# Count pairs with average present in the same array in C++

We are given an array of integers such that each element of the array is in the range [- 1000,1000]. The goal is to find pairs of elements of the array such that their average is also present in that array. If array is arr[]= [1,2,3,4]. Then pairs would be (1,3) and (2,4) as the average of 1,3 is 2 and the average of 2,4 is 3 and both 2 and 3 are present in the array. Count would be 2.

Let us understand with examples.

Input − arr[]= { -1,2,5,-3,8,10 }

Output − Count of pairs with average present in the same array are − 2

Explanation − Pairs such that their average is present in arr[] are − (-1,5) avg is 2, (2,8) avg is 5. Count=2

Input − arr[] = {1,3,2,5,10,6}

Output − Count of pairs with an average present in the same array are − 3

Explanation − Pairs such that their average is present in arr[] are − (1,3) avg is 2, (1,5) avg is 3, (2,10) avg is 6. Count=3.

## The approach used in the below program is as follows

We will first make a frequency array for all elements of the array. The size of the frequency array is double the size of the original array as there could be negative elements as well.

Frequency of negative numbers will start from index 0 to index 1000. And the frequency of positive numbers will start from index 1000 to 2000.

For each non zero frequency do this −

1. Add (freq[i]) * (freq[i]-1)/2 to count, because the average of two same numbers is the number itself. If there are 5 2’s in the array. Then total pairs would be (5*(5-1))/2=10.

2. If the above freq[i] is non zero then start traversing for alternate frequencies as consecutive numbers have average which is floating-point and would not be there in the array.

3. If freq[j] is found which is non-zero and freq[ (i+j)/2 ] is also non-zero. Then add freq[i]*freq[j] to the count( as each number can pair with every other number ).

• Take an integer array arr[]

• Function average_pair(arr, size) takes the array and its size and returns the count of pairs such that the average of elements in pair is also present in the array arr[].

• Take initial count as 0 and initialize N=1000.

• Calculate length of frequency array as size_2=2*N+1 ( for range [-1000 to 1000]

• Initialize frequency array with 0 initially.

• Populate frequency array such that the frequency of negative elements is from 0 to 1000 and positive elements afterward. For this add N to indexes.

• For each element arr[i] update frequency array as arr_freq[arr[i] + N]++;

• Now traverse the frequency array from i=0 to i<size_2.

• For each non-zero frequency add (freq[i]) * (freq[i]-1)/2 to count according to condition 1.

• Now as arr_freq[i] is non-zero, traverse every alternate frequency.

• Calculate temp_2 as arr_freq[(i + j) / 2].

• Now if temp_2 is non-zero and arr_freq[j] is non-zero then condition 3 is met. Update count with product (arr_freq[i] * arr_freq[j]);

• After the end of all iterations, the count will have a total number of such pairs.

• Return count as result.

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int average_pair(int arr[], int size_1){
int count = 0;
int N = 1000;
int size_2 = (2 * N) + 1;
int arr_freq[size_2] = { 0 };
for (int i = 0; i < size_1; i++){
int temp = arr[i];
arr_freq[temp + N]++;
}
for (int i = 0; i < size_2; i++){
if (arr_freq[i] > 0){
int check = (arr_freq[i]) * (arr_freq[i] - 1);
count += check / 2;
for (int j = i + 2; j < 2001; j += 2){
int temp_2 = arr_freq[(i + j) / 2];
if (arr_freq[j] > 0 && temp_2 > 0){
count += (arr_freq[i] * arr_freq[j]);
}
}
}
}
return count;
}
int main(){
int arr[] = { 2, 3, 1, 8, 9, 10 };
int size = sizeof(arr) / sizeof(arr);
cout<<"Count of pairs with average present in the same array are: "<<average_pair(arr, size);
return 0;
}

## Output

If we run the above code it will generate the following output −

Count of pairs with average present in the same array are: 2