Introduction to C++ Programming Language

C++ Program to Find a triplet such that sum of two equals to third element

Last Updated : 23 Jul, 2025

Write a C++ program for a given array of integers, you have to find three numbers such that the sum of two elements equals the third element.
Examples:

Input: {5, 32, 1, 7, 10, 50, 19, 21, 2}
Output: 21, 2, 19

Input: {5, 32, 1, 7, 10, 50, 19, 21, 0}
Output: no such triplet exist

Question source: Arcesium Interview Experience | Set 7 (On campus for Internship)

Simple approach:

Run three loops and check if there exists a triplet such that sum of two elements equals the third element.

Below is the implementation of the above approach:

C++

// CPP program to find three numbers
// such that sum of two makes the
// third element in array
#include <bits/stdc++.h>
using namespace std;

// Utility function for finding
// triplet in array
void findTriplet(int arr[], int n)
{
    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])
                    || (arr[i] + arr[k] == arr[j])
                    || (arr[j] + arr[k] == arr[i])) {
                    // printing out the first triplet
                    cout << "Numbers are: " << arr[i] << " "
                         << arr[j] << " " << arr[k];
                    return;
                }
            }
        }
    }

    // No such triplet is found in array
    cout << "No such triplet exists";
}

// driver program
int main()
{
    int arr[] = { 5, 32, 1, 7, 10, 50, 19, 21, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);

    findTriplet(arr, n);
    return 0;
}

Output

Numbers are: 5 7 2

Time complexity: O(n^3)
Auxiliary Space: O(1)

Efficient approach:

The idea is similar to Find a triplet that sum to a given value.

Step-by-step approach:

Sort the given array first.
Start fixing the greatest element of three from the back and traverse the array to find the other two numbers which sum up to the third element.
Take two pointers j(from front) and k(initially i-1) to find the smallest of the two number and from i-1 to find the largest of the two remaining numbers
If the addition of both the numbers is still less than A[i], then we need to increase the value of the summation of two numbers, thereby increasing the j pointer, so as to increase the value of A[j] + A[k].
If the addition of both the numbers is more than A[i], then we need to decrease the value of the summation of two numbers, thereby decrease the k pointer so as to decrease the overall value of A[j] + A[k].

Below image is a dry run of the above approach:

Below is the implementation of the above approach:

C++

// C++ program to find three numbers
// such that sum of two makes the
// third element in array
#include <bits/stdc++.h>
using namespace std;

// Utility function for finding
// triplet in array
void findTriplet(int arr[], int n)
{
    // Sort the array
    sort(arr, arr + n);

    // For every element in arr check 
    // if a pair exist(in array) whose
    // sum is equal to arr element
    for (int i = n - 1; i >= 0; i--) 
    {
        int j = 0;
        int k = i - 1;

        // Iterate forward and backward to 
        // find the other two elements
        while (j < k) 
        {
            // If the two elements sum is
            // equal to the third element
            if (arr[i] == arr[j] + arr[k]) 
            {
                // Pair found
                cout << "numbers are " << arr[i] << 
                        " " << arr[j] << " " << 
                        arr[k] << endl;
                return;
            }

            // If the element is greater than
            // sum of both the elements, then try
            // adding a smaller number to reach the
            // equality
            else if (arr[i] > arr[j] + arr[k])
                j += 1;

            // If the element is smaller, then
            // try with a smaller number
            // to reach equality, so decrease K
            else
                k -= 1;
        }
    }

    // No such triplet is found in array
    cout << "No such triplet exists";
}

// Driver code
int main()
{
    int arr[] = {5, 32, 1, 7, 10, 
                 50, 19, 21, 2};
    int n = sizeof(arr) / sizeof(arr[0]);
    findTriplet(arr, n);
    return 0;
}

Output:

numbers are 21 2 19

Time complexity: O(N^2)
Auxiliary Space: O(1)