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

Given an array of N elements and an integer K, print all the distinct integers which can be formed by choosing K numbers from the given N numbers. A number from an array can be chosen any number of times.

Examples:

Input: k = 2, a[] = {3, 8, 17, 5}
Output: The 10 distinct integers are:
6 8 10 11 13 16 20 22 25 34
The 2 elements chosen are:
3+3 = 6, 5+3 = 8, 5+5 = 10, 8+3 = 11, 8+5 = 13
8+8 = 16, 17+3 = 20, 17+5 = 22, 17+8 = 25, 17+17 = 34.

Input: k = 3, a[] = {3, 8, 17}
Output: The 10 distinct integers are:
9 14 19 23 24 28 33 37 42 51

Approach: The problem will be solved using recursion. All combinations are to be tried, when the count of elements selected is equal to k, then we keep the num formed in the set so that the repetitive elements are not counted twice. The function generateNumber(int count, int a[], int n, int num, int k) is a recursive function, in which the base case is when the count becomes K which signifies that K elements from the array have been chosen. num in the parameter signifies the number formed by count number of numbers. In the function, iterate over the array and for every element, call the recursive function with count as count+1 and num as num+a[i].

Below is the implementation of the above approach:

## C++

 `// C++ program to print all distinct``// integers that can be formed by K numbers``// from a given array of N numbers.``#include ``using` `namespace` `std;` `// stores all the distinct integers formed``set<``int``> s;` `// Function to generate all possible numbers``void` `generateNumber(``int` `count, ``int` `a[], ``int` `n,``                    ``int` `num, ``int` `k)``{` `    ``// Base case when K elements``    ``// are chosen``    ``if` `(k == count) {``        ``// insert it in set``        ``s.insert(num);``        ``return``;``    ``}` `    ``// Choose every element and call the function``    ``for` `(``int` `i = 0; i < n; i++) {``        ``generateNumber(count + 1, a, n, num + a[i], k);``    ``}``}``// Function to print the distinct integers``void` `printDistinctIntegers(``int` `k, ``int` `a[], ``int` `n)``{``    ``generateNumber(0, a, n, 0, k);``    ``cout << ``"The "` `<< s.size()``         ``<< ``" distinct integers are:\n"``;` `    ``// prints all the elements in the set``    ``while` `(!s.empty()) {``        ``cout << *s.begin() << ``" "``;` `        ``// erase the number after printing it``        ``s.erase(*s.begin());``    ``}``}``// Driver Code``int` `main()``{``    ``int` `a[] = { 3, 8, 17, 5 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);``    ``int` `k = 2;` `    ``// Calling Function``    ``printDistinctIntegers(k, a, n);``    ``return` `0;``}`

## Java

 `// Java program to print all``// distinct integers that can``// be formed by K numbers from``// a given array of N numbers.``import` `java.util.*;``import` `java.lang.*;` `class` `GFG``{``    ``// stores all the distinct``    ``// integers formed``    ``static` `TreeSet set =``                   ``new` `TreeSet();``    ` `    ``// Function to generate``    ``// all possible numbers``    ``public` `static` `void` `generateNumber(``int` `count, ``int` `a[],``                                      ``int` `n, ``int` `num, ``int` `k)``    ``{``        ``// Base case when K``        ``// elements are chosen``        ``if``(count == k)``        ``{``            ``set.add(num);``            ``return``;``        ``}``        ` `        ``// Choose every element``        ``// and call the function``        ``for``(``int` `i = ``0``; i < n; i++)``        ``generateNumber(count + ``1``, a, n,``                       ``num + a[i], k);``    ``}``    ` `    ``// Function to print``    ``// the distinct integers``    ``public` `static` `void` `printDistinctIntegers(``int` `k,``                                             ``int` `a[], ``int` `n)``    ``{``        ``generateNumber(``0``, a, n, ``0``, k);``        ``System.out.print(``"The"` `+ ``" "` `+ set.size() +``                         ``" "` `+ ``"distinct integers are: "``);``        ``System.out.println();``        ``Iterator i = set.iterator();``        ` `        ``// prints all the``        ``// elements in the set``        ``while``(set.isEmpty() == ``false``)``        ``{``            ` `            ``while``(i.hasNext())``            ``{``                ``System.out.print(i.next() + ``" "``);``                ``//set.remove(i.next());``            ``}  ``        ``}``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `arr[] = {``3``, ``8``, ``17``, ``5``};``        ``int` `n = arr.length;``        ``int` `k = ``2``;``        ` `        ``// Calling Function``        ``printDistinctIntegers(k, arr, n);``    ``}``}`

## Python3

 `# Python3 program to print all distinct``# integers that can be formed by K numbers``# from a given array of N numbers.` `# stores all the distinct integers formed``s ``=` `set``()` `# Function to generate all possible numbers``def` `generateNumber(count, a, n, num, k):` `    ``# Base case when K elements are chosen``    ``if` `k ``=``=` `count:``        ` `        ``# insert it in set``        ``s.add(num)``        ``return``    ` `    ``# Choose every element and call the function``    ``for` `i ``in` `range``(``0``, n):``        ``generateNumber(count ``+` `1``, a, n,    ``                         ``num ``+` `a[i], k)` `# Function to print the distinct integers``def` `printDistinctIntegers(k, a, n):` `    ``generateNumber(``0``, a, n, ``0``, k)``    ``print``(``"The"``, ``len``(s),``          ``"distinct integers are:"``)` `    ``# prints all the elements in the set``    ``for` `i ``in` `sorted``(s):``        ``print``(i, end ``=` `" "``)``    ` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``a ``=` `[``3``, ``8``, ``17``, ``5``]``    ``n, k ``=` `len``(a), ``2` `    ``# Calling Function``    ``printDistinctIntegers(k, a, n)``    ` `# This code is contributed by Rituraj Jain`

## C#

 `// C# program to print all``// distinct integers that can``// be formed by K numbers from``// a given array of N numbers.``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ``// stores all the distinct``    ``// integers formed``    ``static` `SortedSet<``int``> ``set` `=``                ``new` `SortedSet<``int``>();``    ` `    ``// Function to generate``    ``// all possible numbers``    ``public` `static` `void` `generateNumber(``int` `count, ``int` `[]a,``                                    ``int` `n, ``int` `num, ``int` `k)``    ``{``        ``// Base case when K``        ``// elements are chosen``        ``if``(count == k)``        ``{``            ``set``.Add(num);``            ``return``;``        ``}``        ` `        ``// Choose every element``        ``// and call the function``        ``for``(``int` `i = 0; i < n; i++)``        ``generateNumber(count + 1, a, n,``                    ``num + a[i], k);``    ``}``    ` `    ``// Function to print``    ``// the distinct integers``    ``public` `static` `void` `printDistinctIntegers(``int` `k,``                                            ``int` `[]a, ``int` `n)``    ``{``        ``generateNumber(0, a, n, 0, k);``        ``Console.Write(``"The"` `+ ``" "` `+ ``set``.Count +``                        ``" "` `+ ``"distinct integers are: "``);``        ``Console.WriteLine();` `        ` `        ``// prints all the``        ``// elements in the set``        ``foreach``(``int` `sets ``in` `set``)``        ``{``                ``Console.Write(sets + ``" "``);` `        ``}``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main (String[] args)``    ``{``        ``int` `[]arr = {3, 8, 17, 5};``        ``int` `n = arr.Length;``        ``int` `k = 2;``        ` `        ``// Calling Function``        ``printDistinctIntegers(k, arr, n);``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Javascript

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Output

```The 10 distinct integers are:
6 8 10 11 13 16 20 22 25 34 ```

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