Numbers of pairs from an array whose average is also present in the array
Last Updated :
23 Jul, 2025
Given an array arr[] consisting of N integers, the task is to count the number of distinct pairs (arr[i], arr[j]) in the array such that the average of pairs is also present in the array.
Note: Consider (arr[i], arr[j]) and (arr[j], arr[i]) as the same pairs.
Examples:
Input: arr[] = {2, 1, 3}
Output: 1
Explanation: The only one pair whose average is present in the given array is (1, 3) (Average = 2).
Input: arr[] = {4, 2, 5, 1, 3, 5}
Output: 7
Naive Approach: Follow the steps below to solve the problem:
- Initialize a variable, say count as 0 to store all the count of pairs whose average exists in the array.
- Insert all the array elements in an set S.
- Traverse over the set S and for every element in the set S, generate all possible pairs of the given array and if the sum of any pairs is same as the current element in the set then increment the value of count by 1.
- After completing the above steps, print the value of count as the resultant count of pairs.
Below is the implementation of the above approach :
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the number of
// pairs from the array having sum S
int getCountPairs(vector<int> arr, int N, int S)
{
// Stores the total count of
// pairs whose sum is 2*S
int count = 0;
// Generate all possible pairs
// and check their sums
for(int i = 0; i < arr.size(); i++)
{
for(int j = i + 1; j < arr.size(); j++)
{
// If the sum is S, then
// increment the count
if ((arr[i] + arr[j]) == S)
count++;
}
}
// Return the total
// count of pairs
return count;
}
// Function to count of pairs having
// whose average exists in the array
int countPairs(vector<int> arr, int N)
{
// Initialize the count
int count = 0;
// Use set to remove duplicates
unordered_set<int> S;
// Add elements in the set
for(int i = 0; i < N; i++)
S.insert(arr[i]);
for(int ele : S)
{
int sum = 2 * ele;
// For every sum, count
// all possible pairs
count += getCountPairs(arr, N, sum);
}
// Return the total count
return count;
}
// Driver Code
int main()
{
vector<int> arr = { 4, 2, 5, 1, 3, 5 };
int N = arr.size();
cout << countPairs(arr, N);
return 0;
}
// This code is contributed by Kingash
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to count the number of
// pairs from the array having sum S
public static int getCountPairs(
int arr[], int N, int S)
{
// Stores the total count of
// pairs whose sum is 2*S
int count = 0;
// Generate all possible pairs
// and check their sums
for (int i = 0;
i < arr.length; i++) {
for (int j = i + 1;
j < arr.length; j++) {
// If the sum is S, then
// increment the count
if ((arr[i] + arr[j]) == S)
count++;
}
}
// Return the total
// count of pairs
return count;
}
// Function to count of pairs having
// whose average exists in the array
public static int countPairs(
int arr[], int N)
{
// Initialize the count
int count = 0;
// Use set to remove duplicates
HashSet<Integer> S = new HashSet<>();
// Add elements in the set
for (int i = 0; i < N; i++)
S.add(arr[i]);
for (int ele : S) {
int sum = 2 * ele;
// For every sum, count
// all possible pairs
count += getCountPairs(
arr, N, sum);
}
// Return the total count
return count;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 4, 2, 5, 1, 3, 5 };
int N = arr.length;
System.out.print(
countPairs(arr, N));
}
}
# Python3 program for the above approach
# Function to count the number of
# pairs from the array having sum S
def getCountPairs(arr, N, S):
# Stores the total count of
# pairs whose sum is 2*S
count = 0
# Generate all possible pairs
# and check their sums
for i in range(len(arr)):
for j in range(i + 1, len(arr)):
# If the sum is S, then
# increment the count
if ((arr[i] + arr[j]) == S):
count += 1
# Return the total
# count of pairs
return count
# Function to count of pairs having
# whose average exists in the array
def countPairs(arr, N):
# Initialize the count
count = 0
# Use set to remove duplicates
S = set([])
# Add elements in the set
for i in range(N):
S.add(arr[i])
for ele in S:
sum = 2 * ele
# For every sum, count
# all possible pairs
count += getCountPairs(arr, N, sum)
# Return the total count
return count
# Driver Code
if __name__ == "__main__":
arr = [ 4, 2, 5, 1, 3, 5 ]
N = len(arr)
print(countPairs(arr, N))
# This code is contributed by ukasp
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG
