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