Count pairs with Bitwise XOR as ODD number
Last Updated :
02 Sep, 2022
Given an array of N integers, the task is to find the number of pairs (i, j) such that A[i] ^ A[j] is odd.
Examples:
Input : N = 5
A[] = { 5, 4, 7, 2, 1}
Output :6
Since pair of A[] =
( 5, 4 ) = 1( 5, 7 ) = 2( 5, 2 ) = 7( 5, 1 ) = 4
( 4, 7 ) = 3( 4, 2 ) = 6( 4, 1 ) = 5
( 7, 2 ) = 5( 7, 1 ) = 6
( 2, 1 ) = 3
Total XOR ODD pair = 6
Input : N = 7
A[] = { 7, 2, 8, 1, 0, 5, 11 }
Output :12
Since pair of A[] =
( 7, 2 ) = 5( 7, 8 ) = 15( 7, 1 ) = 6( 7, 0 ) = 7( 7, 5 ) = 2( 7, 11 ) = 12
( 2, 8 ) = 10( 2, 1 ) = 3( 2, 0 ) = 2( 2, 5 ) = 7( 2, 11 ) = 9
( 8, 1 ) = 9( 8, 0 ) = 8( 8, 5 ) = 13( 8, 11 ) = 3
( 1, 0 ) = 1( 1, 5 ) = 4( 1, 11 ) = 10
( 0, 5 ) = 5( 0, 11 ) = 11
( 5, 11 ) = 14
A naive approach is to check for every pair and print the count of pairs.
Below is the implementation of the above approach:
C++
// C++ program to count pairs
// with XOR giving a odd number
#include <iostream>
using namespace std;
// Function to count number of odd pairs
int findOddPair(int A[], int N)
{
int i, j;
// variable for counting odd pairs
int oddPair = 0;
// find all pairs
for (i = 0; i < N; i++) {
for (j = i + 1; j < N; j++) {
// find XOR operation
// check odd or even
if ((A[i] ^ A[j]) % 2 != 0)
oddPair++;
}
cout << endl;
}
// return number of odd pair
return oddPair;
}
// Driver Code
int main()
{
int A[] = { 5, 4, 7, 2, 1 };
int N = sizeof(A) / sizeof(A[0]);
// calling function findOddPair
// and print number of odd pair
cout << findOddPair(A, N) << endl;
return 0;
}
Java
// Java program to count pairs
// with XOR giving a odd number
class GFG
{
// Function to count
// number of odd pairs
static int findOddPair(int A[],
int N)
{
int i, j;
// variable for counting
// odd pairs
int oddPair = 0;
// find all pairs
for (i = 0; i < N; i++)
{
for (j = i + 1; j < N; j++)
{
// find XOR operation
// check odd or even
if ((A[i] ^ A[j]) % 2 != 0)
oddPair++;
}
}
// return number
// of odd pair
return oddPair;
}
// Driver Code
public static void main(String args[])
{
int A[] = { 5, 4, 7, 2, 1 };
int N = A.length;
// calling function findOddPair
// and print number of odd pair
System.out.println(findOddPair(A, N));
}
}
// This code is contributed
// by Kirti_Mangal
Python3
# Python3 program to count pairs
# with XOR giving a odd number
# Function to count number of odd pairs
def findOddPair(A, N) :
# variable for counting odd pairs
oddPair = 0
# find all pairs
for i in range(0, N) :
for j in range(i+1, N) :
# find XOR operation
# check odd or even
if ((A[i] ^ A[j]) % 2 != 0):
oddPair+=1
# return number of odd pair
return oddPair
# Driver Code
if __name__=='__main__':
A = [5, 4, 7, 2, 1 ]
N = len(A)
# calling function findOddPair
# and print number of odd pair
print(findOddPair(A, N))
# This code is contributed by Smitha Dinesh Semwal
C#
// C# program to count pairs
// with XOR giving a odd number
using System;
class GFG
{
// Function to count
// number of odd pairs
static int findOddPair(int[] A,
int N)
{
int i, j;
// variable for counting
// odd pairs
int oddPair = 0;
// find all pairs
for (i = 0; i < N; i++)
{
for (j = i + 1; j < N; j++)
{
// find XOR operation
// check odd or even
if ((A[i] ^ A[j]) % 2 != 0)
oddPair++;
}
}
// return number
// of odd pair
return oddPair;
}
// Driver Code
public static void Main()
{
int[] A = { 5, 4, 7, 2, 1 };
int N = A.Length;
// calling function findOddPair
// and print number of odd pair
Console.WriteLine(findOddPair(A, N));
}
}
// This code is contributed
// by Akanksha Rai(Abby_akku)
# calling function findOddPair
# and print number of odd pair
print(findOddPair(a, n))
PHP
<?php
//PHP program to count pairs
// with XOR giving a odd number
// Function to count number of odd pairs
function findOddPair($A, $N)
{
$i; $j;
// variable for counting odd pairs
$oddPair = 0;
// find all pairs
for ($i = 0; $i < $N; $i++) {
for ($j = $i + 1; $j < $N; $j++) {
// find XOR operation
// check odd or even
if (($A[$i] ^ $A[$j]) % 2 != 0)
$oddPair++;
}
}
// return number of odd pair
return $oddPair;
}
// Driver Code
$A = array( 5, 4, 7, 2, 1 );
$N = sizeof($A) / sizeof($A[0]);
// calling function findOddPair
// and print number of odd pair
echo findOddPair($A, $N),"\n";
// This Code is Contributed by ajit
?>
JavaScript
<script>
// JavaScript program to count pairs
// with XOR giving a odd number
// Function to count number of odd pairs
function findOddPair(A, N)
{
let i, j;
// variable for counting odd pairs
let oddPair = 0;
// find all pairs
for (i = 0; i < N; i++) {
for (j = i + 1; j < N; j++) {
// find XOR operation
// check odd or even
if ((A[i] ^ A[j]) % 2 != 0)
oddPair++;
}
}
// return number of odd pair
return oddPair;
}
// Driver Code
let A = [ 5, 4, 7, 2, 1 ];
let N = A.length;
// calling function findOddPair
// and print number of odd pair
document.write(findOddPair(A, N) + "<br>");
// This code is contributed by Surbhi Tyagi.
