has a set bit, and now we still have 2 remaining bits we want to set in x. Where should we set them? And the answer is since we want the minimal answer, we should set them at positions, where they contribute as minimum as possible, i.e. LSB positions, and hence we loop from bit position 0.

Case 3 - b == a

This is the simplest case. Just set the bits as placed in num1.

Approach

Count Set Bits in num2: Calculate the number of set bits (1s) in num2 and store it in cnt. We can use methods described in Number of 1 Bits or use standard library function for that.
Set Common Bits in the Result:
- To minimize x XOR num1, align set bits of x with those in num1. Iterate from the most significant bit (MSB) to the least significant bit (LSB) of num1:
  - If a bit in num1 is set and cnt is greater than 0, set the corresponding bit in

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