LeetCode： XOR Operation in an Array

Question

Given an integer n and an integer start.

Define an array nums where nums[i] = start + 2*i (0-indexed) and n == nums.length.

Return the bitwise XOR of all elements of nums.

Example 1:

Input: n = 5, start = 0

Output: 8

Explanation: Array nums is equal to [0, 2, 4, 6, 8] where (0 ^ 2 ^ 4 ^ 6 ^ 8) = 8.

Where "^" corresponds to bitwise XOR operator.

Example 2:

Input: n = 4, start = 3

Output: 8

Explanation: Array nums is equal to [3, 5, 7, 9] where (3 ^ 5 ^ 7 ^ 9) = 8.

Solution #1

def xor_operation(n, start)
    (0..(n-1)).reduce(0) { |r, i| r ^ (start + 2 * i) }
end

Solution #2

def xor_operation(n, start)
  i = 0
  nums = []

  Array.new(n).each_with_index { |_num, i| nums << start + (2 * i) }

  loop do
    start = start ^ nums[i + 1] if nums[i + 1]
    i += 1
    break if i > n
  end
  start
end

Solution #3

def xor_operation(n, start)
  i = 0
  j = 0
  ans = start
  nums = []
  while i < n
    nums[i] = start + (2 * i)
    i += 1
  end

  while j < nums.length-1
    ans = ans ^ nums[j + 1] if nums[j + 1]
    j += 1
  end
  ans
end