Count triplets such that sum of any two number is equal to third

Last Updated : 24 Apr, 2025

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

Examples:

Input: arr[] = [1, 5, 3, 2]
Output: 2
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: 0
Explanation: In the given array there are no such triplets such that sum of two numbers is equal to the third number.

We have already discussed, how to check if there is a triplet such that sum of two is equal to the third. In this article, we are going to focus on approaches to count triplets.

[Naive Approach] Generating all triplets - O(n ^ 3) time and O(1) space

Generate all the triplets of the given array and check the sum of two elements to equal the third element.

C++

#include <iostream>
#include<vector>
using namespace std;

int cntTriplets(vector<int>&arr, int n)
{
    int count = 0;

    // Loop to count for triplets
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {

            for (int k = j + 1; k < n; k++) {
                if (arr[i] + arr[j] == arr[k]) {
                    count++;
                }
                else if (arr[i] + arr[k] == arr[j]) {
                    count++;
                }
                else if (arr[j] + arr[k] == arr[i]) {
                    count++;
                }
            }
        }
    }
    return count;
}

int main()
{
    int n = 4;
    vector<int> arr = { 1, 5, 3, 2 };

    cout << cntTriplets(arr, n);
    return 0;
}

Java

import java.util.List;
import java.util.ArrayList;

public class GfG{
    public static int cntTriplets(List<Integer> arr, int n) {
        int count = 0;

        // Loop to count for triplets
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                for (int k = j + 1; k < n; k++) {
                    if (arr.get(i) + arr.get(j) == arr.get(k)) {
                        count++;
                    } else if (arr.get(i) + arr.get(k) == arr.get(j)) {
                        count++;
                    } else if (arr.get(j) + arr.get(k) == arr.get(i)) {
                        count++;
                    }
                }
            }
        }
        return count;
    }

    public static void main(String[] args) {
        int n = 4;
        List<Integer> arr = new ArrayList<>();
        arr.add(1);
        arr.add(5);
        arr.add(3);
        arr.add(2);

        System.out.println(cntTriplets(arr, n));
    }
}

Python

def cntTriplets(arr, n):
    count = 0

    # Loop to count for triplets
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                if arr[i] + arr[j] == arr[k]:
                    count += 1
                elif arr[i] + arr[k] == arr[j]:
                    count += 1
                elif arr[j] + arr[k] == arr[i]:
                    count += 1
    return count

n = 4
arr = [1, 5, 3, 2]

print(cntTriplets(arr, n))

using System;
using System.Collections.Generic;

class Program {
    public static int cntTriplets(List<int> arr, int n) {
        int count = 0;

        // Loop to count for triplets
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                for (int k = j + 1; k < n; k++) {
                    if (arr[i] + arr[j] == arr[k]) {
                        count++;
                    } else if (arr[i] + arr[k] == arr[j]) {
                        count++;
                    } else if (arr[j] + arr[k] == arr[i]) {
                        count++;
                    }
                }
            }
        }
        return count;
    }

    static void Main() {
        int n = 4;
        List<int> arr = new List<int> { 1, 5, 3, 2 };

        Console.WriteLine(cntTriplets(arr, n));
    }
}

JavaScript

function cntTriplets(arr, n) {
    let count = 0;

    // Loop to count for triplets
    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
            for (let k = j + 1; k < n; k++) {
                if (arr[i] + arr[j] === arr[k]) {
                    count++;
                } else if (arr[i] + arr[k] === arr[j]) {
                    count++;
                } else if (arr[j] + arr[k] === arr[i]) {
                    count++;
                }
            }
        }
    }
    return count;
}

const n = 4;
const arr = [1, 5, 3, 2];

console.log(cntTriplets(arr, n));

Output

[Expected Approach] Using Hash Set - O(n ^ 2) time and O(n) space

The idea is to using a frequency Hash Map or Dictionary to store the count of each element in the array. Then, it iterates over all pairs of elements (arr[i], arr[j]) and checks if their sum exists in the array using the frequency map.

