2429. Minimize XOR

Problem

Given two positive integers num1 and num2, find the 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

Solution (Java)

class Solution {
    public int minimizeXor(int num1, int num2) {

        int []arr = new int[32];

        int count = 0;
        for(int i = 0; i < 32; i++){
            if((num2 & (1 << i)) != 0){
                count++;
            }
        }

        for(int i = 31; i >= 0; i--){
            if((num1 & (1 << i)) != 0 && count > 0){
                arr[i] = 1;
                count--;
            }
        }

      if(count > 0){
            for(int i = 0; i < 32 && count > 0; i++){
                if(arr[i] != 1){
                    arr[i] = 1;
                    count--;
                }
            }
        }

        StringBuilder str = new StringBuilder();
        for(int i = 0; i < 32; i++){
            str.insert(0, arr[i]);
        }

        return Integer.parseInt(str.toString(), 2);
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).