Given an array of N integers, the task is to find the number of pairs (i, j) such that A[i] ^ A[j] is even.
Examples:
Input: A[] = { 5, 4, 7, 2, 1}
Output: 4
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 even pair = 4
Input: A[] = { 7, 2, 8, 1, 0, 5, 11 }
Output: 9
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 that are even.
Below is the implementation of the above approach:
C++
// C++ program to count pairs// with XOR giving a even number#include <iostream>using namespace std;// Function to count number of even pairsint findevenPair(int A[], int N){ int i, j; // variable for counting even pairs int evenPair = 0; // find all pairs for (i = 0; i < N; i++) { for (j = i + 1; j < N; j++) { // find XOR operation // check even or even if ((A[i] ^ A[j]) % 2 == 0) evenPair++; } } // return number of even pair return evenPair;}// Driver Codeint main(){ int A[] = { 5, 4, 7, 2, 1 }; int N = sizeof(A) / sizeof(A[0]); // calling function findevenPair // and print number of even pair cout << findevenPair(A, N) << endl; return 0;} |
C
// C program to count pairs// with XOR giving a even number#include <stdio.h>// Function to count number of even pairsint findevenPair(int A[], int N){ int i, j; // variable for counting even pairs int evenPair = 0; // find all pairs for (i = 0; i < N; i++) { for (j = i + 1; j < N; j++) { // find XOR operation // check even or even if ((A[i] ^ A[j]) % 2 == 0) evenPair++; } } // return number of even pair return evenPair;}// Driver Codeint main(){ int A[] = { 5, 4, 7, 2, 1 }; int N = sizeof(A) / sizeof(A[0]); // calling function findevenPair // and print number of even pair printf("%d\n",findevenPair(A, N)); return 0;}// This code is contributed by kothvvsaakash. |
Java
// Java program to count pairs// with XOR giving a even numberimport java.io.*;class GFG{// Function to count number of even pairsstatic int findevenPair(int []A, int N){ int i, j; // variable for counting even pairs int evenPair = 0; // find all pairs for (i = 0; i < N; i++) { for (j = i + 1; j < N; j++) { // find XOR operation // check even or even if ((A[i] ^ A[j]) % 2 == 0) evenPair++; } } // return number of even pair return evenPair;}// Driver Codepublic static void main (String[] args) { int A[] = { 5, 4, 7, 2, 1 }; int N = A.length; // calling function findevenPair // and print number of even pair System.out.println(findevenPair(A, N));}}// This code is contributed by inder_verma.. |
Python3
# Python3 program to count pairs# with XOR giving a even number # Function to count number of even pairsdef findevenPair(A, N): # variable for counting even pairs evenPair = 0 # find all pairs for i in range(0, N): for j in range(i+1, N): # find XOR operation # check even or even if ((A[i] ^ A[j]) % 2 == 0): evenPair+=1 # return number of even pair return evenPair; # Driver Codedef main(): A = [ 5, 4, 7, 2, 1 ] N = len(A) # calling function findevenPair # and print number of even pair print(findevenPair(A, N)) if __name__ == '__main__': main()# This code is contributed by PrinciRaj1992 |
C#
// C# program to count pairs// with XOR giving a even numberusing System;class GFG{// Function to count number of// even pairsstatic int findevenPair(int []A, int N){ int i, j; // variable for counting even pairs int evenPair = 0; // find all pairs for (i = 0; i < N; i++) { for (j = i + 1; j < N; j++) { // find XOR operation // check even or even if ((A[i] ^ A[j]) % 2 == 0) evenPair++; } } // return number of even pair return evenPair;}// Driver Codepublic static void Main () { int []A = { 5, 4, 7, 2, 1 }; int N = A.Length; // calling function findevenPair // and print number of even pair Console.WriteLine(findevenPair(A, N));}}// This code is contributed// by inder_verma.. |
PHP
<?php// PHP program to count pairs// with XOR giving a even number// Function to count number // of even pairsfunction findevenPair(&$A, $N){ // variable for counting even pairs $evenPair = 0; // find all pairs for ($i = 0; $i < $N; $i++) { for ($j = $i + 1; $j < $N; $j++) { // find XOR operation // check even or even if (($A[$i] ^ $A[$j]) % 2 == 0) $evenPair++; } } // return number of even pair return $evenPair;}// Driver Code$A = array(5, 4, 7, 2, 1 );$N = sizeof($A);// calling function findevenPair// and print number of even pairecho (findevenPair($A, $N)); // This code is contributed// by Shivi_Aggarwal ?> |
Javascript
<script>// Javascript program to count pairs// with XOR giving a even number// Function to count number of even pairsfunction findevenPair(A, N){ let i, j; // variable for counting even pairs let evenPair = 0; // find all pairs for (i = 0; i < N; i++) { for (j = i + 1; j < N; j++) { // find XOR operation // check even or even if ((A[i] ^ A[j]) % 2 == 0) evenPair++; } } // return number of even pair return evenPair;}// Driver Codelet A = [ 5, 4, 7, 2, 1 ];let N = A.length;// calling function findevenPair// and print number of even pairdocument.write(findevenPair(A, N));// This code is contributed by souravmahato348.</script> |
Output
4
Time Complexity: O(n^2)
Auxiliary Space: O(1)
An efficient solution is to Count pairs with Bitwise XOR as ODD number i.e. oddEvenpairs. Then return totalPairs – oddEvenPairs where totalPairs = (N * (N-1) / 2) and oddEvenPairs = count * (N – count).
