1829-maximum-xor-for-each-query.rb

# frozen_string_literal: true

# 1829. Maximum XOR for Each Query
# Medium
# https://leetcode.com/problems/maximum-xor-for-each-query

=begin
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:
1st query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.
2nd query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.
3rd query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3.
4th 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:
1st query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7.
2nd query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7.
3rd query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7.
4th 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.
=end

# @param {Integer[]} nums
# @param {Integer} maximum_bit
# @return {Integer[]}
def get_maximum_xor(nums, maximum_bit)
  mask = (1 << maximum_bit) - 1
  curr = 0
  nums.map do |num|
    curr ^= num
    curr ^ mask
  end.reverse
end

# **************** #
#       TEST       #
# **************** #

require "test/unit"
class Test_get_maximum_xor < Test::Unit::TestCase
  def test_
    assert_equal [0, 3, 2, 3], get_maximum_xor([0, 1, 1, 3], 2)
    assert_equal [5, 2, 6, 5], get_maximum_xor([2, 3, 4, 7], 3)
    assert_equal [4, 3, 6, 4, 6, 7], get_maximum_xor([0, 1, 2, 2, 5, 7], 3)
  end
end