{
// Function to count the number of
// pairs from the array having sum S
public static int getCountPairs(
int []arr, int N, int S)
{
// Stores the total count of
// pairs whose sum is 2*S
int count = 0;
// Generate all possible pairs
// and check their sums
for (int i = 0;
i < arr.Length; i++) {
for (int j = i + 1;
j < arr.Length; j++) {
// If the sum is S, then
// increment the count
if ((arr[i] + arr[j]) == S)
count++;
}
}
// Return the total
// count of pairs
return count;
}
// Function to count of pairs having
// whose average exists in the array
public static int countPairs(
int []arr, int N)
{
// Initialize the count
int count = 0;
// Use set to remove duplicates
HashSet<int> S = new HashSet<int>();
// Add elements in the set
for (int i = 0; i < N; i++)
S.Add(arr[i]);
foreach (int ele in S) {
int sum = 2 * ele;
// For every sum, count
// all possible pairs
count += getCountPairs(
arr, N, sum);
}
// Return the total count
return count;
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 4, 2, 5, 1, 3, 5 };
int N = arr.Length;
Console.Write(
countPairs(arr, N));
}
}
// This code is contributed by Princi Singh
<script>
// JavaScript program for the above approach
// Function to count the number of
// pairs from the array having sum S
function getCountPairs(
arr, N, S)
{
// Stores the total count of
// pairs whose sum is 2*S
let count = 0;
// Generate all possible pairs
// and check their sums
for (let i = 0;
i < arr.length; i++) {
for (let j = i + 1;
j < arr.length; j++) {
// If the sum is S, then
// increment the count
if ((arr[i] + arr[j]) == S)
count++;
}
}
// Return the total
// count of pairs
return count;
}
// Function to count of pairs having
// whose average exists in the array
function countPairs(arr, N)
{
// Initialize the count
let count = 0;
// Use set to remove duplicates
let S = [];
// Add elements in the set
for (let i = 0; i < N; i++)
S.push(arr[i]);
for (let ele in S) {
let sum = 2 * ele;
// For every sum, count
// all possible pairs
count += getCountPairs(
arr, N, sum);
}
// Return the total count
return count;
}
// Driver code
let arr = [ 4, 2, 5, 1, 3, 5 ];
let N = arr.length;
document.write(
countPairs(arr, N));
// This code is contributed by code_hunt.
</script>
Time Complexity: O(N3)
Auxiliary Space: O(N)
Efficient Approach: The above approach can also be optimized by storing the frequency of the sum of all possible pairs in the given array in a HashMap and find the count for each element of the array accordingly. Follow the steps below to solve the problem:
- Initialize a variable, say count as 0 to store all the count of pairs whose average exists in the array.
- Insert all the array elements in an set S.
- Initialize a HashMap, say M that stores the frequency of the sum of all possible pairs in the given array.
- Traverse over the set S and for every element(say X) in the set S update the value of count by the value (M[X]/2).
- After completing the above steps, print the value of count as the resultant count of pairs.
Below is the implementation of the above approach:
C++
// CPP program for the above approach
#include<bits/stdc++.h>
using namespace std;
// Function to count the total count
// of pairs having sum S
int getCountPairs(int arr[],
int N, int S)
{
map<int,int> mp;
// Store the total count of all
// elements in map mp
for (int i = 0; i < N; i++) {
mp[arr[i]]++;
}
// Stores the total count of
// total pairs
int twice_count = 0;
// Iterate through each element
// and increment the count
for (int i = 0; i < N; i++) {
// If the value (S - arr[i])
// exists in the map hm
if (mp.find(S - arr[i]) != mp.end()) {
// Update the twice count
twice_count += mp[S - arr[i]];
}
if (S - arr[i] == arr[i])
twice_count--;
}
// Return the half of twice_count
return twice_count / 2;
}
// Function to count of pairs having
// whose average exists in the array
int countPairs(
int arr[], int N)
{
// Stores the total count of
// pairs
int count = 0;
// Use set to remove duplicates
set<int> S;
// Insert all the element in
// the set S
for (int i = 0; i < N; i++)
S.insert(arr[i]);
for (int ele : S) {
int sum = 2 * ele;
// For every sum find the
// getCountPairs
count += getCountPairs(
arr, N, sum);
}
// Return the total count of
// pairs
return count;
}
// Driver Code
int main()
{
int N = 6;
int arr[] = { 4, 2, 5, 1, 3, 5 };
cout<<(countPairs(arr, N));
}
// This code is contributed by ipg2016107.
