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 tripel “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++

// C/C++ program to find a triplet from three linked lists with // sum equal to a given number #include<stdio.h> #include<stdlib.h> #include<stdbool.h> /* Link list node */ struct Node { int data; struct Node* next; }; /* A utility function to insert a node at the beginning of a linked list*/ void push (struct Node** head_ref, int new_data) { /* allocate node */ struct Node* new_node = (struct Node*) malloc(sizeof(struct Node)); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* A function to chech 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(struct Node *headA, struct Node *headB, struct Node *headC, int givenNumber) { struct Node *a = headA; // Traverse through all nodes of a while (a != NULL) { struct Node *b = headB; struct Node *c = headC; // For every node of list a, prick two nodes // from lists b abd c while (b != NULL && c != NULL) { // If this a triplet with given sum, print // it and return true int sum = a->data + b->data + c->data; if (sum == givenNumber) { printf ("Triplet Found: %d %d %d ", a->data, b->data, c->data); return true; } // If sum of this triplet is smaller, look for // greater values in b else if (sum < givenNumber) b = b->next; else // If sum is greater, look for smaller values in c c = c->next; } a = a->next; // Move ahead in list a } printf ("No such triplet"); return false; } /* Drier program to test above function*/ int main() { /* Start with the empty list */ struct Node* headA = NULL; struct Node* headB = NULL; struct Node* headC = NULL; /*create a linked list 'a' 10->15->5->20 */ push (&headA, 20); push (&headA, 4); push (&headA, 15); push (&headA, 10); /*create a sorted linked list 'b' 2->4->9->10 */ push (&headB, 10); push (&headB, 9); push (&headB, 4); push (&headB, 2); /*create another sorted linked list 'c' 8->4->2->1 */ push (&headC, 1); push (&headC, 2); push (&headC, 4); push (&headC, 8); int givenNumber = 25; isSumSorted (headA, headB, headC, givenNumber); return 0; }

## Java

// Java program to find a triplet from three linked lists with // sum equal to a given number class LinkedList { Node head; // head of list /* Linked list Node*/ class Node { int data; Node next; Node(int d) {data = d; next = null; } } /* A function to chech 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. */ boolean 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) { System.out.println("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; } System.out.println("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; } /* Drier program to test above functions */ 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 is contributed by Rajat Mishra */

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.

This article is compiled by **Abhinav Priyadarshi** and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above