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| 1 | +# frozen_string_literal: true |
| 2 | + |
| 3 | +# 1829. Maximum XOR for Each Query |
| 4 | +# Medium |
| 5 | +# https://leetcode.com/problems/maximum-xor-for-each-query |
| 6 | + |
| 7 | +=begin |
| 8 | +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: |
| 9 | +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. |
| 10 | +2. Remove the last element from the current array nums. |
| 11 | +Return an array answer, where answer[i] is the answer to the ith query. |
| 12 | +
|
| 13 | +Example 1: |
| 14 | +Input: nums = [0,1,1,3], maximumBit = 2 |
| 15 | +Output: [0,3,2,3] |
| 16 | +Explanation: The queries are answered as follows: |
| 17 | +1st query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3. |
| 18 | +2nd query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3. |
| 19 | +3rd query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3. |
| 20 | +4th query: nums = [0], k = 3 since 0 XOR 3 = 3. |
| 21 | +Example 2: |
| 22 | +Input: nums = [2,3,4,7], maximumBit = 3 |
| 23 | +Output: [5,2,6,5] |
| 24 | +Explanation: The queries are answered as follows: |
| 25 | +1st query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7. |
| 26 | +2nd query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7. |
| 27 | +3rd query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7. |
| 28 | +4th query: nums = [2], k = 5 since 2 XOR 5 = 7. |
| 29 | +Example 3: |
| 30 | +Input: nums = [0,1,2,2,5,7], maximumBit = 3 |
| 31 | +Output: [4,3,6,4,6,7] |
| 32 | +
|
| 33 | +Constraints: |
| 34 | +* nums.length == n |
| 35 | +* 1 <= n <= 105 |
| 36 | +* 1 <= maximumBit <= 20 |
| 37 | +* 0 <= nums[i] < 2maximumBit |
| 38 | +* nums is sorted in ascending order. |
| 39 | +=end |
| 40 | + |
| 41 | +# @param {Integer[]} nums |
| 42 | +# @param {Integer} maximum_bit |
| 43 | +# @return {Integer[]} |
| 44 | +def get_maximum_xor(nums, maximum_bit) |
| 45 | + mask = (1 << maximum_bit) - 1 |
| 46 | + curr = 0 |
| 47 | + nums.map do |num| |
| 48 | + curr ^= num |
| 49 | + curr ^ mask |
| 50 | + end.reverse |
| 51 | +end |
| 52 | + |
| 53 | +# **************** # |
| 54 | +# TEST # |
| 55 | +# **************** # |
| 56 | + |
| 57 | +require "test/unit" |
| 58 | +class Test_get_maximum_xor < Test::Unit::TestCase |
| 59 | + def test_ |
| 60 | + assert_equal [0, 3, 2, 3], get_maximum_xor([0, 1, 1, 3], 2) |
| 61 | + assert_equal [5, 2, 6, 5], get_maximum_xor([2, 3, 4, 7], 3) |
| 62 | + assert_equal [4, 3, 6, 4, 6, 7], get_maximum_xor([0, 1, 2, 2, 5, 7], 3) |
| 63 | + end |
| 64 | +end |
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