# Append two elements to make the array satisfy the given condition

Given an array arr[] of non-negative integers, let’s define X as the XOR of all the array elements and S as the sum of all the array elements. The task is to find two elements such that when they are appended to the array S = 2 * X is satisfied for the updated array.

Examples:

Input: arr[] = {1, 7}
Output: 6 14
Initially S = 8, and X = 6. After appending 6
and 14, S_NEW = (8 + 6 + 14) = 28
and X_NEW = (6 ^ 6 ^ 14) = 14
Clearly, S_NEW = 2 * X_NEW

Input: arr[] = {1, 3}
Output: 2 6

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

Naive approach: Run two nested loops from 1 to S and check for each pair whether it satisfies the condition or not. This will take O(S2) time.

Efficient approach: It can be observed that if X and S + X are appended to the array then S_NEW = 2 * (S + X) and X_NEW = S + X which satisfy the given condition.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the required numbers ` `void` `findNums(``int` `arr[], ``int` `n) ` `{ ` ` `  `    ``// Find the sum and xor ` `    ``int` `S = 0, X = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``S += arr[i]; ` `        ``X ^= arr[i]; ` `    ``} ` ` `  `    ``// Print the required elements ` `    ``cout << X << ``" "` `<< (X + S); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 7 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` ` `  `    ``findNums(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG  ` `{ ` `     `  `    ``// Function to find the required numbers  ` `    ``static` `void` `findNums(``int` `arr[], ``int` `n)  ` `    ``{  ` `     `  `        ``// Find the sum and xor  ` `        ``int` `S = ``0``, X = ``0``;  ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{  ` `            ``S += arr[i];  ` `            ``X ^= arr[i];  ` `        ``}  ` `     `  `        ``// Print the required elements  ` `        ``System.out.println(X + ``" "` `+ (X + S));  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `arr[] = { ``1``, ``7` `};  ` `        ``int` `n = arr.length;  ` `     `  `        ``findNums(arr, n);  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to find the required numbers  ` `def` `findNums(arr, n) :  ` ` `  `    ``# Find the sum and xor  ` `    ``S ``=` `0``; X ``=` `0``;  ` `    ``for` `i ``in` `range``(n) :  ` `        ``S ``+``=` `arr[i];  ` `        ``X ^``=` `arr[i]; ` ` `  `    ``# Print the required elements  ` `    ``print``(X, X ``+` `S);  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``arr ``=` `[ ``1``, ``7` `];  ` `    ``n ``=` `len``(arr);  ` ` `  `    ``findNums(arr, n);  ` `     `  `# This code is contributed by AnkiRai01 `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Function to find the required numbers  ` `    ``static` `void` `findNums(``int` `[]arr, ``int` `n)  ` `    ``{  ` `     `  `        ``// Find the sum and xor  ` `        ``int` `S = 0, X = 0;  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{  ` `            ``S += arr[i];  ` `            ``X ^= arr[i];  ` `        ``}  ` `     `  `        ``// Print the required elements  ` `        ``Console.WriteLine(X + ``" "` `+ (X + S));  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main() ` `    ``{  ` `        ``int` `[]arr = { 1, 7 };  ` `        ``int` `n = arr.Length;  ` `     `  `        ``findNums(arr, n);  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```6 14
```

Time Complexity: O(n)

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