Given an array arr[] consisting of N positive integers and an integer K, the task is to count all pairs possible from the given array with Bitwise OR equal to K.
Examples:
Input: arr[] = {2, 38, 44, 29, 62}, K = 46
Output: 2
Explanation: Only the following two pairs are present in the array whose Bitwise OR is 46:
- 2 OR 44 = 46
- 38 OR 44 = 46
Input: arr[] = {1, 5, 20, 15, 14}, K = 20
Output: 5
Explanation:
There are only 5 pairs whose Bitwise OR is 20:
- 1 OR 15 = 15
- 1 OR 14 = 15
- 5 OR 15 = 15
- 5 OR 14 = 15
- 15 OR 14 = 15
Approach: To solve the problem, the idea is to generate all possible pairs from the given array and count those pairs whose Bitwise OR is equal to K. After checking for all the pairs, print the count stored.
Below is the implementation of the above approach:
// C++ program for the above approach #include <iostream> using namespace std;
// Function that counts the pairs from // the array whose Bitwise OR is K void countPairs( int arr[], int k, int size)
{ // Stores the required
// count of pairs
int count = 0, x;
// Generate all possible pairs
for ( int i = 0; i < size - 1; i++) {
for ( int j = i + 1; j < size; j++) {
// Perform OR operation
x = arr[i] | arr[j];
// If Bitwise OR is equal
// to K, increment count
if (x == k)
count++;
}
}
// Print the total count
cout << count;
} // Driver Code int main()
{ int arr[] = { 2, 38, 44, 29, 62 };
int K = 46;
int N = sizeof (arr) / sizeof (arr[0]);
// Function Call
countPairs(arr, K, N);
return 0;
} |
// Java program for the above approach import java.io.*;
import java.util.*;
class GFG{
// Function that counts the pairs from // the array whose Bitwise OR is K static void countPairs( int [] arr, int k,
int size)
{ // Stores the required
// count of pairs
int count = 0 , x;
// Generate all possible pairs
for ( int i = 0 ; i < size - 1 ; i++)
{
for ( int j = i + 1 ; j < size; j++)
{
// Perform OR operation
x = arr[i] | arr[j];
// If Bitwise OR is equal
// to K, increment count
if (x == k)
count++;
}
}
// Print the total count
System.out.println(count);
} // Driver Code public static void main(String[] args)
{ int [] arr = { 2 , 38 , 44 , 29 , 62 };
int K = 46 ;
int N = arr.length;
// Function Call
countPairs(arr, K, N);
} } // This code is contributed by code_hunt |
# Python3 program for the above approach # Function that counts the pairs from # the array whose Bitwise OR is K def countPairs(arr, k, size):
# Stores the required
# count of pairs
count = 0
# Generate all possible pairs
for i in range (size - 1 ):
for j in range (i + 1 , size):
# Perform OR operation
x = arr[i] | arr[j]
# If Bitwise OR is equal
# to K, increment count
if (x = = k):
count + = 1
# Print the total count
print (count)
# Driver Code arr = [ 2 , 38 , 44 , 29 , 62 ]
K = 46
N = len (arr)
# Function Call countPairs(arr, K, N) # This code is contributed by sanjoy_62 |
// C# program for the above approach using System;
class GFG{
// Function that counts the pairs from // the array whose Bitwise OR is K static void countPairs( int [] arr, int k,
int size)
{ // Stores the required
// count of pairs
int count = 0, x;
// Generate all possible pairs
for ( int i = 0; i < size - 1; i++)
{
for ( int j = i + 1; j < size; j++)
{
// Perform OR operation
x = arr[i] | arr[j];
// If Bitwise OR is equal
// to K, increment count
if (x == k)
count++;
}
}
// Print the total count
Console.WriteLine(count);
} // Driver Code public static void Main()
{ int [] arr = { 2, 38, 44, 29, 62 };
int K = 46;
int N = arr.Length;
// Function Call
countPairs(arr, K, N);
} } // This code is contributed by susmitakundugoaldanga |
<script> // JavaScript program for the above approach // Function that counts the pairs from // the array whose Bitwise OR is K function countPairs(arr, k, size)
{ // Stores the required
// count of pairs
let count = 0, x;
// Generate all possible pairs
for (let i = 0; i < size - 1; i++)
{
for (let j = i + 1; j < size; j++)
{
// Perform OR operation
x = arr[i] | arr[j];
// If Bitwise OR is equal
// to K, increment count
if (x == k)
count++;
}
}
// Print the total count
document.write(count);
} // Driver Code let arr = [ 2, 38, 44, 29, 62 ];
let K = 46;
let N = arr.length;
// Function Call
countPairs(arr, K, N);
// This code is contributed by avijitmondal998.
</script> |
2
Time Complexity: O(N2)
Auxiliary Space: O(1)