Lonely Integer (C# Solution)

✅ C# Solution using XOR (Efficient and Clean)

`csharp using System; using System.Collections.Generic; using System.Linq;

class Result { // XOR-based solution: all duplicates cancel out public static int lonelyinteger(List a) { int result = 0; foreach (int num in a) { result ^= num; } return result; } }

class Solution { public static void Main(string[] args) { int n = Convert.ToInt32(Console.ReadLine().Trim());

    List<int> a = Console.ReadLine()
                         .Trim()
                         .Split(' ')
                         .Select(int.Parse)
                         .ToList();

    int result = Result.lonelyinteger(a);
    Console.WriteLine(result);
}

}

🔍 Explanation

We use the bitwise XOR (^) operator because:

• x ^ x = 0 (duplicate elements cancel each other)
• x ^ 0 = x

Since every element except one appears twice, XOR-ing all elements leaves the unique one.

⏱️ Time and Space Complexity Time Complexity: O(n) – Single loop through the list.

Space Complexity: O(1) – No additional data structures needed.

def lonelyinteger(a): unique = 0 for num in a: unique ^= num return unique

My Java 8 Solution:

public static int lonelyinteger(List<Integer> a) {
        Set<Integer> set = new HashSet<>();
        for (int number : a) {
            boolean isAdded = set.add(number);
            if (!isAdded) {
                set.remove(number);
            }
        }
        
        return set.iterator().next();
    }

c# linq

return a.Select(x=>x).Where(y=>a.Where(x=>x == y).Count() == 1).First();