Print elements that can be added to form a given sum

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 
needs to be added. 

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 <bits/stdc++.h>
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 included 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[0]);
    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 included 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

<?php
// PHP implementation of above approach

// Function that finds whether an
// element will be included or not
function includeElement(&$a, $n, $sum)
{
    for ($i = 0; $i < $n; $i++)
    {
        // Check if the current element
        // will be included or not
        if (($sum - $a[$i]) >= 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

<script>

// Javascript implementation of above approach

// Function that finds whether an element
// will be included or not
function includeElement(a, n, sum)
{
    for(var i = 0; i < n; i++)
    {

        // Check if the current element
        // will be included or not
        if ((sum - a[i]) >= 0)
        {
            sum = sum - a[i];
            document.write( a[i] + " ");
        }
    }
}

// Driver Code
var arr = [ 3, 5, 3, 2, 1 ];
var n = arr.length;
var sum = 10;

includeElement(arr, n, sum);

// This code is contributed by itsok

</script>

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.