Being Code Expert: Sum of sum of digits

Monk likes math problems very much. His math teacher taught him about the sum of digits of a number. He decided to experiment a little with them.

Given a number, he recursively finds the sum of its digits till it becomes a single-digit number. He calls this as Digit-Value of a number "

It can be written as :

sumOfDigits(n):
    if n is a single digit number:
        return n
    else:
        x = sum of the digits of n
        return sumOfDigits(x)

After seeing his interest in this concept, his teacher gave him an interesting problem, that uses the above function defined by him. She gave him an array A of N different numbers. Then she asks him Q queries. In each query, he has to form a set of K numbers from the array and find the sum of Digit-Values of those K numbers. This sum is called the value of that set.

The queries are of the following type :
1 K : Monk must output the maximum value of a set of size K, that can be obtained, as described above.

2 K : Monk must output the minimum value of a set of size K, that can be obtained, as described above.

Monk needs your help to complete this task.

Example:

Input:  n = 5, arr ={13, 345, 193, 44444, 100303}, x =1, y = 4
Output: 18

Approach

C++

#include <bits/stdc++.h>
using namespace std;
long long singleDigit(long long n)
{
    long long res = 0;
    while (n >= 10)
    {
        res = 0;
        while (n)
        {
            res += n % 10;
            n = n / 10;
        }
        n = res;
    }
    return n;
}
int main()
{

    long long n = 5;

    long long arr[n] = {13, 345, 193, 44444, 100303};

    for (long long i = 0; i < n; i++)
        arr[i] = singleDigit(arr[i]);
    sort(arr, arr + n);
    long long sum1[n], sum2[n];
    sum1[n - 1] = arr[n - 1];
    for (long long i = n - 2; i >= 0; i--)
        sum1[i] = sum1[i + 1] + arr[i];
    sum2[0] = arr[0];
    for (long long i = 1; i < n; i++)
        sum2[i] = arr[i] + sum2[i - 1];

    long long x = 1, y = 4;

    if (x == 1)
        cout << sum1[n - y] << "\n";
    else
        cout << sum2[y - 1] << "\n";
}