LeetCode in Kotlin

3681. Maximum XOR of Subsequences

Hard

You are given an integer array nums of length n where each element is a non-negative integer.

Select two subsequences of nums (they may be empty and are allowed to overlap), each preserving the original order of elements, and let:

• X be the bitwise XOR of all elements in the first subsequence.
• Y be the bitwise XOR of all elements in the second subsequence.

Return the maximum possible value of X XOR Y.

Note: The XOR of an empty subsequence is 0.

Example 1:

Input: nums = [1,2,3]

Output: 3

Explanation:

Choose subsequences:

• First subsequence [2], whose XOR is 2.
• Second subsequence [2,3], whose XOR is 1.

Then, XOR of both subsequences = 2 XOR 1 = 3.

This is the maximum XOR value achievable from any two subsequences.

Example 2:

Input: nums = [5,2]

Output: 7

Explanation:

Choose subsequences:

• First subsequence [5], whose XOR is 5.
• Second subsequence [2], whose XOR is 2.

Then, XOR of both subsequences = 5 XOR 2 = 7.

This is the maximum XOR value achievable from any two subsequences.

Constraints:

• 2 <= nums.length <= 105
• 0 <= nums[i] <= 109

Solution

import kotlin.math.max
import kotlin.math.min

class Solution {
    fun maxXorSubsequences(nums: IntArray): Int {
        val n = nums.size
        if (n == 0) {
            return 0
        }
        var x = 0
        while (true) {
            var y = 0
            for (v in nums) {
                if (v > y) {
                    y = v
                }
            }
            if (y == 0) {
                return x
            }
            x = max(x, x xor y)
            for (i in 0..<n) {
                val v = nums[i]
                nums[i] = min(v, v xor y)
            }
        }
    }
}

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