Java 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.

Below is the implementation of the above approach:

Java

import java.util.*;
 
public class Main {
 
    // Utility function for finding triplet in array
    public static 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
                        System.out.println(
                            "Numbers are: " + arr[i] + " "
                            + arr[j] + " " + arr[k]);
                        return;
                    }
                }
            }
        }
        // No such triplet is found in array
        System.out.println("No such triplet exists");
    }
 
    // Driver program
    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: 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:

Java

// Java program to find three numbers
// such that sum of two makes the
// third element in array
import java.util.Arrays;
 
public class GFG 
{
    // Utility function for finding
    // triplet in array
    static void findTriplet(int arr[], int n)
    {
        // Sort the array
        Arrays.sort(arr);
 
        // 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;
            while (j < k) {
                if (arr[i] == arr[j] + arr[k]) 
                {
                    // Pair found
                    System.out.println("numbers are " + arr[i] + 
                                       " " + arr[j] + " " + arr[k]);
                    return;
                }
                else if (arr[i] > arr[j] + arr[k])
                    j += 1;
                else
                    k -= 1;
            }
        }
 
        // No such triplet is found in array
        System.out.println("No such triplet exists");
    }
 
    // 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);
    }
}
// This code is contributed by Sumit Ghosh

Output

numbers are 21 2 19

Time complexity: O(N^2)
Auxiliary Space: O(1) as no extra space has been used.