2429. Minimize XOR

#Problem

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.

https://leetcode.com/problems/minimize-xor/

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 <= 10⁹

#Test Cases

class Solution:
    def minimizeXor(self, num1: int, num2: int) -> int:

import pytest

from solution import Solution


@pytest.mark.parametrize('num1, num2, expected', [
    (3, 5, 3),
    (1, 12, 3),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, num1, num2, expected):
    assert sol.minimizeXor(num1, num2) == expected

#Thoughts

先统计 num1 和 num2 各自二进制表示中的 1 的个数（可以用 191. Number of 1 Bits 里提到的各种方法，或者直接用语言自带的方法如 Python 中 int.bit_count）。

如果它俩的二进制表示中 1 的个数一致，那么 num1 就是所求结果，它与自己的 XOR 结果为 0，是最小的。

如果 num1 的 1 的个数少一些，就需要把一些二进制位的 1 改成 0，为了让 XOR 的结果最小，显然应该先改低位的 1。

如果 num1 的 1 的个数多一些，就需要把一些二进制为的 0 改成 1，同理可知为了让 XOR 的结果最小，应该先改低位的 0。

时间复杂度 O(log num1 + log num2)，空间复杂度 O(1)。

#Code

class Solution:
    def minimizeXor(self, num1: int, num2: int) -> int:
        b2 = num2.bit_count()
        b1 = num1.bit_count()

        mask = 1
        while b1 > b2:
            if num1 & mask:
                num1 ^= mask
                b1 -= 1
            mask <<= 1

        while b1 < b2:
            if num1 & mask == 0:
                num1 |= mask
                b1 += 1
            mask <<= 1

        return num1