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

  • Last Updated : 11 Jan, 2022

Given an 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.
Time complexity: O(n^3)
Efficient approach: The idea is similar to Find a triplet that sum to a given value.

  • 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)

Another Approach: The idea is similar to previous approach.

  1. Sort the given array.
  2. Start a nested loop, fixing the first element i(from 0 to n-1) and moving the other one j (from i+1 to n-1).
  3. Take the sum of both the elements and search it in the remaining array using Binary Search.

C++




// C++ program to find three numbers
// such that sum of two makes the
// third element in array
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
  
// Function to perform binary search
bool search(int sum, int start, 
            int end, int arr[])
{
    while (start <= end) 
    {
        int mid = (start + end) / 2;
        if (arr[mid] == sum) 
        {
            return true;
        }
        else if (arr[mid] > sum) 
        {
            end = mid - 1;
        }
        else 
        {
            start = mid + 1;
        }
    }
    return false;
}
  
// Function to find the triplets
void findTriplet(int arr[], int n)
{
    // Sorting the array
    sort(arr, arr + n);
  
    // Initialising nested loops
    for (int i = 0; i < n; i++) 
    {
        for (int j = i + 1; j < n; j++) 
        {
            // Finding the sum of the numbers
            if (search((arr[i] + arr[j]), 
                j, n - 1, arr)) 
            {
                // Printing out the first triplet
                cout << "Numbers are: " << arr[i] << 
                        " " << arr[j] << " " << 
                        (arr[i] + arr[j]);
                return;
            }
        }
    }
    // If no such triplets are found
    cout << "No such numbers exist" << endl;
}
  
// 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;
}
// This code is contributed by Sarthak Delori

Time Complexity: O(N^2*log N)

Space Complexity:  O(1)

Please refer complete article on Find a triplet such that sum of two equals to third element for more details!


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