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:

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:

👁 Image

Below is the implementation of the above approach:

numbers are 21 2 19

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

Java Program to Find a triplet such that sum of two equals to third element using Binary Search:

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.

Below is the implementation of the above approach:

Numbers are: 2 5 7

Time Complexity: O(N^2*log N)
Auxiliary Space: O(1)

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