Given an array A[] of size N. The task is to find the how many pair(i, j) exists such that A[i] OR A[j] is even.
Examples:
Input : N = 4
A[] = { 5, 6, 2, 8 }
Output :3
Explanation :
Since pair of A[] = ( 5, 6 ), ( 5, 2 ), ( 5, 8 ),
( 6, 2 ), ( 6, 8 ), ( 2, 8 )
5 OR 6 = 7, 5 OR 2 = 7, 5 OR 8 = 13
6 OR 2 = 6, 6 OR 8 = 14, 2 OR 8 = 10
Total pair A( i, j ) = 6 and Even = 3
Input : N = 7
A[] = {8, 6, 2, 7, 3, 4, 9}
Output :6
Implementation: A simple solution is to check every pair.
C++
// C++ program to count pairs with even OR#include <iostream>using namespace std;int findEvenPair(int A[], int N){ int evenPair = 0; for (int i = 0; i < N; i++) { for (int j = i + 1; j < N; j++) { // find OR operation // check odd or even if ((A[i] | A[j]) % 2 == 0) evenPair++; } } // return count of even pair return evenPair;}// Driver mainint main(){ int A[] = { 5, 6, 2, 8 }; int N = sizeof(A) / sizeof(A[0]); cout << findEvenPair(A, N) << endl; return 0;} |
Java
// Java program to count// pairs with even ORimport java.io.*;class GFG {static int findEvenPair(int A[], int N){ int evenPair = 0; for (int i = 0; i < N; i++) { for (int j = i + 1; j < N; j++) { // find OR operation // check odd or even if ((A[i] | A[j]) % 2 == 0) evenPair++; } } // return count of even pair return evenPair;}// Driver Codepublic static void main (String[] args){ int A[] = { 5, 6, 2, 8 }; int N = A.length; System.out.println(findEvenPair(A, N));}}// This code is contributed// by inder_verma. |
Python 3
# Python 3 program to count pairs# with even ORdef findEvenPair(A, N) : evenPair = 0 for i in range(N) : for j in range(i+1, N): # find OR operation # check odd or even if (A[i] | A[j]) % 2 == 0 : evenPair += 1 # return count of even pair return evenPair# Driver Codeif __name__ == "__main__" : A = [ 5, 6, 2, 8] N = len(A) # function calling print(findEvenPair(A, N))# This code is contributed by ANKITRAI1 |
C#
// C# program to count// pairs with even ORusing System;class GFG {static int findEvenPair(int []A, int N){ int evenPair = 0; for (int i = 0; i < N; i++) { for (int j = i + 1; j < N; j++) { // find OR operation // check odd or even if ((A[i] | A[j]) % 2 == 0) evenPair++; } } // return count of even pair return evenPair;}// Driver Codepublic static void Main (){ int []A = { 5, 6, 2, 8 }; int N = A.Length; Console.WriteLine(findEvenPair(A, N));}}// This code is contributed// by inder_verma. |
PHP
<?php// PHP program to count // pairs with even ORfunction findEvenPair($A, $N){ $evenPair = 0; for ($i = 0; $i < $N; $i++) { for ($j = $i + 1; $j < $N; $j++) { // find OR operation // check odd or even if (($A[$i] | $A[$j]) % 2 == 0) $evenPair++; } } // return count of even pair return $evenPair;}// Driver Code$A = array(5, 6, 2, 8);$N = count($A);echo findEvenPair($A, $N);// This code is contributed// by inder_verma.?> |
Javascript
<script>// JavaScript program to count pairs with even ORfunction findEvenPair(A, N){ let evenPair = 0; for (let i = 0; i < N; i++) { for (let j = i + 1; j < N; j++) { // find OR operation // check odd or even if ((A[i] | A[j]) % 2 == 0) evenPair++; } } // return count of even pair return evenPair;}// Driver main let A = [5, 6, 2, 8 ]; let N = A.length; document.write(findEvenPair(A, N)); // This code contributed by Rajput-Ji </script> |
Output
3
Complexity Analysis:
Time Complexity: O(N^2), since two nested loops are used.
Auxiliary Space: O(1), as constant extra space used by the algorithm.
Implementation: A efficient solution is to count numbers with last bit as 0. Then return count * (count – 1)/2. Note that OR of two numbers can be even only if their last bits are 0.
C++
// C++ program to count pairs with even OR#include <iostream>using namespace std;int findEvenPair(int A[], int N){ // Count total even numbers in // array. int count = 0; for (int i = 0; i < N; i++) if (!(A[i] & 1)) count++; // return count of even pair return count * (count - 1) / 2;}// Driver mainint main(){ int A[] = { 5, 6, 2, 8 }; int N = sizeof(A) / sizeof(A[0]); cout << findEvenPair(A, N) << endl; return 0;} |
Java
// Java program to count// pairs with even ORimport java.io.*;class GFG {static int findEvenPair(int A[], int N){ // Count total even numbers in // array. int count = 0; for (int i = 0; i < N; i++) if ((!((A[i] & 1) > 0))) count++; // return count of even pair return count * (count - 1) / 2;}// Driver Codepublic static void main (String[] args) { int A[] = { 5, 6, 2, 8 }; int N = A.length; System.out.println(findEvenPair(A, N));}}// This code is contributed// by inder_verma. |
Python 3
# Python 3 program to count # pairs with even ORdef findEvenPair(A, N): # Count total even numbers # in array. count = 0 for i in range(N): if (not (A[i] & 1)): count += 1 # return count of even pair return count * (count - 1) // 2# Driver Codeif __name__ == "__main__": A = [ 5, 6, 2, 8 ] N = len(A) print(findEvenPair(A, N))# This code is contributed# by ChitraNayal |
C#
// C# program to count// pairs with even ORusing System;class GFG {static int findEvenPair(int []A, int N){ // Count total even numbers // in array. int count = 0; for (int i = 0; i < N; i++) if ((!((A[i] & 1) > 0))) count++; // return count of even pair return count * (count - 1) / 2;}// Driver Codepublic static void Main (String[] args) { int []A = { 5, 6, 2, 8 }; int N = A.Length; Console.WriteLine(findEvenPair(A, N));}}// This code is contributed // by Kirti_Mangal |
PHP
<?php // PHP program to count pairs// with even ORfunction findEvenPair(&$A, $N){ // Count total even numbers in // array. $count = 0; for ($i = 0; $i < $N; $i++) if (!($A[$i] & 1)) $count++; // return count of even pair return $count * ($count - 1) / 2;}// Driver Code$A = array(5, 6, 2, 8 );$N = sizeof($A);echo findEvenPair($A, $N)."\n";// This code is contributed // by ChitraNayal?> |
Javascript
<script>// Javascript program to count// pairs with even OR function findEvenPair(A,N) { // Count total even numbers in // array. let count = 0; for (let i = 0; i < N; i++) if ((!((A[i] & 1) > 0))) count++; // return count of even pair return count * (count - 1) / 2; } // Driver Code let A=[5, 6, 2, 8 ]; let N = A.length; document.write(findEvenPair(A, N)); // This code is contributed by rag2127</script> |
Output
3
Complexity Analysis:
Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(1), as constant extra space used by the algorithm.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
