Open In App

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

Last Updated : 23 Jul, 2025
Comments
Improve
Suggest changes
Like Article
Like
Report

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++ 
#include <bits/stdc++.h>
using namespace std;

// CPP program to find a triplet  from three linked lists
// with sum equal to a given number

/* Linked list Node*/
class Node {
  public:
  int data;
  Node* next;
  Node(int d)
  {
    this->data = d;
    this->next = NULL;
  }
};

Node* head; // head of list

/* 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(Node*& la, Node* lb, Node* lc,
                 int givenNumber)
{
  Node* a = la;

  // Traverse all nodes of la
  while (a != NULL) {
    Node* b = lb;
    Node* c = lc;

    // 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) {
        cout << "Triplet Found: " << a->data << " "
          << b->data << " " << c->data << endl;
        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;
  }
  cout << "No Triplet found" << endl;
  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(Node*& list, 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 = list;

  /* 4. Move the head to point to new Node */
  list = new_node;
}

int main()
{

  Node* list1 = new Node(20);
  Node* list2 = new Node(10);
  Node* list3 = new Node(1);

  /* Create Linked List llist1 100->15->5->20 */
  // list1.push(20);
  push(list1, 5);
  push(list1, 15);
  push(list1, 100);

  /*create a sorted linked list 'b' 2->4->9->10 */
  // list2.push(10);
  push(list2, 9);
  push(list2, 4);
  push(list2, 2);

  /*create another sorted linked list 'c' 8->4->2->1
     */
  // list3.push(1);
  push(list3, 2);
  push(list3, 4);
  push(list3, 8);

  int givenNumber = 25;
  isSumSorted(list1, list2, list3, givenNumber);
  return 0;
}

// This code is contributed by akashish__
Java 
// Java program to find a triplet  from three linked lists
// with sum equal to a given number
public class LinkedList {

    /* Linked list Node*/
    public class Node {
        int data;
        Node next;
        public Node(int d)
        {
            this.data = d;
            next = null;
        }
    }

    public Node head; // head of list

    /* 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.
     */
    public 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. */
    public 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;
    }

    public static void main(String[] args)
    {

        LinkedList list1 = new LinkedList();
        LinkedList list2 = new LinkedList();
        LinkedList list3 = new LinkedList();

        /* Create Linked List llist1 100->15->5->20 */
        list1.push(20);
        list1.push(5);
        list1.push(15);
        list1.push(100);

        /*create a sorted linked list 'b' 2->4->9->10 */
        list2.push(10);
        list2.push(9);
        list2.push(4);
        list2.push(2);

        /*create another sorted linked list 'c' 8->4->2->1
         */
        list3.push(1);
        list3.push(2);
        list3.push(4);
        list3.push(8);

        int givenNumber = 25;
        list1.isSumSorted(list1, list2, list3, givenNumber);
    }
}

// This code is contributed by lokesh (lokeshmvs21).
Python3 
# Python program to find a triplet 
# from three linked lists with
# sum equal to a given number

# Node class
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

# Linked List class
class LinkedList:
    def __init__(self):
        self.head = None

    # function to add new node at the beginning
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

    # 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.
    def isSumSorted(self, la, lb, lc, givenNumber):
        a = la.head

        # Traverse all nodes of la
        while a is not None:
            b = lb.head
            c = lc.head

            # for every node in la pick 
            # 2 nodes from lb and lc
            while b is not None and c is not None:
                sum = a.data + b.data + c.data
                if sum == givenNumber:
                    print("Triplet found", a.data, b.data, c.data)
                    return True

                # If sum is smaller then 
                # look for greater value of b
                elif sum < givenNumber:
                    b = b.next

                else:
                    c = c.next
            a = a.next
        print("No Triplet found")
        return False

# Driver code
if __name__ == '__main__':
    llist1 = LinkedList()
    llist2 = LinkedList()
    llist3 = 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)

    givenNumber = 25
    llist1.isSumSorted(llist1, llist2, llist3, givenNumber)
    
# This code is contributed by akashish__
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
JavaScript 
// Javascript code for the above approach
class Node {
  constructor(data) {
    this.data = data;
    this.next = null;
  }
}

class LinkedList {
  constructor() {
    this.head = null;
  }

  // function to add new node at the beginning
  push(newData) {
    const newNode = new Node(newData);
    newNode.next = this.head;
    this.head = newNode;
  }

  // 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.
  isSumSorted(la, lb, lc, givenNumber) {
    let a = la.head;

    // Traverse all nodes of la
    while (a !== null) {
      let b = lb.head;
      let c = lc.head;

      // for every node in la pick 
      // 2 nodes from lb and lc
      while (b !== null && c !== null) {
        const sum = a.data + b.data + c.data;
        if (sum === givenNumber) {
          console.log("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.log("No Triplet found");
    return false;
  }
}

// Driver code
if (typeof exports !== "undefined") {
  exports.LinkedList = LinkedList;
}

// usage
const llist1 = new LinkedList();
const llist2 = new LinkedList();
const 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);

const givenNumber = 25;
llist1.isSumSorted(llist1, llist2, llist3, givenNumber);

// This code is contributed by lokeshpotta20.

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!


K
kartik
Improve
Article Tags :

Explore

Lightbox