Скачать или смотреть leetcode 268 - Find the Missing Number Using XOR | Optimal Approach in JAVA.

✅ Problem Name:
Find the Missing Number Using XOR

📝 Problem Description:
You are given an array nums containing n distinct numbers in the range [0, n].
That means:

The array should have n + 1 numbers total.

But one number is missing from it.

👉 Your task is to find and return the missing number.

📥 Input:
An integer array nums of length n

All elements are unique and range from 0 to n

📤 Output:
An integer representing the missing number in the range [0, n]

💡 Approach Used:
Bit Manipulation using XOR

🧠 Key Logic:
XOR of a number with itself is 0: a ^ a = 0

XOR of any number with 0 is the number itself: a ^ 0 = a

So, if you XOR all numbers from 0 to n, and also XOR all elements in nums,
then all matching numbers will cancel out, and you’ll be left with the missing number.

⛓️ Steps:
Initialize two variables: xor1 = 0, xor2 = 0

Traverse the array:

XOR xor1 with numbers from 1 to n

XOR xor2 with elements in the array

Return xor1 ^ xor2 → This gives the missing number

⏱️ Time Complexity: O(n)
🗂️ Space Complexity: O(1)