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

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;
}

Java

import java.util.Arrays;

public class Triplet {
    // Utility function for finding triplet in array
    public static void findTriplet(int[] arr, int n)
    {
        // Sort the array
        Arrays.sort(arr);
        // For every element in arr check
        // if a pair exists (in array) whose
        // sum is equal to the third 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
                    System.out.println(
                        "numbers are " + arr[i] + " "
                        + arr[j] + " " + arr[k]);
                    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
        System.out.println("No such triplet exists");
    }

    public static void main(String[] args)
    {
        int[] arr = { 5, 32, 1, 7, 10, 50, 19, 21, 2 };
        int n = arr.length;
        findTriplet(arr, n);
    }
}

Output:

numbers are 21 2 19

Time complexity: O(N^2)

Auxiliary Space: O(1)

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

Java

import java.util.Arrays;

// Class to find three numbers such that
// the sum of two makes the third element in array
public class TripletSum {
    // Function to perform binary search
    public static boolean search(int sum, int start,
                                 int end, int arr[])
    {
        // Perform binary search
        while (start <= end) {
            int mid = (start + end) / 2;
            if (arr[mid] == sum) {
                // Return true if sum is found
                return true;
            }
            else if (arr[mid] > sum) {
                end = mid - 1;
            }
            else {
                start = mid + 1;
            }
        }
        // Return false if sum is not found
        return false;
    }
    // Function to find the triplets
    public static void findTriplet(int arr[], int n)
    {
        // Sorting the array
        Arrays.sort(arr);

        // Initializing 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
                    System.out.println(
                        "Numbers are: " + arr[i] + " "
                        + arr[j] + " " + (arr[i] + arr[j]));
                    return;
                }
            }
        }
        // If no such triplets are found
        System.out.println("No such numbers exist");
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 5, 32, 1, 7, 10, 50, 19, 21, 2 };
        int n = arr.length;
        findTriplet(arr, n);
    }
}

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

Space Complexity:  O(1)

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