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Count pairs with average present in the same array

  • Difficulty Level : Easy
  • Last Updated : 29 Apr, 2021
Geek Week

Given an array arr[] of N integers where |arr[i] ≤ 1000| for all valid i. The task is to count the pairs of integers from the array whose average is also present in the same array i.e. for (arr[i], arr[j]) to be a valid pair (arr[i] + arr[j]) / 2 must also be present in the array.
 

Examples: 

Input: arr[] = {2, 1, 3} 
Output:
Only valid pair is (1, 3) as (1 + 3) / 2 = 2 is also present in the array.
Input: arr[] = {4, 2, 5, 1, 3, 5} 
Output: 7  

Approach: Make a frequency array storing frequencies of every array element. Remember if the array contains negative numbers also then we have to take the size of the frequency array just double the original size. After updating the frequency array, there are two cases:  

  1. If freq[i] > 0 then the total number of required pairs will be count = (freq[i] * (freq[i] – 1)) / 2.
  2. And for every pair (freq[i], freq[j]) where freq[i] > 0, freq[j] > 0 and freq[(i + j) / 2] > 0 then the total number of required pairs will be count = (freq[i] * freq[j]).

Below is the implementation of the above approach:  



C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
const int N = 1000;
 
// Function to return the count
// of valid pairs
int countPairs(int arr[], int n)
{
 
    // Frequency array
    // Twice the original size to hold
    // negative elements as well
    int size = (2 * N) + 1;
    int freq[size] = { 0 };
 
    // Update the frequency of each
    // of the array element
    for (int i = 0; i < n; i++) {
        int x = arr[i];
 
        // If say x = -1000 then we will place
        // the frequency of -1000 at
        // (-1000 + 1000 = 0) a[0] index
        freq[x + N]++;
    }
 
    // To store the count of valid pairs
    int ans = 0;
 
    // Remember we will check only for (even, even)
    // or (odd, odd) pairs of indexes as the average
    // of two consecutive elements is
    // a floating point number
    for (int i = 0; i < size; i++) {
 
        if (freq[i] > 0) {
 
            ans += ((freq[i]) * (freq[i] - 1)) / 2;
 
            for (int j = i + 2; j < 2001; j += 2) {
                if (freq[j] > 0 && (freq[(i + j) / 2] > 0)) {
                    ans += (freq[i] * freq[j]);
                }
            }
        }
    }
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 4, 2, 5, 1, 3, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << countPairs(arr, n);
 
    return 0;
}

Java




// Java implementation of the approach
 
class GFG
{
 
    static int N = 1000;
 
    // Function to return the count
    // of valid pairs
    static int countPairs(int arr[], int n)
    {
 
        // Frequency array
        // Twice the original size to hold
        // negative elements as well
        int size = (2 * N) + 1;
        int freq[] = new int[size];
 
        // Update the frequency of each
        // of the array element
        for (int i = 0; i < n; i++)
        {
            int x = arr[i];
 
            // If say x = -1000 then we will place
            // the frequency of -1000 at
            // (-1000 + 1000 = 0) a[0] index
            freq[x + N]++;
        }
 
        // To store the count of valid pairs
        int ans = 0;
 
        // Remember we will check only for (even, even)
        // or (odd, odd) pairs of indexes as the average
        // of two consecutive elements is
        // a floating point number
        for (int i = 0; i < size; i++)
        {
 
            if (freq[i] > 0)
            {
 
                ans += ((freq[i]) * (freq[i] - 1)) / 2;
 
                for (int j = i + 2; j < 2001; j += 2)
                {
                    if (freq[j] > 0 && (freq[(i + j) / 2] > 0))
                    {
                        ans += (freq[i] * freq[j]);
                    }
                }
            }
        }
        return ans;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {4, 2, 5, 1, 3, 5};
        int n = arr.length;
 
        System.out.println(countPairs(arr, n));
    }
}
 
// This code has been contributed by 29AjayKumar

Python3




# Python 3 implementation of the approach
N = 1000
 
# Function to return the count
# of valid pairs
def countPairs(arr, n):
     
    # Frequency array
    # Twice the original size to hold
    # negative elements as well
    size = (2 * N) + 1
    freq = [0 for i in range(size)]
 