Algorithm:

Declare a Hash Set to store all numbers.
Run two loops to choose two different indexes of the matrix and check if the sum of the elements at those indices is present in t or not.

C++

#include <iostream>
#include <vector>
#include <unordered_set>
using namespace std;

int cntTriplets(vector<int>& arr) {
    int n = arr.size();
    unordered_set<int> s;

    // Add all elements to set
    for (int i = 0; i < n; i++) {
        s.insert(arr[i]);
    }

    int count = 0;

    // Check for each pair if their sum exists 
    // in the set
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            int sum = arr[i] + arr[j];
            
            // If sum is present. Please note that
            // this works because the question says
            // all elements are distinct
            if (s.find(sum) != s.end()) {
                count++;
            }
        }
    }

    return count;
}

int main() {
    vector<int> arr = {1, 5, 3, 2};
    cout << cntTriplets(arr);
    return 0;
}

Java

import java.util.HashSet;
import java.util.Set;

public class Main {
    public static int cntTriplets(int[] arr) {
        int n = arr.length;
        Set<Integer> s = new HashSet<>();

        // Add all elements to set
        for (int i = 0; i < n; i++) {
            s.add(arr[i]);
        }

        int count = 0;

        // Check for each pair if their sum exists 
        // in the set
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int sum = arr[i] + arr[j];
                
                // If sum is present. Please note that
                // this works because the question says
                // all elements are distinct
                if (s.contains(sum)) {
                    count++;
                }
            }
        }

        return count;
    }

    public static void main(String[] args) {
        int[] arr = {1, 5, 3, 2};
        System.out.println(cntTriplets(arr));
    }
}

Python

def cnt_triplets(arr):
    n = len(arr)
    s = set()

    # Add all elements to set
    for i in range(n):
        s.add(arr[i])

    count = 0

    # Check for each pair if their sum exists 
    # in the set
    for i in range(n):
        for j in range(i + 1, n):
            sum_ = arr[i] + arr[j]
            
            # If sum is present. Please note that
            # this works because the question says
            # all elements are distinct
            if sum_ in s:
                count += 1

    return count

arr = [1, 5, 3, 2]
print(cnt_triplets(arr))

using System;
using System.Collections.Generic;

class Program {
    public static int CntTriplets(int[] arr) {
        int n = arr.Length;
        HashSet<int> s = new HashSet<int>();

        // Add all elements to set
        for (int i = 0; i < n; i++) {
            s.Add(arr[i]);
        }

        int count = 0;

        // Check for each pair if their sum exists 
        // in the set
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int sum = arr[i] + arr[j];
                
                // If sum is present. Please note that
                // this works because the question says
                // all elements are distinct
                if (s.Contains(sum)) {
                    count++;
                }
            }
        }

        return count;
    }

    static void Main() {
        int[] arr = {1, 5, 3, 2};
        Console.WriteLine(CntTriplets(arr));
    }
}

JavaScript

function cntTriplets(arr) {
    const n = arr.length;
    const s = new Set();

    // Add all elements to set
    for (let i = 0; i < n; i++) {
        s.add(arr[i]);
    }

    let count = 0;

    // Check for each pair if their sum exists 
    // in the set
    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
            const sum = arr[i] + arr[j];
            
            // If sum is present. Please note that
            // this works because the question says
            // all elements are distinct
            if (s.has(sum)) {
                count++;
            }
        }
    }

    return count;
}

const arr = [1, 5, 3, 2];
console.log(cntTriplets(arr));

Output

Count prime triplets upto N having sum of first two elements equal to the third number

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Article Tags :

Practice Tags :

Count triplets such that sum of any two number is equal to third

[Naive Approach] Generating all triplets - O(n ^ 3) time and O(1) space

[Expected Approach] Using Hash Set - O(n ^ 2) time and O(n) space

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