:

The first approach involves iterating through the bits of the binary representation of the number and counting the set bits. Here how it works:

• We repeatedly right-shift the number by 1 bit and add the rightmost bit to the counter variable cnt if it is 1 counter increased . If the bit is 0, the counter remains unchanged.

Example

   if [n = 18 (10010)]
   - 10010     , cnt = 0
   - 1001      , cnt = 1
   - 100       , cnt = 1
   - 10        , cnt = 1
   - 1         , cnt = 2

Time and Auxiliary Space Complexity

• Time Complexity : O(log(n)), as we iterate through all the bits of n, which totals to log(n).

• Auxiliary Space Complexity : constant as it does not depend on variables

Code (C++)

class Solution {
public:
    int setSetBit(int x, int y, int l, int r) {
        int maskLen = r - l + 1;
        int mask = (1 << maskLen) - 1;
        mask = mask << (l - 1);
        y = y & mask;
        return (x | y);  
    }
};

Approach 2:

The second approach utilizes the standard library function __builtin_popcount(n) in C++ to count the number of set bits. This approach involves a concise one-liner code using the built-in function.

Code (C++):

class Solution {
  public:
    int setBits(int n) {
        return __builtin_popcount(n);
    }
};

Time and Auxiliary Space Complexity

• Time Complexity : O(k), where k is the number of bits.

• Auxiliary Space Complexity : constant as it does not depend on variables

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