How to return a unique element from a C# array

That's not what Distinct does.

It'll return distinct elements from a sequence, thus removing the duplicates.

You only want the items that have no duplicates to begin with;

var numbers = new int[] { 11, 11, 11, 14, 11 };

foreach (var i in numbers.GroupBy(i => i).Where(g => g.Count() == 1).Select(g => g.Key))
    Console.WriteLine(i);

You obviously want the first non-duplicate value, so you'd get something like this;

var numbers = new int[] { 11, 11, 11, 14, 11 };
    
int? firstNonDuplicate = numbers.GroupBy(i => i).Where(g => g.Count() == 1).Select(g => g.Key).FirstOrDefault();
    
Console.WriteLine(firstNonDuplicate);

Be vary of your null checks here though, this just proves a point.

using distinct it will return all distinct element. Get a unique value in following way

int[] numbers = { 11, 11, 11, 14, 11 };
            Console.WriteLine(numbers.GroupBy(i => i).Where(g => g.Count() == 1).Select(g => g.Key).First());

Generally speaking you should group your collection by a key and then filter that by the number of containing elements:

var groupings=  numbers.GroupBy(x => x)
                       .Where(x => x.Count() == 1)
                       .Select(x => x.Key);

Now that gives you all unique numbers. Your question is a bit vague, because you want "a" unique number - not "the" unique number, so there is room for interpretation, what should happen, if there are multiple numbers.

Option 1: Just take the first result:

var uniques = numbers.GroupBy(x => x)
                       .Where(x => x.Count() == 1)
                       .Select(x => x.Key)
                       .First();

Option 1.1 Order results and take the smallest (or largest)

var uniques = numbers.GroupBy(x => x)
                       .Where(x => x.Count() == 1)
                       .Select(x => x.Key)
                       .OrderBy(x => x)
                       .First();

Option 2: Ensure there is only one unique number and throw otherwise:

var uniques = numbers.GroupBy(x => x)
                       .Where(x => x.Count() == 1)
                       .Select(x => x.Key)
                       .Single();

Be aware: If there are no unique numbers, Single() and First() will throw, where as SingleOrDefault() and FirstOrDefault() will return the default value of int, which is 0, which can lead to false results. You can consider changing it to int? to allow for null to be returned, if there are no unique numbers.

You can try below code.

    var numbers = new int[] { 11, 11, 11, 14, 11, 11 };
    var uniqueList = numbers.GroupBy(n => n).Where(item => item.Count() == 1).Select(item => item.Key);

    foreach (var item in uniqueList)
        Console.WriteLine(item);

I made a special method for you. I approached the subject somewhat primitively. Thanks to this method, you can easily find unique variables.

public static int[] getUniqiue(int[] vs)
        {
            List<int> vs1 = new List<int>(vs);
            List<int> vs2 = new List<int>(vs);
            List<int> ee = new List<int>();
            List<int> vs3 = new List<int>();
            int i = 0;
            foreach (var item in vs1)
            {
                vs2.Remove(item);
                if(vs3.Contains(item) || vs2.Contains(item))
                {
                    vs3.Add(item);
                }
                else
                {
                    ee.Add(item);
                }
                i++;
            }
            return ee.ToArray();
        }