LeetCode in Kotlin

1829. Maximum XOR for Each Query

Medium

You are given a sorted array nums of n non-negative integers and an integer maximumBit. You want to perform the following query n times:

1. Find a non-negative integer k < 2maximumBit such that nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k is maximized. k is the answer to the ith query.
2. Remove the last element from the current array nums.

Return an array answer, where answer[i] is the answer to the ith query.

Example 1:

Input: nums = [0,1,1,3], maximumBit = 2

Output: [0,3,2,3]

Explanation: The queries are answered as follows:

1^st query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.

2^nd query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.

3^rd query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3.

4^th query: nums = [0], k = 3 since 0 XOR 3 = 3.

Example 2:

Input: nums = [2,3,4,7], maximumBit = 3

Output: [5,2,6,5]

Explanation: The queries are answered as follows:

1^st query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7.

2^nd query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7.

3^rd query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7.

4^th query: nums = [2], k = 5 since 2 XOR 5 = 7.

Example 3:

Input: nums = [0,1,2,2,5,7], maximumBit = 3

Output: [4,3,6,4,6,7]

Constraints:

• nums.length == n
• 1 <= n <= 105
• 1 <= maximumBit <= 20
• 0 <= nums[i] < 2maximumBit
• nums is sorted in ascending order.

Solution

class Solution {
    fun getMaximumXor(nums: IntArray, maximumBit: Int): IntArray {
        val result = IntArray(nums.size)
        var `val` = nums[0]
        val target = (1 shl maximumBit) - 1
        for (i in 1 until nums.size) {
            `val` = `val` xor nums[i]
        }
        for (i in result.indices) {
            result[i] = target xor `val`
            `val` = `val` xor nums[nums.size - i - 1]
        }
        return result
    }
}

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