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to count the total count
// of pairs having sum S
static int getCountPairs(int arr[],
int N, int S)
{
HashMap<Integer, Integer> mp
= new HashMap<>();
// Store the total count of all
// elements in map mp
for (int i = 0; i < N; i++) {
// Initialize value to 0,
// if key not found
if (!mp.containsKey(arr[i]))
mp.put(arr[i], 0);
mp.put(arr[i],
mp.get(arr[i]) + 1);
}
// Stores the total count of
// total pairs
int twice_count = 0;
// Iterate through each element
// and increment the count
for (int i = 0; i < N; i++) {
// If the value (S - arr[i])
// exists in the map hm
if (mp.get(S - arr[i])
!= null) {
// Update the twice count
twice_count += mp.get(
S - arr[i]);
}
if (S - arr[i] == arr[i])
twice_count--;
}
// Return the half of twice_count
return twice_count / 2;
}
// Function to count of pairs having
// whose average exists in the array
public static int countPairs(
int arr[], int N)
{
// Stores the total count of
// pairs
int count = 0;
// Use set to remove duplicates
HashSet<Integer> S = new HashSet<>();
// Insert all the element in
// the set S
for (int i = 0; i < N; i++)
S.add(arr[i]);
for (int ele : S) {
int sum = 2 * ele;
// For every sum find the
// getCountPairs
count += getCountPairs(
arr, N, sum);
}
// Return the total count of
// pairs
return count;
}
// Driver Code
public static void main(String[] args)
{
int N = 6;
int arr[] = { 4, 2, 5, 1, 3, 5 };
System.out.println(
countPairs(arr, N));
}
}
# Python program for the above approach
# Function to count the total count
# of pairs having sum S
def getCountPairs(arr,N,S):
mp = {}
# Store the total count of all
# elements in map mp
for i in range(N):
# Initialize value to 0,
# if key not found
if (arr[i] not in mp):
mp[arr[i]] = 0
mp[arr[i]] += 1
# Stores the total count of
# total pairs
twice_count = 0
# Iterate through each element
# and increment the count
for i in range(N):
# If the value (S - arr[i])
# exists in the map hm
if ((S - arr[i]) in mp):
# Update the twice count
twice_count += mp[S - arr[i]]
if (S - arr[i] == arr[i]):
twice_count -= 1
# Return the half of twice_count
return (twice_count // 2)
# Function to count of pairs having
# whose average exists in the array
def countPairs(arr,N):
# Stores the total count of
# pairs
count = 0
# Use set to remove duplicates
S = set()
# Insert all the element in
# the set S
for i in range(N):
S.add(arr[i])
for ele in S:
sum = 2 * ele
# For every sum find the
# getCountPairs
count += getCountPairs(arr, N, sum)
# Return the total count of
# pairs
return count
# Driver Code
N = 6
arr = [4, 2, 5, 1, 3, 5 ]
print(countPairs(arr, N))
# This code is contributed by shinjanpatra
using System;
using System.Collections.Generic;
class GFG
{
// Function to count the total count
// of pairs having sum S
static int GetCountPairs(int[] arr, int N, int S)
{
Dictionary<int, int> mp = new Dictionary<int, int>();
// Store the total count of all
// elements in map mp
for (int i = 0; i < N; i++)
{
// Initialize value to 0,
// if key not found
if (!mp.ContainsKey(arr[i]))
mp[arr[i]] = 0;
mp[arr[i]] = mp[arr[i]] + 1;
}
// Stores the total count of
// total pairs
int twice_count = 0;
// Iterate through each element
// and increment the count
for (int i = 0; i < N; i++)
{
// If the value (S - arr[i])
// exists in the map hm
if (mp.ContainsKey(S - arr[i]))
{
// Update the twice count
twice_count += mp[S - arr[i]];
}
if (S - arr[i] == arr[i])
twice_count--;
}
// Return the half of twice_count
return twice_count / 2;
}
// Function to count of pairs having
// whose average exists in the array
public static int CountPairs(int[] arr, int N)
{
// Stores the total count of
// pairs
int count = 0;
// Use set to remove duplicates
HashSet<int> S = new HashSet<int>();
// Insert all the element in
// the set S
for (int i = 0; i < N; i++)
S.Add(arr[i]);
foreach (int ele in S)
{
int sum = 2 * ele;
// For every sum find the
// getCountPairs
count += GetCountPairs(
arr, N, sum);
}
// Return the total count of
// pairs
return count;
}
// Driver Code
public static void Main(string[] args)
{
int N = 6;
int[] arr = { 4, 2, 5, 1, 3, 5 };
Console.WriteLine(
CountPairs(arr, N));
}
}
// This code is contributed by phasing17.
<script>
// JavaScript program for the above approach
// Function to count the total count
// of pairs having sum S
function getCountPairs(arr,N,S)
{
let mp = new Map();
// Store the total count of all
// elements in map mp
for (let i = 0; i < N; i++) {
// Initialize value to 0,
// if key not found
if (!mp.has(arr[i]))
mp.set(arr[i], 0);
mp.set(arr[i],
mp.get(arr[i]) + 1);
}
// Stores the total count of
// total pairs
let twice_count = 0;
// Iterate through each element
// and increment the count
for (let i = 0; i < N; i++) {
// If the value (S - arr[i])
// exists in the map hm
if (mp.get(S - arr[i])
!= null) {
// Update the twice count
twice_count += mp.get(
S - arr[i]);
}
if (S - arr[i] == arr[i])
twice_count--;
}
// Return the half of twice_count
return Math.floor(twice_count / 2);
}
// Function to count of pairs having
// whose average exists in the array
function countPairs(arr,N)
{
// Stores the total count of
// pairs
let count = 0;
// Use set to remove duplicates
let S = new Set();
// Insert all the element in
// the set S
for (let i = 0; i < N; i++)
S.add(arr[i]);
for (let ele of S.values()) {
let sum = 2 * ele;
// For every sum find the
// getCountPairs
count += getCountPairs(
arr, N, sum);
}
// Return the total count of
// pairs
return count;
}
// Driver Code
let N = 6;
let arr=[4, 2, 5, 1, 3, 5 ];
document.write(countPairs(arr, N));
// This code is contributed by avanitrachhadiya2155
</script>
Time Complexity: O(N2)
Auxiliary Space: O(N)
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