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++ 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 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[0]);
insert_element(a, n);
} // This code is contributed by chitranayal |
// 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 |
# 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# 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 |
<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> |
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.