Given an array of n integers. Find out number of pairs in array whose XOR is odd.
Examples :
Input : arr[] = { 1, 2, 3 }
Output : 2
All pairs of array
1 ^ 2 = 3
1 ^ 3 = 2
2 ^ 3 = 1
Input : arr[] = { 1, 2, 3, 4 }
Output : 4
Naive Approach: We can find pairs whose XOR is odd by running two loops. If XOR of two number is odd increase count of pairs.
Implementation:
C++
#include <iostream>
using namespace std;
int countXorPair( int arr[], int n)
{
int count = 0;
for ( int i = 0; i < n; i++) {
for ( int j = i + 1; j < n; j++)
if ((arr[i] ^ arr[j]) % 2 == 1)
count++;
}
return count;
}
int main()
{
int arr[] = { 1, 2, 3 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << countXorPair(arr, n);
return 0;
}
|
Java
public class CountXor {
static int countXorPair( int arr[], int n)
{
int count = 0 ;
for ( int i = 0 ; i < n; i++) {
for ( int j = i + 1 ; j < n; j++)
if ((arr[i] ^ arr[j]) % 2 == 1 )
count++;
}
return count;
}
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 3 };
System.out.println(countXorPair(arr, arr.length));
}
}
|
Python 3
def countXorPair(arr, n):
count = 0
for i in range (n):
for j in range (i + 1 , n):
if ((arr[i] ^ arr[j]) % 2 = = 1 ):
count + = 1
return count
if __name__ = = "__main__" :
arr = [ 1 , 2 , 3 ]
n = len (arr)
print (countXorPair(arr, n))
|
C#
using System;
public class CountXor {
static int countXorPair( int [] arr, int n)
{
int count = 0;
for ( int i = 0; i < n; i++) {
for ( int j = i + 1; j < n; j++)
if ((arr[i] ^ arr[j]) % 2 == 1)
count++;
}
return count;
}
public static void Main()
{
int [] arr = {1, 2, 3};
Console.WriteLine(countXorPair(arr, arr.Length));
}
}
|
PHP
<?php
function countXorPair( $arr , $n )
{
$count = 0;
for ( $i = 0; $i < $n ; $i ++)
{
for ( $j = $i + 1; $j < $n ; $j ++)
if (( $arr [ $i ] ^ $arr [ $j ]) % 2 == 1)
$count ++;
}
return $count ;
}
$arr = array (1, 2, 3);
$n = count ( $arr );
echo countXorPair( $arr , $n );
?>
|
Javascript
<script>
function countXorPair(arr, n)
{
let count = 0;
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++)
if ((arr[i] ^ arr[j]) % 2 == 1)
count++;
}
return count;
}
let arr = [1, 2, 3];
document.write(countXorPair(arr, arr.length));
</script>
|
Time Complexity : O(n*n)
Space Complexity – O(1)
Efficient Approach: We can observe that:
odd ^ odd = even
odd ^ even = odd
even ^ odd = odd
even ^ even = even
Therefore total pairs in array whose XOR is odd will be equal to count of odd numbers multiplied by count of even numbers.
Implementation:
C++
#include <iostream>
using namespace std;
int countXorPair( int arr[], int n)
{
int odd = 0, even = 0;
for ( int i = 0; i < n; i++) {
if (arr[i] % 2 == 0)
even++;
else
odd++;
}
return odd * even;
}
int main()
{
int arr[] = { 1, 2, 3 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << countXorPair(arr, n);
return 0;
}
|
Java
public class CountXor {
static int countXorPair( int arr[], int n)
{
int odd = 0 , even = 0 ;
for ( int i = 0 ; i < n; i++) {
if (arr[i] % 2 == 0 )
even++;
else
odd++;
}
return odd * even;
}
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 3 };
System.out.println(countXorPair(arr, arr.length));
}
}
|
Python 3
def countXorPair(arr, n):
odd = 0
even = 0
for i in range (n):
if arr[i] % 2 = = 0 :
even + = 1
else :
odd + = 1
return odd * even
if __name__ = = "__main__" :
arr = [ 1 , 2 , 3 ]
n = len (arr)
print (countXorPair(arr, n))
|
C#
using System;
public class CountXor {
static int countXorPair( int [] arr, int n)
{
int odd = 0, even = 0;
for ( int i = 0; i < n; i++) {
if (arr[i] % 2 == 0)
even++;
else
odd++;
}
return odd * even;
}
public static void Main()
{
int [] arr = {1, 2, 3};
Console.WriteLine(countXorPair(arr, arr.Length));
}
}
|
PHP
<?php
function countXorPair( $arr , $n )
{
$odd = 0;
$even = 0;
for ( $i = 0; $i < $n ; $i ++)
{
if ( $arr [ $i ] % 2 == 0)
$even ++;
else
$odd ++;
}
return $odd * $even ;
}
$arr = array ( 1, 2, 3 );
$n = sizeof( $arr );
echo countXorPair( $arr , $n );
?>
|
Javascript
<script>
function countXorPair(arr, n)
{
let odd = 0, even = 0;
for (let i = 0; i < n; i++)
{
if (arr[i] % 2 == 0)
even++;
else
odd++;
}
return odd * even;
}
let arr=[ 1, 2, 3 ];
document.write(countXorPair(arr, arr.length));
</script>
|
Time Complexity : O(n)
Space Complexity : O(1)
Last Updated :
19 Sep, 2023
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