Java 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.

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 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. */    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;     }       /* Driver 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.

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