1486. XOR Operation in an Array.md

1486. XOR Operation in an Array

1. Description

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.

Example 3:

Input: n = 1, start = 7
Output: 7

Example 4:

Input: n = 10, start = 5
Output: 2

Constraints:

• 1 <= n <= 1000
• 0 <= start <= 1000
• n == nums.length

2. Solutions

Solution 1: Language: C Boring Solution for Loop Using

The XOR part might be helpful and valuable in mathematics and computer science, but not this question. It is a loop-using question, no connection with the XOR.

• Sunday, 25 October, 2020
• Time Complexity: $O(n)$
• Space Complexity: $O(1)$
• Runtime: 0 ms, faster than 100.00% of C online submissions for XOR Operation in an Array.
• Memory Usage: 5.6 MB, less than 35.15% of C online submissions for XOR Operation in an Array.

int xorOperation(int n, int start) {
    int xor = 0;
    for (int i = 0; i < n; i++) {
        xor ^= (start + 2 * i);
    }
    return xor;
}

3. TODO and Bonus

• Solve this question by a deeper way, not only for the loop using.