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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
Updated on: 26-Oct-2021
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