Given an array arr containing integers of size N, the task is to check if the XOR of this array is even or odd
Examples:
Input: arr[] = { 2, 4, 7}
Output: Odd
Explanation:
XOR of array = 2 ^ 4 ^ 7 = 1, which is oddInput: arr[] = { 3, 9, 12, 13, 15 }
Output: Even
Naive Solution: First find the XOR of the given array of integers, and then check if this XOR is even or odd.
Time Complexity: O(N)
Efficient Solution: A better Solution is based on bit manipulation fact, that:
- Bitwise XOR of any two even or any two odd numbers is always even.
- Bitwise XOR of an even and an odd number is always odd.
Therefore if the count of odd numbers in the array is odd, then the final XOR will be odd and if it is even, then final XOR will be even.
Below is the implementation of the above approach:
// C++ program to check if the XOR // of an array is Even or Odd #include <bits/stdc++.h> using namespace std;
// Function to check if the XOR of // an array of integers is Even or Odd string check( int arr[], int n)
{ int count = 0;
for ( int i = 0; i < n; i++) {
// Count the number
// of odd elements
if (arr[i] & 1)
count++;
}
// If count of odd elements
// is odd, then XOR will be odd
if (count & 1)
return "Odd" ;
// Else even
else
return "Even" ;
} // Driver Code int main()
{ int arr[] = { 3, 9, 12, 13, 15 };
int n = sizeof (arr) / sizeof (arr[0]);
// Function call
cout << check(arr, n) << endl;
return 0;
} |
// Java program to check if the XOR // of an array is Even or Odd import java.util.*;
class GFG{
// Function to check if the XOR of // an array of integers is Even or Odd static String check( int []arr, int n)
{ int count = 0 ;
for ( int i = 0 ; i < n; i++) {
// Count the number
// of odd elements
if ((arr[i] & 1 )!= 0 )
count++;
}
// If count of odd elements
// is odd, then XOR will be odd
if ((count & 1 )!= 0 )
return "Odd" ;
// Else even
else
return "Even" ;
} // Driver Code public static void main(String args[])
{ int []arr = { 3 , 9 , 12 , 13 , 15 };
int n = arr.length;
// Function call
System.out.println(check(arr, n));
} } // This code is contributed by Surendra_Gangwar |
# Python3 program to check if the XOR # of an array is Even or Odd # Function to check if the XOR of # an array of integers is Even or Odd def check(arr, n):
count = 0 ;
for i in range (n):
# Count the number
# of odd elements
if (arr[i] & 1 ):
count = count + 1 ;
# If count of odd elements
# is odd, then XOR will be odd
if (count & 1 ):
return "Odd" ;
# Else even
else :
return "Even" ;
# Driver Code if __name__ = = '__main__' :
arr = [ 3 , 9 , 12 , 13 , 15 ]
n = len (arr)
# Function call
print (check(arr, n))
# This code is contributed by Princi Singh |
// C# program to check if the XOR // of an array is Even or Odd using System;
using System.Collections.Generic;
using System.Linq;
class GFG
{ // Function to check if the XOR of // an array of integers is Even or Odd static String check( int []arr, int n)
{ int count = 0;
for ( int i = 0; i < n; i++) {
// Count the number
// of odd elements
if (arr[i] == 1)
count++;
}
// If count of odd elements
// is odd, then XOR will be odd
if (count == 1)
return "Odd" ;
// Else even
else
return "Even" ;
} // Driver Code public static void Main(String[] args)
{
int []arr= { 3, 9, 12, 13, 15 };
int n = arr.Length;
// Function call
Console.Write(check(arr, n));
} } // This code is contributed by shivanisinghss2110 |
<script> // Javascript program to check if the XOR // of an array is Even or Odd // Function to check if the XOR of // an array of integers is Even or Odd function check(arr, n)
{ let count = 0;
for (let i = 0; i < n; i++)
{
// Count the number
// of odd elements
if (arr[i] & 1)
count++;
}
// If count of odd elements
// is odd, then XOR will be odd
if (count & 1)
return "Odd" ;
// Else even
else
return "Even" ;
} // Driver Code let arr = [ 3, 9, 12, 13, 15 ]; let n = arr.length; // Function call document.write(check(arr, n)); // This code is contributed by subham348 </script> |
Even
Time Complexity: O(N)
Auxiliary Space: O(1)