13. Reverse Bits

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

My Approach

For this problem, I'm using a straightforward approach. I iterate through each bit of the given number x. In each iteration, I extract the least significant bit of x using the bitwise AND operator (x & 1). I add this bit to the result variable out. Then, I left shift out by one bit to make space for the next bit. Finally, I right shift x by one bit to move to the next bit position. I repeat this process for 31 iterations, as we are considering the 32-bit representation of x (assuming a 32-bit integer).

Time and Auxiliary Space Complexity

• Time Complexity: O(1) - Since the number of iterations (31) is fixed and independent of the input size.

• Auxiliary Space Complexity: O(1) - No extra space is used; the space complexity remains constant.

Code (C++)

class Solution {
  public:
    long long reversedBits(long long x) {
        long long out = 0;
        for(int i = 0; i < 31; ++i)
            out += x & 1,
            out <<= 1,
            x >>= 1;
            
        return out;
    }
};

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