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Count pairs with Bitwise-AND as even number in C++
We are given an integer array and the task is to count the total number of pairs that can be formed using the given array values such that the AND operation on the pairs will result in an even number.
The truth table for AND operation is given below
| A | B | A^B |
| 0 | 0 | 0 |
| 1 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 1 | 1 |
Input − int arr[] = {2, 5, 1, 8, 9}
Output − Count of pairs with Bitwise AND as EVEN number are − 7
Explanation −
| a1 | a2 | a1^a2 |
| 2 | 5 | 0 |
| 2 | 1 | 0 |
| 2 | 8 | 0 |
| 2 | 9 | 0 |
| 5 | 1 | 1 |
| 5 | 8 | 0 |
| 5 | 9 | 1 |
| 1 | 8 | 0 |
| 1 | 9 | 1 |
| 8 | 9 | 8 |
Approach used in the below program is as follows
Input an array of integer elements to form an pair
Calculate the size of an array pass the data to the function for further processing
Create a temporary variable count to store the pairs formed with AND operation as an even value.
Start loop FOR from i to 0 till the size of an array
Inside the loop, check IF arr[i] % 2 == FALSE then increment the count by 1
Set the count as count * (count - 1) / 2
Create a temporary variable to store the total number of pairs formed and set it to (size * (size - 1) / 2)
Store the odd number as subtracting count from total of pairs formed
Return the odd value
Print the result.
Example
#include <iostream>
using namespace std;
//Count pairs with Bitwise AND as EVEN number
int count_pair(int arr[], int size){
int count = 0;
for (int i = 0; i < size; i++){
if (arr[i] % 2 != 0){
count++;
}
}
count = count * (count - 1) / 2;
int total_pair = (size * (size - 1) / 2);
int odd = total_pair - count;
return odd;
}
int main(){
int arr[] = {2, 5, 1, 8, 3 };
int size = sizeof(arr) / sizeof(arr[0]);
cout<<"Count of pairs with Bitwise-AND as even number are: "<<count_pair(arr, size) << endl;
return 0;
}Output
If we run the above code it will generate the following output −
Count of pairs with Bitwise-AND as even number are: 7
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