# Program For Finding A Triplet From Three Linked Lists With Sum Equal To A Given Number

• Last Updated : 21 Dec, 2021

Given three linked lists, say a, b and c, find one node from each list such that the sum of the values of the nodes is equal to a given number.
For example, if the three linked lists are 12->6->29, 23->5->8, and 90->20->59, and the given number is 101, the output should be triple “6 5 90”.
In the following solutions, size of all three linked lists is assumed same for simplicity of analysis. The following solutions work for linked lists of different sizes also.

A simple method to solve this problem is to run three nested loops. The outermost loop picks an element from list a, the middle loop picks an element from b and the innermost loop picks from c. The innermost loop also checks whether the sum of values of current nodes of a, b and c is equal to given number. The time complexity of this method will be O(n^3).
Sorting can be used to reduce the time complexity to O(n*n). Following are the detailed steps.
1) Sort list b in ascending order, and list c in descending order.
2) After the b and c are sorted, one by one pick an element from list a and find the pair by traversing both b and c. See isSumSorted() in the following code. The idea is similar to Quadratic algorithm of 3 sum problem.

Following code implements step 2 only. The solution can be easily modified for unsorted lists by adding the merge sort code discussed here

## C#

 `// C# program to find a triplet ``// from three linked lists with``// sum equal to a given number``using` `System;`` ` `public` `class` `LinkedList``{``    ``public` `Node head; ``// head of list`` ` `    ``/* Linked list Node*/``    ``public` `class` `Node``    ``{``        ``public` `int` `data;``        ``public` `Node next;``        ``public` `Node(``int` `d) ``        ``{``            ``data = d; next = ``null``;``        ``}``    ``}`` ` `    ``/* A function to check if there ``    ``are three elements in a, b``    ``and c whose sum is equal to ``    ``givenNumber. The function``    ``assumes that the list b is ``    ``sorted in ascending order and``    ``c is sorted in descending order. */``bool` `isSumSorted(LinkedList la, LinkedList lb, ``                ``LinkedList lc, ``int` `givenNumber)``{``    ``Node a = la.head;`` ` `    ``// Traverse all nodes of la``    ``while` `(a != ``null``)``    ``{``        ``Node b = lb.head;``        ``Node c = lc.head;`` ` `        ``// for every node in la pick ``        ``// 2 nodes from lb and lc``        ``while` `(b != ``null` `&& c!=``null``)``        ``{``            ``int` `sum = a.data + b.data + c.data;``            ``if` `(sum == givenNumber)``            ``{``                ``Console.WriteLine(``"Triplet found "` `+ a.data +``                                    ``" "` `+ b.data + ``" "` `+ c.data);``                ``return` `true``;``            ``}`` ` `            ``// If sum is smaller then ``            ``// look for greater value of b``            ``else` `if` `(sum < givenNumber)``                ``b = b.next;`` ` `            ``else``                ``c = c.next;``        ``}``        ``a = a.next;``    ``}``    ``Console.WriteLine(``"No Triplet found"``);``    ``return` `false``;``}`` ` ` ` `    ``/* Given a reference (pointer to pointer) to the head``    ``of a list and an int, push a new node on the front``    ``of the list. */``    ``void` `push(``int` `new_data)``    ``{``        ``/* 1 & 2: Allocate the Node &``                ``Put in the data*/``        ``Node new_node = ``new` `Node(new_data);`` ` `        ``/* 3. Make next of new Node as head */``        ``new_node.next = head;`` ` `        ``/* 4. Move the head to point to new Node */``        ``head = new_node;``    ``}`` ` `    ``/* Driver code*/``    ``public` `static` `void` `Main(String []args)``    ``{``        ``LinkedList llist1 = ``new` `LinkedList();``        ``LinkedList llist2 = ``new` `LinkedList();``        ``LinkedList llist3 = ``new` `LinkedList();`` ` `        ``/* Create Linked List llist1 100->15->5->20 */``        ``llist1.push(20);``        ``llist1.push(5);``        ``llist1.push(15);``        ``llist1.push(100);`` ` `        ``/*create a sorted linked list 'b' 2->4->9->10 */``        ``llist2.push(10);``        ``llist2.push(9);``        ``llist2.push(4);``        ``llist2.push(2);`` ` `        ``/*create another sorted linked list 'c' 8->4->2->1 */``        ``llist3.push(1);``        ``llist3.push(2);``        ``llist3.push(4);``        ``llist3.push(8);`` ` `        ``int` `givenNumber = 25;``        ``llist1.isSumSorted(llist1,llist2,llist3,givenNumber);``    ``}``}`` ` `// This code contributed by Rajput-Ji`

Output:

`Triplet Found: 15 2 8`

Time complexity: The linked lists b and c can be sorted in O(nLogn) time using Merge Sort (See this). The step 2 takes O(n*n) time. So the overall time complexity is O(nlogn) + O(nlogn) + O(n*n) = O(n*n).
In this approach, the linked lists b and c are sorted first, so their original order will be lost. If we want to retain the original order of b and c, we can create copy of b and c.

Please refer complete article on Find a triplet from three linked lists with sum equal to a given number for more details!

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