leetcode 268 - Find the Missing Number Using XOR | Optimal Approach in JAVA. скачать в хорошем качестве

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)