# Count Triplets such that one of the numbers can be written as sum of the other two

• Difficulty Level : Hard
• Last Updated : 30 Jun, 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#includeusing namespace std;     // Function 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[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    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 twoimport math as mt # Function to count the number of ways# to choose the triplesdef 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 codearr = [ 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 twousing System; class GFG{ // Function to count the number// of ways to choose the triplesstatic 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 codepublic 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..



## Javascript



Output:

4

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