Bit Manipulation - Missing Number

Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

Input: nums = [3, 0, 1]
Output: 2
Explanation: n = 3, the range is [0,3], and 2 is missing.

Input: nums = [9, 6, 4, 2, 3, 5, 7, 0, 1]
Output: 8
Explanation: n = 9, the range is [0,9], and 8 is missing.

XOR all indices (0 to n) with all values in nums. Numbers that appear in both cancel out (x ^ x = 0), leaving only the missing number (x ^ 0 = x).

Start with result = n (the length of the array).
For each index i, XOR result with both i and nums[i].
Paired values cancel; the missing number is left.

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

Complexity Analysis:

Time complexity: O(n) — single pass through the array.
Space complexity: O(1) — constant extra space.