LeetCode 2429: Minimize XOR

Explanation

To solve this problem, we need to find a number x that has the same number of set bits as num2, and when XORed with num1, gives the minimal result. We can achieve this by iteratively trying out different values of x that have the same number of set bits as num2 and calculating the XOR value with num1 to find the minimum.

1. Count the number of set bits in num2.
2. Iterate from x = 0 to x = 2^32 - 1 (as we are dealing with 32-bit integers) and for each x:
   • Count the number of set bits in x.
   • If the count matches the count from step 1, calculate the XOR of x and num1.
   • Update the minimum XOR value and corresponding x if the calculated XOR is less than the current minimum.
3. Return the x that gives the minimum XOR value.

By iterating through all possible values of x with the same number of set bits as num2, we can find the solution to this problem.

Time complexity: O(1) as we iterate through all possible 32-bit integers Space complexity: O(1)

class Solution {
    public int minimizeXOR(int num1, int num2) {
        int countSetBitsNum2 = Integer.bitCount(num2);
        int min = Integer.MAX_VALUE;
        int result = 0;
        
        for (int x = 0; x <= Integer.MAX_VALUE; x++) {
            if (Integer.bitCount(x) == countSetBitsNum2) {
                int xor = x ^ num1;
                if (xor < min) {
                    min = xor;
                    result = x;
                }
            }
        }
        
        return result;
    }
}