# Count of elements to be inserted to make Array sum twice the XOR of Array

• Difficulty Level : Hard
• Last Updated : 18 Jun, 2022

Given an array arr[] of size N, the task is to find the minimum number of elements that need to be inserted into the array so that the sum of the array is equal to two times the XOR of the array
Examples:

Input: arr[] = {1, 2, 3, 6}
Output:
Explanation:
Xor = (1 ^ 2 ^ 3 ^ 6) = 6 and sum of the array = 12.
The required condition (sum = 2 * Xor) satisfies. So no need to insert more elements.
Input: arr[] = {1, 2, 3}
Output:
Explanation:
Xor = (1 ^ 2 ^ 3) = 0 and sum of the array = (1 + 2 + 3) = 6.
Here, we insert one element {6} into the array. Now, NewXor = (0 ^ 6) = 6 and NewSum = (6 + 6) = 12.
Hence, NewSum = 2*NewXor.
So, the given condition satisfies.

Approach: The idea is to compute the following steps in order to find the answer:

• Initially, we find the sum of the array and the XOR of the array.
• Now, we check if the given condition satisfies or not. If it satisfies, then print 0 as we need not insert any element.
• Now, check if the XOR is equal to 0 or not. If it is, then the element that needs to be inserted into the array is the sum of all the elements of the array.
• This is because, by inserting the sum, the new XOR becomes (0 ^ sum = sum) and the sum of the array becomes sum + sum = 2 * sum. Therefore the condition satisfies.
• Else, we add two more elements which are XOR and (sum + XOR). This is because:

NewXor = (Xor ^ (sum + Xor) ^ Xor) = Sum + Xor.
NewSum = (Sum + (Sum + Xor) + Xor) = 2 * (Sum + Xor) = 2 * NewXor

Below is the implementation of the above approach:

## C++

 `// C++ program to find the count``// of elements to be inserted to``// make Array sum twice the XOR of Array``#include``using` `namespace` `std;` `// Function to find the minimum``// number of elements that need to be``// inserted such that the sum of the``// elements of the array is twice``// the XOR of the array``void` `insert_element(``int` `a[], ``int` `n)``{``    ` `    ``// Variable to store the``    ``// Xor of all the elements``    ``int` `Xor = 0;` `    ``// Variable to store the``    ``// sum of all elements``    ``int` `Sum = 0;``    ` `    ``// Loop to find the Xor``    ``// and the sum of the array``    ``for``(``int` `i = 0; i < n; i++)``    ``{ ``        ``Xor ^= a[i];``        ``Sum += a[i];``    ``}``    ` `    ``// If sum = 2 * Xor``    ``if``(Sum == 2 * Xor)``    ``{``      ` `        ``// No need to insert``        ``// more elements``        ``cout << ``"0"` `<< endl;``        ``return``;``     ``}``  ` `    ``// We insert one more element``    ``// which is Sum``    ``if``(Xor == 0)``    ``{``        ``cout << ``"1"` `<< endl;``        ``cout << Sum << endl;``        ``return``;``     ``}` `    ``// We insert two more elements``    ``// Sum + Xor and Xor.``    ``int` `num1 = Sum + Xor;``    ` `    ``int` `num2 = Xor;` `    ``// Print the number of elements``    ``// inserted in the array``    ``cout << ``"2"``;` `    ``// Print the elements that are``    ``// inserted in the array``    ``cout << num1 << ``" "``         ``<< num2 << endl; ``}` `// Driver code``int` `main()``{   ``    ``int` `a[] = {1, 2, 3};``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);``  ` `    ``insert_element(a, n);``}` `// This code is contributed by chitranayal`

## Java

 `// Java program to find the count``// of elements to be inserted to``// make Array sum twice the XOR of Array``class` `GFG{`` ` `// Function to find the minimum``// number of elements that need to be``// inserted such that the sum of the``// elements of the array is twice``// the XOR of the array``static` `void` `insert_element(``int` `a[], ``int` `n)``{``     ` `    ``// Variable to store the``    ``// Xor of all the elements``    ``int` `Xor = ``0``;`` ` `    ``// Variable to store the``    ``// sum of all elements``    ``int` `Sum = ``0``;``     ` `    ``// Loop to find the Xor``    ``// and the sum of the array``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{ ``        ``Xor ^= a[i];``        ``Sum += a[i];``    ``}``     ` `    ``// If sum = 2 * Xor``    ``if``(Sum == ``2` `* Xor)``    ``{``       ` `        ``// No need to insert``        ``// more elements``        ``System.out.println(``"0"``);``        ``return``;``     ``}``   ` `    ``// We insert one more element``    ``// which is Sum``    ``if``(Xor == ``0``)``    ``{``        ``System.out.println(``"1"``);``        ``System.out.println(Sum);``        ``return``;``     ``}`` ` `    ``// We insert two more elements``    ``// Sum + Xor and Xor.``    ``int` `num1 = Sum + Xor;``     ` `    ``int` `num2 = Xor;`` ` `    ``// Print the number of elements``    ``// inserted in the array``    ``System.out.print(``"2"``);`` ` `    ``// Print the elements that are``    ``// inserted in the array``    ``System.out.println(num1 + ``" "` `+ num2);``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{   ``    ``int` `a[] = {``1``, ``2``, ``3``};``    ``int` `n = a.length;``   ` `    ``insert_element(a, n);``}``}` `// This code is contributed by rock_cool`

