# Element equal to the sum of all the remaining elements

Given an array of N positive elements. The task is to find an element which is equal to the sum of all elements of array except itself.

Examples:

```Input: arr[] = {1, 2, 3, 6}
Output: 6
6 is the element which is equal to the sum of all
remaining elements i.e. 1 + 2+ 3 = 6

Input: arr[] = {2, 2, 2, 2}
Output: -1
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: First of all, find the sum of all elements of an array. Then iterate over each element and check the condition that if (a[i] == sum-a[i] ). If true then print that a[i], else print “-1” if no such element found.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `#define ll long long int ` `using` `namespace` `std; ` ` `  `// Function to find the element ` `int` `findEle(``int` `arr[], ``int` `n) ` `{ ` `    ``// sum is use to store ` `    ``// sum of all elements ` `    ``// of array ` `    ``ll sum = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``sum += arr[i]; ` ` `  `    ``// iterate over all elements ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``if` `(arr[i] == sum - arr[i]) ` `            ``return` `arr[i]; ` ` `  `    ``return` `-1; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3, 6 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``cout << findEle(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach ` ` `  `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  ` `  `// Function to find the element ` `static` `int` `findEle(``int` `arr[], ``int` `n) ` `{ ` `    ``// sum is use to store ` `    ``// sum of all elements ` `    ``// of array ` `    ``int` `sum = ``0``; ` ` `  `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``sum += arr[i]; ` ` `  `    ``// iterate over all elements ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``if` `(arr[i] == sum - arr[i]) ` `            ``return` `arr[i]; ` ` `  `    ``return` `-``1``; ` `} ` ` `  `// Driver code ` ` `  `    ``public` `static` `void` `main (String[] args) { ` `        ``int` `arr[] = { ``1``, ``2``, ``3``, ``6` `}; ` `    ``int` `n = arr.length; ` `    ``System.out.print(findEle(arr, n)); ` `    ``} ` `} ` `// This code is contributed by shs.. `

## Python3

 `# Python 3 implementation of  ` `# the above approach ` ` `  `# Function to find the element ` `def` `findEle(arr, n) : ` `     `  `    ``# sum is use to store  ` `    ``# sum of all elements  ` `    ``# of array  ` `    ``sum` `=` `0` `     `  `    ``for` `i ``in` `range``(n) : ` `        ``sum` `+``=` `arr[i] ` `     `  `    ``# iterate over all elements ` `    ``for` `i ``in` `range``(n) : ` `        ``if` `arr[i] ``=``=` `sum` `-` `arr[i] : ` `            ``return` `arr[i] ` `     `  `    ``return` `-``1` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"` `: ` `     `  `    ``arr ``=` `[``1``, ``2``, ``3``, ``6` `] ` `    ``n ``=` `len``(arr) ` `    ``print``(findEle(arr, n)) ` ` `  `# This code is contributed by Ryuga `

## C#

 `// C# implementation of the  ` `// above approach  ` `using` `System;  ` ` `  `class` `GFG  ` `{  ` ` `  `// Function to find the element  ` `static` `int` `findEle(``int` `[]arr, ``int` `n)  ` `{  ` `    ``// sum is use to store  ` `    ``// sum of all elements  ` `    ``// of array  ` `    ``int` `sum = 0;  ` ` `  `    ``for` `(``int` `i = 0; i < n; i++)  ` `        ``sum += arr[i];  ` ` `  `    ``// iterate over all elements  ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `        ``if` `(arr[i] == sum - arr[i])  ` `            ``return` `arr[i];  ` ` `  `    ``return` `-1;  ` `}  ` ` `  `// Driver code  ` `static` `public` `void` `Main (String []args) ` `{  ` `    ``int` `[]arr = { 1, 2, 3, 6 };  ` `    ``int` `n = arr.Length;  ` `    ``Console.WriteLine(findEle(arr, n));  ` `}  ` `}  ` ` `  `// This code is contributed  ` `// by Arnab Kundu `

## PHP

 ` `

Output:

```6
```

Note: Above problem can be solved with the concept used in Check if the array has an element which is equal to sum of all the remaining elements.

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