As, pairs that will give Even Bitwise XOR are :
Even, Even
Odd, Odd
So, find the count of pairs with both odd and even elements and subtract from total no. of pairs.
Below is the implementation of the above approach:
C++
// C++ program to count pairs// with XOR giving a even number#include <iostream>using namespace std;// Function to count number of even pairsint findEvenPair(int A[], int N){ int count = 0; // find all pairs for (int i = 0; i < N; i++) { if (A[i] % 2 != 0) count++; } int totalPairs = (N * (N - 1) / 2); int oddEvenPairs = count * (N - count); // return number of even pair return totalPairs - oddEvenPairs;}// Driver Codeint main(){ int a[] = { 5, 4, 7, 2, 1 }; int n = sizeof(a) / sizeof(a[0]); // calling function findEvenPair // and print number of even pair cout << findEvenPair(a, n) << endl; return 0;} |
C
// C program to count pairs// with XOR giving a even number#include <stdio.h>// Function to count number of even pairsint findEvenPair(int A[], int N){ int count = 0; // find all pairs for (int i = 0; i < N; i++) { if (A[i] % 2 != 0) count++; } int totalPairs = (N * (N - 1) / 2); int oddEvenPairs = count * (N - count); // return number of even pair return totalPairs - oddEvenPairs;}// Driver Codeint main(){ int a[] = { 5, 4, 7, 2, 1 }; int n = sizeof(a) / sizeof(a[0]); // calling function findEvenPair // and print number of even pair printf("%d\n",findEvenPair(a, n)); return 0;}// This code is contributed by kothvvsaakash. |
Java
// Java program to count pairs // with XOR giving a even numberimport java.io.*;class GFG { // Function to count number of even pairs static int findEvenPair(int A[], int N) { int count = 0; // find all pairs for (int i = 0; i < N; i++) { if (A[i] % 2 != 0) count++; } int totalPairs = (N * (N - 1) / 2); int oddEvenPairs = count * (N - count); // return number of even pair return totalPairs - oddEvenPairs; } // Driver Code public static void main (String[] args) { int a[] = { 5, 4, 7, 2, 1 }; int n = a.length; // calling function findEvenPair // and print number of even pair System.out.println(findEvenPair(a, n)); }//This code is contributed by akt_mit } |
Python3
# python program to count pairs# with XOR giving a even number# Function to count number of even pairsdef findEvenPair(A, N): count = 0 # find all pairs for i in range(0,N): if (A[i] % 2 != 0): count+=1 totalPairs = (N * (N - 1) / 2) oddEvenPairs = count * (N - count) # return number of even pair return (int)(totalPairs - oddEvenPairs)# Driver Codedef main(): a = [ 5, 4, 7, 2, 1 ] n = len(a) # calling function findEvenPair # and print number of even pair print(findEvenPair(a, n)) if __name__ == '__main__': main() # This code is contributed by 29AjayKumar |
C#
// C# program to count pairs // with XOR giving a even number using System; public class GFG { // Function to count number of even pairs static int findEvenPair(int []A, int N) { int count = 0; // find all pairs for (int i = 0; i < N; i++) { if (A[i] % 2 != 0) count++; } int totalPairs = (N * (N - 1) / 2); int oddEvenPairs = count * (N - count); // return number of even pair return totalPairs - oddEvenPairs; } // Driver Code public static void Main() { int []a = { 5, 4, 7, 2, 1 }; int n = a.Length; // calling function findEvenPair // and print number of even pair Console.Write(findEvenPair(a, n)); }}// This code is contributed by 29AjayKumar |
PHP
<?php// PHP program to count pairs// with XOR giving a even number// Function to count number of even pairsfunction findEvenPair($A, $N){ $count = 0; // find all pairs for ($i = 0; $i < $N; $i++) { if ($A[$i] % 2 != 0) $count++; } $totalPairs = ($N * ($N - 1) / 2); $oddEvenPairs = $count * ($N - $count); // return number of even pair return $totalPairs - $oddEvenPairs;}// Driver Code$a = array(5, 4, 7, 2, 1);$n = sizeof($a);// calling function findEvenPair// and print number of even pairecho findEvenPair($a, $n) . "\n";// This code is contributed // by Akanksha Rai?> |
Javascript
<script>// Javascript program to count pairs// with XOR giving a even number// Function to count number of even pairsfunction findEvenPair(A, N){ let count = 0; // find all pairs for (let i = 0; i < N; i++) { if (A[i] % 2 != 0) count++; } let totalPairs = parseInt(N * (N - 1) / 2); let oddEvenPairs = count * (N - count); // return number of even pair return totalPairs - oddEvenPairs;}// Driver Code let a = [ 5, 4, 7, 2, 1 ]; let n = a.length; // calling function findEvenPair // and print number of even pair document.write(findEvenPair(a, n)); </script> |
Output
4
Time Complexity: O(n)
Auxiliary Space: O(1)
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