LeetCode in Kotlin

2429. Minimize XOR

Medium

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 <= 109

Solution

class Solution {
    fun minimizeXor(num1: Int, num2: Int): Int {
        var bits = Integer.bitCount(num2)
        var result = 0
        run {
            var i = 30
            while (i >= 0 && bits > 0) {
                if (1 shl i and num1 != 0) {
                    --bits
                    result = result xor (1 shl i)
                }
                --i
            }
        }
        var i = 0
        while (i <= 30 && bits > 0) {
            if (1 shl i and num1 == 0) {
                --bits
                result = result xor (1 shl i)
            }
            ++i
        }
        return result
    }
}

This site is open source. Improve this page.