Write an algorithm to find a pair of 2 elements to sum defined (user input) Eg Input array [1,2,3,6,8,1,6] and Sum = 5 so Expected output = [2,3] Explanation: only 2 and 3 forms a pair to sum = 5 in the given array. The answer should be generic for any array and sum input

   static  bool findPairFromArray(int[] v,int suma,int n) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n && j != i; j++) {
                if (v[i] + v[j] == suma) {
                    Console.WriteLine("The pair is :[{0},{1}]", v[i], v[j]);
                    return true;
                }
            }
        }
        return false;
    }
    static void Main(string[] args)
    {
        int[] v ={ 1, 2, 4, 7, 8,6,3 };
        //findPairFromArray(v, 8, 7);
        if (findPairFromArray(v, 10, 7))
        {
            Console.WriteLine("I find a pair!");
        }
        Console.ReadKey();
    }
}

public static void addaraay()
{
int[] addnum = { 2, 1, 6, 4, 8, 3, 2, 1 };
string sum = Console.ReadLine();
int totalsum = Convert.ToInt32(sum);
int[] addnum1 = addnum;
for (int i = 0; i <= addnum.Length - 1; i++)
{
int j = i + 1;
for (; j < addnum1.Length - 1; j++)
{
 int howmuch = addnum[i] + addnum1[j];
 if (howmuch == totalsum)
 {
 Console.WriteLine("pair is " + addnum[i] +", " +addnum1[j]);
 }
 }
 }

 public static int[] ArraySum(int[] arr,int sum)
        {
            int len = arr.Length;
            int[] sumArr = null;
            for (int i = 0; i <; len; i++)
            {
                for (int j = 0; j <; len; j++)
                {
                    if (i != j)
                    {
                        if (arr[i] + arr[j] == sum)
                        {
                            sumArr = new int[2];
                            sumArr[0] = arr[i];
                            sumArr[1] = arr[j];
                            break;
                        }
                    }
                }
                if (sumArr != null) break;               
            }
            return sumArr;
        }