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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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