Algorithm to find two repeated numbers in an array, without sorting

Question

There is an array of size n (numbers are between 0 and n - 3) and only 2 numbers are repeated. Elements are placed randomly in the array.

E.g. in {2, 3, 6, 1, 5, 4

Answer 1

You can use simple nested for loop

 int[] numArray = new int[] { 1, 2, 3, 4, 5, 7, 8, 3, 7 };

        for (int i = 0; i < numArray.Length; i++)
        {
            for (int j = i + 1; j < numArray.Length; j++)
            {
                if (numArray[i] == numArray[j])
                {
                   //DO SOMETHING
                }
            }

*OR you can filter the array and use recursive function if you want to get the count of occurrences*

int[] array = { 1, 2, 3, 4, 5, 4, 4, 1, 8, 9, 23, 4, 6, 8, 9, 1,4 };
int[] myNewArray = null;
int a = 1;

 void GetDuplicates(int[] array)
    for (int i = 0; i < array.Length; i++)
            {
                for (int j = i + 1; j < array.Length; j++)
                {
                    if (array[i] == array[j])
                    {
                          a += 1;
                    }
                }
                Console.WriteLine(" {0} occurred {1} time/s", array[i], a);

                IEnumerable num = from n in array where n != array[i] select n;
                 myNewArray = null;
                 a = 1;
                 myNewArray = num.ToArray() ;

                 break;

            }
             GetDuplicates(myNewArray);