    # Update the frequency of each
    # of the array element
    for i in range(n):
        x = arr[i]
 
        # If say x = -1000 then we will place
        # the frequency of -1000 at
        # (-1000 + 1000 = 0) a[0] index
        freq[x + N] += 1
 
    # To store the count of valid pairs
    ans = 0
 
    # Remember we will check only for (even, even)
    # or (odd, odd) pairs of indexes as the average
    # of two consecutive elements is
    # a floating point number
    for i in range(size):
        if (freq[i] > 0):
            ans += int(((freq[i]) * (freq[i] - 1)) / 2)
 
            for j in range(i + 2, 2001, 2):
                if (freq[j] > 0 and
                   (freq[int((i + j) / 2)] > 0)):
                    ans += (freq[i] * freq[j])
                 
    return ans
 
# Driver code
if __name__ == '__main__':
    arr = [4, 2, 5, 1, 3, 5]
    n = len(arr)
 
    print(countPairs(arr, n))
     
# This code is contributed by
# Surendra_Gangwar

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    static int N = 1000;
 
    // Function to return the count
    // of valid pairs
    static int countPairs(int []arr, int n)
    {
 
        // Frequency array
        // Twice the original size to hold
        // negative elements as well
        int size = (2 * N) + 1;
        int []freq = new int[size];
 
        // Update the frequency of each
        // of the array element
        for (int i = 0; i < n; i++)
        {
            int x = arr[i];
 
            // If say x = -1000 then we will place
            // the frequency of -1000 at
            // (-1000 + 1000 = 0) a[0] index
            freq[x + N]++;
        }
 
        // To store the count of valid pairs
        int ans = 0;
 
        // Remember we will check only for (even, even)
        // or (odd, odd) pairs of indexes as the average
        // of two consecutive elements is
        // a floating point number
        for (int i = 0; i < size; i++)
        {
 
            if (freq[i] > 0)
            {
 
                ans += ((freq[i]) * (freq[i] - 1)) / 2;
 
                for (int j = i + 2; j < 2001; j += 2)
                {
                    if (freq[j] > 0 && (freq[(i + j) / 2] > 0))
                    {
                        ans += (freq[i] * freq[j]);
                    }
                }
            }
        }
        return ans;
    }
 
    // Driver code
    public static void Main()
    {
        int []arr = {4, 2, 5, 1, 3, 5};
        int n = arr.Length;
 
        Console.WriteLine(countPairs(arr, n));
    }
}
 
// This code is contributed by AnkitRai01

Javascript




<script>
// javascript implementation of the approach
    var N = 1000;
 
    // Function to return the count
    // of valid pairs
    function countPairs(arr , n) {
 
        // Frequency array
        // Twice the original size to hold
        // negative elements as well
        var size = (2 * N) + 1;
        var freq = Array(size).fill(0);
 
        // Update the frequency of each
        // of the array element
        for (i = 0; i < n; i++) {
            var x = arr[i];
 
            // If say x = -1000 then we will place
            // the frequency of -1000 at
            // (-1000 + 1000 = 0) a[0] index
            freq[x + N]++;
        }
 
        // To store the count of valid pairs
        var ans = 0;
 
        // Remember we will check only for (even, even)
        // or (odd, odd) pairs of indexes as the average
        // of two consecutive elements is
        // a floating povar number
        for (i = 0; i < size; i++) {
 
            if (freq[i] > 0) {
 
                ans += ((freq[i]) * (freq[i] - 1)) / 2;
 
                for (j = i + 2; j < 2001; j += 2) {
                    if (freq[j] > 0 && (freq[(i + j) / 2] > 0)) {
                        ans += (freq[i] * freq[j]);
                    }
                }
            }
        }
        return ans;
    }
 
    // Driver code
     
        var arr = [ 4, 2, 5, 1, 3, 5 ];
        var n = arr.length;
 
        document.write(countPairs(arr, n));
 
// This code contributed by Rajput-Ji
</script>
Output: 
7

 

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