</script>
OUTPUT:
6
Time Complexity: O(N^2) as two nested loops are being used.
Auxiliary Space: O(1), as constant space is being used by the algorithm.
An efficient solution is to count the even numbers. Then return count * (N - count).
C++
// C++ program to count pairs
// with XOR giving a odd number
#include <iostream>
using namespace std;
// Function to count number of odd pairs
int findOddPair(int A[], int N)
{
int i, count = 0;
// find all pairs
for (i = 0; i < N; i++) {
if (A[i] % 2 == 0)
count++;
}
// return number of odd pair
return count * (N - count);
}
// Driver Code
int main()
{
int a[] = { 5, 4, 7, 2, 1 };
int n = sizeof(a) / sizeof(a[0]);
// calling function findOddPair
// and print number of odd pair
cout << findOddPair(a, n) << endl;
return 0;
}
Java
// Java program to count pairs
// with XOR giving a odd number
class GFG
{
// Function to count
// number of odd pairs
static int findOddPair(int A[],
int N)
{
int i, count = 0;
// find all pairs
for (i = 0; i < N; i++)
{
if (A[i] % 2 == 0)
count++;
}
// return number of odd pair
return count * (N - count);
}
// Driver Code
public static void main(String[] arg)
{
int a[] = { 5, 4, 7, 2, 1 };
int n = a.length ;
// calling function findOddPair
// and print number of odd pair
System.out.println(findOddPair(a, n));
}
}
// This code is contributed
// by Smitha
Python3
# Python3 program to count pairs
# with XOR giving a odd number
# Function to count number of odd pairs
def findOddPair(A, N) :
count = 0
# find all pairs
for i in range(0 , N) :
if (A[i] % 2 == 0) :
count+=1
# return number of odd pair
return count * (N - count)
# Driver Code
if __name__=='__main__':
a = [5, 4, 7, 2, 1]
n = len(a)
print(findOddPair(a,n))
# this code is contributed by Smitha Dinesh Semwal
C#
// C# program to count pairs
// with XOR giving a odd number
using System;
class GFG
{
// Function to count
// number of odd pairs
static int findOddPair(int []A,
int N)
{
int i, count = 0;
// find all pairs
for (i = 0; i < N; i++)
{
if (A[i] % 2 == 0)
count++;
}
// return number of odd pair
return count * (N - count);
}
// Driver Code
public static void Main()
{
int []a = { 5, 4, 7, 2, 1 };
int n = a.Length ;
// calling function findOddPair
// and print number of odd pair
Console.Write(findOddPair(a, n));
}
}
// This code is contributed
// by Smitha
PHP
<?php
// PHP program to count pairs
// with XOR giving a odd number
// Function to count number
// of odd pairs
function findOddPair($A, $N)
{
$count = 0;
// find all pairs
for ($i = 0; $i < $N; $i++)
{
if ($A[$i] % 2 == 0)
$count++;
}
// return number of odd pair
return $count * ($N - $count);
}
// Driver Code
$a = array( 5, 4, 7, 2, 1 );
$n = count($a);
// calling function findOddPair
// and print number of odd pair
echo( findOddPair($a, $n));
// This code is contributed
// by Smitha
?>
JavaScript
<script>
// Javascript program to count pairs
// with XOR giving a odd number
// Function to count number of odd pairs
function findOddPair(A, N)
{
let i, count = 0;
// find all pairs
for (i = 0; i < N; i++) {
if (A[i] % 2 == 0)
count++;
}
// return number of odd pair
return count * (N - count);
}
// Driver Code
let a = [ 5, 4, 7, 2, 1 ];
let n = a.length;
// calling function findOddPair
// and print number of odd pair
document.write(findOddPair(a, n));
</script>
Output:
6
Time Complexity: O(N), since one traversal of the array is required to complete all operations.
Auxiliary Space: O(1), as constant space is being used.
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