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: 0
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: 1
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<bits/stdc++.h>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 arrayvoid 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 codeint main(){ int a[] = {1, 2, 3}; int n = sizeof(a) / sizeof(a[0]); 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 Arrayclass 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 arraystatic 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 codepublic 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 arraydef 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 codeif __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 Arrayusing 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 arraystatic 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 codepublic 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
<script> // Javascript 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 function insert_element(a, n) { // Variable to store the // Xor of all the elements let Xor = 0; // Variable to store the // sum of all elements let Sum = 0; // Loop to find the Xor // and the sum of the array for(let 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 document.write("0" + "</br>"); return; } // We insert one more element // which is Sum if(Xor == 0) { document.write("1" + "</br>"); document.write(Sum + "</br>"); return; } // We insert two more elements // Sum + Xor and Xor. let num1 = Sum + Xor; let num2 = Xor; // Print the number of elements // inserted in the array document.write("2" + "</br>"); // Print the elements that are // inserted in the array document.write(num1 + " " + num2 + "</br>"); } let a = [1, 2, 3]; let n = a.length; insert_element(a, n); </script> |
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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