1486. XOR Operation in an Array.md

1486. XOR Operation in an Array

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the first one in Weekly Contest 194.

leetcode-cn Daily Challenge on May 7th, 2021.

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Difficulty : Easy

Related Topics : Array、Bit Manipulation

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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

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Solution

• mine

  • Java

    Runtime: 0 ms, faster than 100.00%, Memory Usage: 36.2 MB, less than 100.00% of Java online submissions

    // O(N)time O(1)space
    public int xorOperation(int n, int start) {
        int res = start;
        for(int i = 1; i < n; i++){
            res ^= start + 2 * i; 
        }
        return res;
    }
    

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