MINIMIZE XOR

/*

Given two positive integers num1 and num2, find the positive integer x such that:

x has the same number of set bits as num2, and
The value x XOR num1 is minimal.
Note that XOR is the bitwise XOR operation.

Return the integer x. The test cases are generated such that x is uniquely determined.

The number of set bits of an integer is the number of 1's in its binary representation.



Example 1:

Input: num1 = 3, num2 = 5
Output: 3
Explanation:
The binary representations of num1 and num2 are 0011 and 0101, respectively.
The integer 3 has the same number of set bits as num2, and the value 3 XOR 3 = 0 is minimal.
Example 2:

Input: num1 = 1, num2 = 12
Output: 3
Explanation:
The binary representations of num1 and num2 are 0001 and 1100, respectively.
The integer 3 has the same number of set bits as num2, and the value 3 XOR 1 = 2 is minimal.


Constraints:

1 <= num1, num2 <= 10^9

*/

class Solution {
public:
    int minimizeXor(int num1, int num2) {
        int count=__builtin_popcount(num2);
        int res=0;
        for(int i=31;i>=0 && count>0;i--){   /* 2^31 to 2^0, first try to set bits in res which are already set in num1 at highest powers of 1. Because when finally res and num1 will xor, 1 and 1 will give 0, so we try to cancel out the largest powers positions */
            int bit_state=(num1 >> i) & 1;
            if(bit_state==1){
                res=res | (1 << i);
                count--;
            }
        }

        for(int i=0;i<32 && count>0;i++){   // 2^0 to 2^31, if you still have some count left, turn on bits of small powers of 2, so start from power 0
            int bit_state=(res >> i) & 1;
            if(bit_state==1){
                continue;
            }
            res=res | (1 << i);
            count--;
        }
        return res;
    }
};