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Count Triplets such that one of the numbers can be written as sum of the other two
• Difficulty Level : Hard
• Last Updated : 31 May, 2021

Given an array A[] of N integers. The task is to find the number of triples (i, j, k) , where i, j, k are indices and (1 <= i < j < k <= N), such that in the set {   } at least one of the numbers can be written as the sum of the other two.
Examples

```Input : A[] = {1, 2, 3, 4, 5}
Output : 4
The valid triplets are:
(1, 2, 3), (1, 3, 4), (1, 4, 5), (2, 3, 5)

Input : A[] = {1, 1, 1, 2, 2}
Output : 6```

This is a counting problem. Let’s say f(x) represents the frequency of number in our array.
There exist four cases:

1. All three numbers are equal to 0. The number of ways = f(0)C3 (where pCq is the number of ways of choosing q numbers from p numbers).
2. One number is equal to 0, the other two are equal to some x > 0: f(0) * f(x)C2.
3. Two numbers are equal to some x>0, the third is 2*x: f(x)C2 * f(2 * x).
4. The three numbers are x, y and x + y, 0 < x, y: f(x) * f(y) * f(x + y).

Below is the implementation of the above approach:

## C++

 `// C++ program to count Triplets such that at``// least one of the numbers can be written``// as sum of the other two``#include``using` `namespace` `std;` `    ``// Functoin to count the number of ways``    ``// to choose the triples``    ``int` `countWays(``int` `arr[], ``int` `n)``    ``{``        ``// compute the max value in the array``        ``// and create frequency array of size``        ``// max_val + 1.``        ``// We can also use HashMap to store``        ``// frequencies. We have used an array``        ``// to keep remaining code simple.``        ``int` `max_val = 0;``        ``for` `(``int` `i = 0; i < n; i++)``            ``max_val = max(max_val, arr[i]);``        ``int` `freq[max_val + 1]={0};``        ``for` `(``int` `i = 0; i < n; i++)``            ``freq[arr[i]]++;` `        ``int` `ans = 0; ``// stores the number of ways` `        ``// Case 1: 0, 0, 0``        ``ans += freq * (freq - 1) * (freq - 2) / 6;` `        ``// Case 2: 0, x, x``        ``for` `(``int` `i = 1; i <= max_val; i++)``            ``ans += freq * freq[i] * (freq[i] - 1) / 2;` `        ``// Case 3: x, x, 2*x``        ``for` `(``int` `i = 1; 2 * i <= max_val; i++)``            ``ans += freq[i] * (freq[i] - 1) / 2 * freq[2 * i];` `        ``// Case 4: x, y, x + y``        ``// iterate through all pairs (x, y)``        ``for` `(``int` `i = 1; i <= max_val; i++) {``            ``for` `(``int` `j = i + 1; i + j <= max_val; j++)``                ``ans += freq[i] * freq[j] * freq[i + j];``        ``}` `        ``return` `ans;``    ``}` `    ``// Driver code``    ``int` `main()``    ``{``        ``int` `arr[]={ 1, 2, 3, 4, 5 };``        ``int` `n = ``sizeof``(arr)/``sizeof``(``int``);``        ``cout<<(countWays(arr, n));``        ``return` `0;``    ``}` `//contributed by Arnab Kundu`

## Java

 `// Java program to count Triplets such that at``// least one of the numbers can be written``// as a sum of the other two` `class` `GFG {` `    ``// Function to count the number of ways``    ``// to choose the triples``    ``static` `int` `countWays(``int``[] arr, ``int` `n)``    ``{``        ``// compute the max value in the array``        ``// and create frequency array of size``        ``// max_val + 1.``        ``// We can also use HashMap to store``        ``// frequencies. We have used an array``        ``// to keep remaining code simple.``        ``int` `max_val = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``max_val = Math.max(max_val, arr[i]);``        ``int``[] freq = ``new` `int``[max_val + ``1``];``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``freq[arr[i]]++;` `        ``int` `ans = ``0``; ``// stores the number of ways` `        ``// Case 1: 0, 0, 0``        ``ans += freq[``0``] * (freq[``0``] - ``1``) * (freq[``0``] - ``2``) / ``6``;` `        ``// Case 2: 0, x, x``        ``for` `(``int` `i = ``1``; i <= max_val; i++)``            ``ans += freq[``0``] * freq[i] * (freq[i] - ``1``) / ``2``;` `        ``// Case 3: x, x, 2*x``        ``for` `(``int` `i = ``1``; ``2` `* i <= max_val; i++)``            ``ans += freq[i] * (freq[i] - ``1``) / ``2` `* freq[``2` `* i];` `        ``// Case 4: x, y, x + y``        ``// iterate through all pairs (x, y)``        ``for` `(``int` `i = ``1``; i <= max_val; i++) {``            ``for` `(``int` `j = i + ``1``; i + j <= max_val; j++)``                ``ans += freq[i] * freq[j] * freq[i + j];``        ``}` `        ``return` `ans;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] arr = ``new` `int``[] { ``1``, ``2``, ``3``, ``4``, ``5` `};``        ``int` `n = arr.length;``        ``System.out.println(countWays(arr, n));``    ``}``}`

