LeetCode in Kotlin

3513. Number of Unique XOR Triplets I

Medium

You are given an integer array nums of length n, where nums is a permutation of the numbers in the range [1, n].

A XOR triplet is defined as the XOR of three elements nums[i] XOR nums[j] XOR nums[k] where i <= j <= k.

Return the number of unique XOR triplet values from all possible triplets (i, j, k).

A permutation is a rearrangement of all the elements of a set.

Example 1:

Input: nums = [1,2]

Output: 2

Explanation:

The possible XOR triplet values are:

• (0, 0, 0) → 1 XOR 1 XOR 1 = 1
• (0, 0, 1) → 1 XOR 1 XOR 2 = 2
• (0, 1, 1) → 1 XOR 2 XOR 2 = 1
• (1, 1, 1) → 2 XOR 2 XOR 2 = 2

The unique XOR values are {1, 2}, so the output is 2.

Example 2:

Input: nums = [3,1,2]

Output: 4

Explanation:

The possible XOR triplet values include:

• (0, 0, 0) → 3 XOR 3 XOR 3 = 3
• (0, 0, 1) → 3 XOR 3 XOR 1 = 1
• (0, 0, 2) → 3 XOR 3 XOR 2 = 2
• (0, 1, 2) → 3 XOR 1 XOR 2 = 0

The unique XOR values are {0, 1, 2, 3}, so the output is 4.

Constraints:

• 1 <= n == nums.length <= 105
• 1 <= nums[i] <= n
• nums is a permutation of integers from 1 to n.

Solution

class Solution {
    fun uniqueXorTriplets(nums: IntArray): Int {
        val n = nums.size
        return if (n < 3) n else Integer.highestOneBit(n) shl 1
    }
}

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