# Count triplets such that sum of any two number is equal to third | Set 2

• Difficulty Level : Medium
• Last Updated : 13 Aug, 2021

Given an array of distinct positive integers arr[] of length N, the task is to count all the triplets such that sum of two elements equals the third element.

Examples:

Input: arr[] = {1, 5, 3, 2}
Output:
Explanation:
In the given array, there are two such triplets such that sum of the two numbers is equal to the third number, those are –
(1, 2, 3), (3, 2, 5)
Input: arr[] = {3, 2, 7}
Output:
Explanation:
In the given array there are no such triplets such that sum of two numbers is equal to the third number.

Approach: The idea is to create a frequency array of the numbers which are present in the array and then check for each pair of the element that the sum of the pair elements is present in the array or not with the help of frequency array in O(1) time.
Algorithm:

• Declare a frequency array to store the frequency of the numbers.
• Iterate over the elements of the array and increment the count of that number in the frequency array.
• Run two loops to choose two different indexes of the matrix and check if the sum of the elements at those indices has a frequency more than 0 in the frequency array.
```If frequency of the sum is greater than 0:
Increment the count of the triplets.```

Note: We have assumed in the program that the value of array elements lies in the range [1, 100].
Below is the implementation of the above approach:

## C++

 `// C++ implementation to count the``// triplets such that the sum of the``// two numbers is equal to third number` `#include ``using` `namespace` `std;` `// Function to find the count of the``// triplets such that sum of two``// numbers is equal to the third number``int` `countTriplets(``int` `arr[], ``int` `n){``    ``int` `freq = {0};``    ` `    ``// Loop to count the frequency``    ``for` `(``int` `i=0; i < n; i++){``        ``freq[arr[i]]++;``    ``}``    ``int` `count = 0;``    ` `    ``// Loop to count for triplets``    ``for``(``int` `i = 0;i < n; i++){``        ``for``(``int` `j = i+1; j < n; j++){``            ``if``(freq[arr[i] + arr[j]]){``                ``count++;``            ``}``        ``}``    ``}``    ``return` `count;``}` `// Driver Code``int` `main()``{``    ``int` `n = 4;``    ``int` `arr[] = {1, 5, 3, 2};``    ` `    ``// Function Call``    ``cout << countTriplets(arr, n);``    ``return` `0;``}`

## Java

 `// Java implementation to count the``// triplets such that the sum of the``// two numbers is equal to third number``class` `GFG{` `// Function to find the count of the``// triplets such that sum of two``// numbers is equal to the third number``static` `int` `countTriplets(``int` `arr[], ``int` `n){``    ``int` `[]freq = ``new` `int``[``100``];``    ` `    ``// Loop to count the frequency``    ``for` `(``int` `i = ``0``; i < n; i++){``        ``freq[arr[i]]++;``    ``}``    ``int` `count = ``0``;``    ` `    ``// Loop to count for triplets``    ``for``(``int` `i = ``0``; i < n; i++){``        ``for``(``int` `j = i + ``1``; j < n; j++){``            ``if``(freq[arr[i] + arr[j]] > ``0``){``                ``count++;``            ``}``        ``}``    ``}``    ``return` `count;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``4``;``    ``int` `arr[] = {``1``, ``5``, ``3``, ``2``};``    ` `    ``// Function Call``    ``System.out.print(countTriplets(arr, n));``}``}` `// This code is contributed by Rajput-Ji`

## Python 3

 `# Python 3 implementation to count the``# triplets such that the sum of the``# two numbers is equal to third number` `# Function to find the count of the``# triplets such that sum of two``# numbers is equal to the third number``def` `countTriplets(arr, n):``    ``freq ``=` `[``0` `for` `i ``in` `range``(``100``)]``    ` `    ``# Loop to count the frequency``    ``for` `i ``in` `range``(n):``        ``freq[arr[i]] ``+``=` `1``    ``count ``=` `0``    ` `    ``# Loop to count for triplets``    ``for` `i ``in` `range``(n):``        ``for` `j ``in` `range``(i ``+` `1``, n, ``1``):``            ``if``(freq[arr[i] ``+` `arr[j]]):``                ``count ``+``=` `1``    ``return` `count` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `4``    ``arr ``=` `[``1``, ``5``, ``3``, ``2``]``    ` `    ``# Function Call``    ``print``(countTriplets(arr, n))` `# This code is contributed by Surendra_Gangwar`

## C#

 `// C# implementation to count the``// triplets such that the sum of the``// two numbers is equal to third number``using` `System;` `class` `GFG{` `// Function to find the count of the``// triplets such that sum of two``// numbers is equal to the third number``static` `int` `countTriplets(``int` `[]arr, ``int` `n){``    ``int` `[]freq = ``new` `int``;``    ` `    ``// Loop to count the frequency``    ``for` `(``int` `i = 0; i < n; i++){``        ``freq[arr[i]]++;``    ``}``    ``int` `count = 0;``    ` `    ``// Loop to count for triplets``    ``for``(``int` `i = 0; i < n; i++){``        ``for``(``int` `j = i + 1; j < n; j++){``            ``if``(freq[arr[i] + arr[j]] > 0){``                ``count++;``            ``}``        ``}``    ``}``    ``return` `count;``}` `// Driver Code``public` `static` `void` `Main(``string``[] args)``{``    ``int` `n = 4;``    ``int` `[]arr = {1, 5, 3, 2};``    ` `    ``// Function Call``    ``Console.WriteLine(countTriplets(arr, n));``}``}` `// This code is contributed by Yahs_R`

## Javascript

 ``

Output:

`2`

Performance Analysis:

• Time Complexity: O(N2).
• Auxiliary Space: O(N).

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