## Python3

 `# Python3 program to count Triplets such``# that at least one of the numbers can be``# written as sum of the other two``import` `math as mt` `# Functoin to count the number of ways``# to choose the triples``def` `countWays(arr, n):` `    ``# compute the max value in the array``    ``# and create frequency array of size``    ``# max_val + 1.``    ``# We can also use HashMap to store``    ``# frequencies. We have used an array``    ``# to keep remaining code simple.``    ``max_val ``=` `0``    ``for` `i ``in` `range``(n):``        ``max_val ``=` `max``(max_val, arr[i])` `    ``freq ``=` `[``0` `for` `i ``in` `range``(max_val ``+` `1``)]` `    ``for` `i ``in` `range``(n):``        ``freq[arr[i]] ``+``=` `1` `    ``ans ``=` `0` `# stores the number of ways` `    ``# Case 1: 0, 0, 0``    ``ans ``+``=` `(freq[``0``] ``*` `(freq[``0``] ``-` `1``) ``*``           ``(freq[``0``] ``-` `2``) ``/``/` `6``)` `    ``# Case 2: 0, x, x``    ``for` `i ``in` `range``(``1``, max_val ``+` `1``):``        ``ans ``+``=` `(freq[``0``] ``*` `freq[i] ``*``               ``(freq[i] ``-` `1``) ``/``/` `2``)` `    ``# Case 3: x, x, 2*x``    ``for` `i ``in` `range``(``1``, (max_val ``+` `1``) ``/``/` `2``):``        ``ans ``+``=` `(freq[i] ``*``               ``(freq[i] ``-` `1``) ``/``/` `2` `*` `freq[``2` `*` `i])` `    ``# Case 4: x, y, x + y``    ``# iterate through all pairs (x, y)``    ``for` `i ``in` `range``(``1``, max_val ``+` `1``):``        ``for` `j ``in` `range``(i ``+` `1``, max_val ``-` `i ``+` `1``):``            ``ans ``+``=` `freq[i] ``*` `freq[j] ``*` `freq[i ``+` `j]` `    ``return` `ans` `# Driver code``arr ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5``]``n ``=` `len``(arr)``print``(countWays(arr, n))` `# This code is contributed by``# mohit kumar 29`

## C#

 `// C# program to count Triplets``// such that at least one of the``// numbers can be written as sum``// of the other two``using` `System;` `class` `GFG``{` `// Function to count the number``// of ways to choose the triples``static` `int` `countWays(``int``[] arr, ``int` `n)``{``    ``// compute the max value in the array``    ``// and create frequency array of size``    ``// max_val + 1.``    ``// We can also use HashMap to store``    ``// frequencies. We have used an array``    ``// to keep remaining code simple.``    ``int` `max_val = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``max_val = Math.Max(max_val, arr[i]);``        ` `    ``int``[] freq = ``new` `int``[max_val + 1];``    ``for` `(``int` `i = 0; i < n; i++)``        ``freq[arr[i]]++;` `    ``int` `ans = 0; ``// stores the number of ways` `    ``// Case 1: 0, 0, 0``    ``ans += freq * (freq - 1) *``                     ``(freq - 2) / 6;` `    ``// Case 2: 0, x, x``    ``for` `(``int` `i = 1; i <= max_val; i++)``        ``ans += freq * freq[i] *``                        ``(freq[i] - 1) / 2;` `    ``// Case 3: x, x, 2*x``    ``for` `(``int` `i = 1;``             ``2 * i <= max_val; i++)``        ``ans += freq[i] * (freq[i] - 1) /``                      ``2 * freq[2 * i];` `    ``// Case 4: x, y, x + y``    ``// iterate through all pairs (x, y)``    ``for` `(``int` `i = 1; i <= max_val; i++)``    ``{``        ``for` `(``int` `j = i + 1;``                 ``i + j <= max_val; j++)``            ``ans += freq[i] * freq[j] *``                             ``freq[i + j];``    ``}` `    ``return` `ans;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int``[] arr = { 1, 2, 3, 4, 5 };``    ``int` `n = arr.Length;``    ``Console.WriteLine(countWays(arr, n));``}``}` `// This code is contributed by shs..`

## PHP

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## Javascript

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Output:
`4`

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