Count pairs with Bitwise-AND as even number
Given an array of 
integers. The task is to find the number of pairs (i, j) such that A[i] & A[j] is even.
Examples:
Input: N = 4, A[] = { 5, 1, 3, 2 }
Output: 3
Since pair of A[] are:
( 5, 1 ), ( 5, 3 ), ( 5, 2 ), ( 1, 3 ), ( 1, 2 ), ( 3, 2 )
5 AND 1 = 1, 5 AND 3 = 1, 5 AND 2 = 0,
1 AND 3 = 1, 1 AND 2 = 0,
3 AND 2 = 2
Total even pair A( i, j ) = 3
Input : N = 6, A[] = { 5, 9, 0, 6, 7, 3 }
Output : 9
Since pair of A[] =
( 5, 9 ) = 1, ( 5, 0 ) = 0, ( 5, 6 ) = 4, ( 5, 7 ) = 5, ( 5, 3 ) = 1,
( 9, 0 ) = 0, ( 9, 6 ) = 0, ( 9, 7 ) = 1, ( 9, 3 ) = 1,
( 0, 6 ) = 0, ( 0, 7 ) = 0, ( 0, 3 ) = 0,
( 6, 7 ) = 6, ( 6, 3 ) = 2,
( 7, 3 ) = 3
A Naive Approach is to check for every pair and print the count of pairs which are even.
Below is the implementation of the above approach:
C++
// C++ program to count pair with // bitwise-AND as even number #include <iostream> using namespace std; // Function to count number of pairs EVEN bitwise AND 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 AND operation // to check evenpair if ((A[i] & A[j]) % 2 == 0) evenPair++; } } // return number of even pair return evenPair; } // Driver Code int main() { int a[] = { 5, 1, 3, 2 }; int n = sizeof(a) / sizeof(a[0]); cout << findevenPair(a, n) << endl; return 0; } |
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Java
// Java program to count pair with // bitwise-AND as even number import java.io.*; class GFG { // Function to count number of // pairs EVEN bitwise AND static 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 AND operation // to check evenpair if ((A[i] & A[j]) % 2 == 0) evenPair++; } } // return number of even pair return evenPair; } // Driver Code public static void main (String[] args) { int []a = { 5, 1, 3, 2 }; int n = a.length; System.out.println(findevenPair(a, n)); } } // This code is contributed by anuj_67.. |
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Python3
# Python 3 program to count pair with # bitwise-AND as even number # Function to count number of # pairs EVEN bitwise AND def 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 AND operation to # check evenpair if ((A[i] & A[j]) % 2 == 0): evenPair += 1 # return number of even pair return evenPair # Driver Code a = [ 5, 1, 3, 2 ] n = len(a) print(findevenPair(a, n)) # This code is contributed # by PrinciRaj1992 |
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C#
// C# program to count pair with // bitwise-AND as even number using System; class GFG { // Function to count number of // pairs EVEN bitwise AND static 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 AND operation // to check evenpair if ((A[i] & A[j]) % 2 == 0) evenPair++; } } // return number of even pair return evenPair; } // Driver Code public static void Main () { int []a = { 5, 1, 3, 2 }; int n = a.Length; Console.WriteLine(findevenPair(a, n)); } } // This code is contributed by anuj_67.. |
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PHP
<?php // PHP program to count pair with // bitwise-AND as even number // Function to count number of // pairs EVEN bitwise AND function 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; $j++) { // find AND operation // to check evenpair if (($A[$i] & $A[$j]) % 2 == 0) $evenPair++; } } // return number of even pair return $evenPair; } // Driver Code $a = array(5, 1, 3, 2 ); $n = sizeof($a); echo findevenPair($a, $n); // This code is contributed by akt_mit ?> |
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Output:
3
An Efficient Approach will be to observe that the bitwise AND of two numbers will be even if atleast one of the two numbers is even. So, count all the odd numbers in the array say count. Now, number of pairs with both odd elements will be count*(count-1)/2.
Therefore, number of elements with atleast one even element will be:
Total Pairs - Count of pair with both odd elements
Below is the implementation of the above approach:
C++
// C++ program to count pair with // bitwise-AND as even number #include <iostream> using namespace std; // Function to count number of pairs // with EVEN bitwise AND int findevenPair(int A[], int N) { int count = 0; // count odd numbers for (int i = 0; i < N; i++) if (A[i] % 2 != 0) count++; // count odd pairs int oddCount = count * (count - 1) / 2; // return number of even pair return (N * (N - 1) / 2) - oddCount; } // Driver Code int main() { int a[] = { 5, 1, 3, 2 }; int n = sizeof(a) / sizeof(a[0]); cout << findevenPair(a, n) << endl; return 0; } |
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Java
// Java program to count pair with // bitwise-AND as even number import java.io.*; class GFG { // Function to count number of pairs // with EVEN bitwise AND static int findevenPair(int A[], int N) { int count = 0; // count odd numbers for (int i = 0; i < N; i++) if (A[i] % 2 != 0) count++; // count odd pairs int oddCount = count * (count - 1) / 2; // return number of even pair return (N * (N - 1) / 2) - oddCount; } // Driver Code public static void main (String[] args) { int a[] = { 5, 1, 3, 2 }; int n =a.length; System.out.print( findevenPair(a, n)); } } // This code is contributed by anuj_67.. |
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Python3
# Python 3 program to count pair with # bitwise-AND as even number # Function to count number of pairs # with EVEN bitwise AND def findevenPair(A, N): count = 0 # count odd numbers for i in range(0, N): if (A[i] % 2 != 0): count += 1 # count odd pairs oddCount = count * (count - 1) / 2 # return number of even pair return (int)((N * (N - 1) / 2) - oddCount) # Driver Code a = [5, 1, 3, 2 ] n = len(a) print(findevenPair(a, n)) # This code is contributed # by PrinciRaj1992 |
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C#
// C# program to count pair with // bitwise-AND as even number using System; public class GFG{ // Function to count number of pairs // with EVEN bitwise AND static int findevenPair(int []A, int N) { int count = 0; // count odd numbers for (int i = 0; i < N; i++) if (A[i] % 2 != 0) count++; // count odd pairs int oddCount = count * (count - 1) / 2; // return number of even pair return (N * (N - 1) / 2) - oddCount; } // Driver Code static public void Main (){ int []a = { 5, 1, 3, 2 }; int n =a.Length; Console.WriteLine( findevenPair(a, n)); } } // This code is contributed by ajit |
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PHP
<?php // PHP program to count pair with // bitwise-AND as even number // Function to count number of pairs // with EVEN bitwise AND function findevenPair($A, $N) { $count = 0; // count odd numbers for ($i = 0; $i < $N; $i++) if ($A[$i] % 2 != 0) $count++; // count odd pairs $oddCount = $count * ($count - 1) / 2; // return number of even pair return ($N * ($N - 1) / 2) - $oddCount; } // Driver Code $a = array(5, 1, 3, 2); $n = sizeof($a); echo findevenPair($a, $n) . "\n"; // This code is contributed // by Akanksha Rai ?> |
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Output:
3
Time Complexity: O(N)
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