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# Number of pairs whose sum is a power of 2 in C++

Given an array, we have to find the number of pairs whose sum is a power of 2. Let's see the example.

**Input**

arr = [1, 2, 3]

**Output**

1

There is only one pair whose sum is a power of 2. And the pair is (1, 3).

## Algorithm

- Initialise array with random numbers.
- Initialise the count to 0.
- Write two loops to get all the pairs of the array.
- Compute the sum of every pair.
- Check whether the sum is a power of 2 or not using bitwise AND.
- Increment the count if the count is a power of 2.

- Return the count.

## Implementation

Following is the implementation of the above algorithm in C++

#include <bits/stdc++.h> using namespace std; int get2PowersCount(int arr[], int n) { int count = 0; for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { int sum = arr[i] + arr[j]; if ((sum & (sum - 1)) == 0) { count++; } } } return count; } int main() { int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; int n = 10; cout << get2PowersCount(arr, n) << endl; return 0; }

## Output

If you run the above code, then you will get the following result.

6

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