[LeetCode] 268. Missing Number

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.
For example,
Given nums = [0, 1, 3] return 2.
Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

Thought process:
Use the XOR operation. a ^ a = 0. Since the numbers run from 0 to n, nums[i] ^ i = 0. If there's no number missing, nums[0] ^ 0 ^ nums[1] ^ 1 ^ ... ^ nums[n] ^ n = 0. Now that there's one number missing, the result of the above XOR calculation will give me the missing number.

Solution:
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class Solution {
    public int missingNumber(int[] nums) {
        int i = 0;
        int xor = 0;
        
        for (; i < nums.length; i++) {
            xor ^= i ^ nums[i];
        }
        return xor ^ i;
    }
}

Time complexity: O(n).
Space complexity: O(1).
Location: Philadelphia, PA, USA

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