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Append two elements to make the array satisfy the given condition

  • Last Updated : 08 Nov, 2021

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

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 <bits/stdc++.h>
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

Javascript




<script>
// javascript implementation of the approach    
// Function to find the required numbers
    function findNums(arr , n) {
 
        // Find the sum and xor
        var S = 0, X = 0;
        for (i = 0; i < n; i++) {
            S += arr[i];
            X ^= arr[i];
        }
 
        // Prvar the required elements
        document.write(X + " " + (X + S));
    }
 
    // Driver code
     
        var arr = [ 1, 7 ];
        var n = arr.length;
 
        findNums(arr, n);
 
// This code contributed by gauravrajput1
</script>
Output: 
6 14

 

Time Complexity: O(n)

Auxiliary Space: O(1)
 




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