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Count pairs with average present in the same array
• Last Updated : 13 May, 2019

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: 1
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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 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); ` ` `  `    ``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 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 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 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 `

Output:

```7
```

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