Program For Finding A Triplet From Three Linked Lists With Sum Equal To A Given Number
Last Updated :
18 Jan, 2023
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 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).
# 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# 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 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!
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