## Python3

 `# Python program to find the count``# of elements to be inserted to``# make Array sum twice the XOR of Array` `# Function to find the minimum``# number of elements that need to be``# inserted such that the sum of the``# elements of the array is twice``# the XOR of the array``def` `insert_element(a, n):``    ` `    ``# Variable to store the``    ``# Xor of all the elements``    ``Xor ``=` `0` `    ``# Variable to store the``    ``# sum of all elements``    ``Sum` `=` `0``    ` `    ``# Loop to find the Xor``    ``# and the sum of the array``    ``for` `i ``in` `range``(n):``        ` `        ``Xor^``=` `a[i]``        ``Sum``+``=` `a[i]``    ` `    ``# If sum = 2 * Xor``    ``if``(``Sum` `=``=` `2` `*` `Xor):` `        ``# No need to insert``        ``# more elements``        ``print``(``0``)``        ``return` `    ``# We insert one more element``    ``# which is Sum``    ``if``(Xor ``=``=` `0``):``        ``print``(``1``)``        ``print``(``Sum``)``        ``return` `    ``# We insert two more elements``    ``# Sum + Xor and Xor.``    ``num1 ``=` `Sum` `+` `Xor``    ` `    ``num2 ``=` `Xor` `    ``# Print the number of elements``    ``# inserted in the array``    ``print``(``2``)` `    ``# Print the elements that are``    ``# inserted in the array``    ``print``(num1, num2)``    ` `        ` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``a ``=` `[``1``, ``2``, ``3``]``    ``n ``=` `len``(a)``    ``insert_element(a, n)`

## C#

 `// C# program to find the count``// of elements to be inserted to``// make Array sum twice the XOR of Array``using` `System;``class` `GFG{``  ` `// Function to find the minimum``// number of elements that need to be``// inserted such that the sum of the``// elements of the array is twice``// the XOR of the array``static` `void` `insert_element(``int``[] a, ``int` `n)``{``      ` `    ``// Variable to store the``    ``// Xor of all the elements``    ``int` `Xor = 0;``  ` `    ``// Variable to store the``    ``// sum of all elements``    ``int` `Sum = 0;``      ` `    ``// Loop to find the Xor``    ``// and the sum of the array``    ``for``(``int` `i = 0; i < n; i++)``    ``{ ``        ``Xor ^= a[i];``        ``Sum += a[i];``    ``}``      ` `    ``// If sum = 2 * Xor``    ``if``(Sum == 2 * Xor)``    ``{``        ` `        ``// No need to insert``        ``// more elements``        ``Console.Write(``"0"``);``        ``return``;``     ``}``    ` `    ``// We insert one more element``    ``// which is Sum``    ``if``(Xor == 0)``    ``{``        ``Console.Write(``"1"` `+ ``'\n'``);``        ``Console.Write(Sum);``        ``return``;``     ``}``  ` `    ``// We insert two more elements``    ``// Sum + Xor and Xor.``    ``int` `num1 = Sum + Xor;``      ` `    ``int` `num2 = Xor;``  ` `    ``// Print the number of elements``    ``// inserted in the array``    ``Console.Write(``"2"``);``  ` `    ``// Print the elements that are``    ``// inserted in the array``    ``Console.Write(num1 + ``" "` `+ num2);``}``  ` `// Driver code``public` `static` `void` `Main(``string``[] args)``{   ``    ``int``[] a = {1, 2, 3};``    ``int` `n = a.Length;``    ` `    ``insert_element(a, n);``}``}`` ` `// This code is contributed by Ritik Bansal`

## Javascript

 ``

Output:

```1
6```

Time Complexity : O(n) ,as we are traversing once in an array.

Space Complexity : O(1) ,as we are not using any extra space.

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