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# Print all distinct integers that can be formed by K numbers from a given array of N numbers in C++

In this problem, we are given an array of N integers and a number K. Our task is to print all distinct numbers that can be created by adding any K elements from the array. While choosing any number can be repeated K times.

Let’s take an example to understand the problem −

Input: array = {2, 5, 13, 9} K = 2 Output: 2,7,15,11,10,18,14,26,22 Explaination: 2 elements added : 2+2=4, 2+5=7, 2+13=15, 2+9=11, 5+5=10, 5+13=18, 5+9=14, 13+13=26, 13+9=22, 9+9=18

To solve this problem, we will find all combinations of the k element from the array. For this, we will use recursion which will be called recursively to generate numbers. To avoid duplicate values, we will store the numbers in a set.

## Example

The code will show the implementation of our solution −

#include <bits/stdc++.h> using namespace std; set<int> distNumbers; void generateNumberFromArray(int count, int arr[], int n, int num, int k) { if (k == count) { distNumbers.insert(num); return; } for (int i = 0; i < n; i++) { generateNumberFromArray(count + 1, arr, n, num + arr[i], k); } } void printDistinctIntegers(int k, int arr[], int n) { generateNumberFromArray(0, arr, n, 0, k); cout<<"The "<<distNumbers.size()<<" distinct integers are:\n"; while (!distNumbers.empty()) { cout << *distNumbers.begin() <<"\t"; distNumbers.erase(*distNumbers.begin()); } } int main() { int arr[]={ 2, 5, 13, 9 }; int n=4; int k=2; printDistinctIntegers(k, arr, n); return 0; }

## Output

The 9 distinct integers are − 4 7 10 11 14 15 18 22 26

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