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# Print elements that can be added to form a given sum

• Difficulty Level : Medium
• Last Updated : 23 Dec, 2022

Given an array arr[] of positive integers and a sum, the task is to print the elements that will be included to get the given sum.

Note:

1. Consider the elements in the form of queue i.e. Elements to be added from starting and up to the sum of elements is lesser or becomes equal to the given sum.
2. Also, it is not necessary that the sum of array elements should be equal to the given sum.

As the task is to check that element can be included or not.

Examples:

Input: arr[] = {3, 5, 3, 2, 1}, Sum = 10
Output: 3 5 2
By adding 3, 5 and 3, sum becomes 11 so remove last 3.
Then on adding 2, sum becomes 10. So no other element

Input: arr[] = {7, 10, 6, 4}, Sum = 12
Output: 7 4
As, 7+10 and 7+6 sums to a higher value than 12
but 7+4 = 11 which is smaller than 12.
So, 7 and 4 can be included

Approach:

1. Check if on adding the current element, the sum is less than the given sum.
2. If yes, the add it.
3. Else go to next element and repeat the same until their sum is less than or equals to the given sum.

Below is the implementation of above approach:

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `// Function that finds whether an element``// will be included or not``void` `includeElement(``int` `a[], ``int` `n, ``int` `sum)``{``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Check if the current element``        ``// will be incuded or not``        ``if` `((sum - a[i]) >= 0) {``            ``sum = sum - a[i];``            ``cout << a[i]<< ``" "``;``        ``}``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 3, 5, 3, 2, 1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `sum = 10;` `    ``includeElement(arr, n, sum);``    ``return` `0;``}`

## Java

 `// Java implementation``// of above approach``class` `GFG``{` `// Function that finds whether``// an element will be included``// or not``static` `void` `includeElement(``int` `a[],``                           ``int` `n, ``int` `sum)``{``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{` `        ``// Check if the current element``        ``// will be included or not``        ``if` `((sum - a[i]) >= ``0``)``        ``{``            ``sum = sum - a[i];``            ``System.out.print(a[i] + ``" "``);``        ``}``    ``}``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``3``, ``5``, ``3``, ``2``, ``1` `};``    ``int` `n = arr.length;``    ``int` `sum = ``10``;` `    ``includeElement(arr, n, sum);``}``}` `// This code is contributed by Bilal`

## Python3

 `# Python 3 implementation of above approach` `# Function that finds whether an element``# will be included or not``def` `includeElement(a, n, ``sum``) :` `    ``for` `i ``in` `range``(n) :` `        ``# Check if the current element``        ``# will be incuded or not``        ``if` `sum` `-` `a[i] >``=` `0` `:` `            ``sum` `=` `sum` `-` `a[i]` `            ``print``(a[i],end ``=` `" "``)` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr ``=` `[ ``3``, ``5``, ``3``, ``2``, ``1``]``    ``n ``=` `len``(arr)``    ``sum` `=` `10` `    ``includeElement(arr, n, ``sum``)``                       ` `# This code is contributed by ANKITRAI1`

## C#

 `// C# implementation``// of above approach``using` `System;` `class` `GFG``{``// Function that finds whether``// an element will be included``// or not``static` `void` `includeElement(``int``[] a,``                           ``int` `n, ``int` `sum)``{``    ``for` `(``int` `i = 0; i < n; i++)``    ``{` `        ``// Check if the current element``        ``// will be included or not``        ``if` `((sum - a[i]) >= 0)``        ``{``            ``sum = sum - a[i];``            ``Console.Write(a[i] + ``" "``);``        ``}``    ``}``}` `// Driver code``static` `void` `Main()``{``    ``int``[] arr = ``new` `int``[]{ 3, 5, 3, 2, 1 };``    ``int` `n = arr.Length;``    ``int` `sum = 10;` `    ``includeElement(arr, n, sum);``}``}` `// This code is contributed by mits`

## PHP

 `= 0)``        ``{``            ``\$sum` `= ``\$sum` `- ``\$a``[``\$i``];``            ``echo` `\$a``[``\$i``] . ``" "``;``        ``}``    ``}``}` `// Driver Code``\$arr` `= ``array``( 3, 5, 3, 2, 1 );``\$n` `= sizeof(``\$arr``);``\$sum` `= 10;` `includeElement(``\$arr``, ``\$n``, ``\$sum``);` `// This code is contributed``// by ChitraNayal``?>`

## Javascript

 ``

Output

`3 5 2 `

Time Complexity: O(n), As we are traversing the array only once.
Auxiliary Space: O(1), As constant extra space is used.

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