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Append two elements to make the array satisfy the given condition
  • Last Updated : 06 Jan, 2020

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
Output:
6 14

Time Complexity: O(n)

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