Find a triplet from three linked lists with sum equal to a given number
Last Updated :
28 Feb, 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++
// C++ program to find a triplet
// from three linked lists with
// sum equal to a given number
#include <bits/stdc++.h>
using namespace std;
/* Link list node */
class Node
{
public:
int data;
Node* next;
};
/* A utility function to insert
a node at the beginning of a
linked list*/
void push (Node** head_ref, int new_data)
{
/* allocate node */
Node* new_node = new Node();
/* put in the data */
new_node->data = new_data;
/* link the old list of the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = 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. */
bool isSumSorted(Node *headA, Node *headB,
Node *headC, int givenNumber)
{
Node *a = headA;
// Traverse through all nodes of a
while (a != NULL)
{
Node *b = headB;
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)
{
cout << "Triplet Found: " << 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
}
cout << "No such triplet";
return false;
}
/* Driver code*/
int main()
{
/* Start with the empty list */
Node* headA = NULL;
Node* headB = NULL;
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;
}
// This code is contributed by rathbhupendra
// 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 of the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = 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. */
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;
}
/* Driver 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 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 */
# Python program to find a triplet
# from three linked lists with
# sum equal to a given number
# Link list node
class Node:
def __init__(self, new_data):
self.data = new_data
self.next = None
# A utility function to insert
# a node at the beginning of a
# linked list
def push ( head_ref, new_data) :
# allocate node
new_node = Node(0)
# put in the data
new_node.data = new_data
# link the old list of the new node
new_node.next = (head_ref)
# move the head to point to the new node
(head_ref) = new_node
return head_ref;
# 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(headA, headB,headC, givenNumber) :
a = headA
# Traverse through all nodes of a
while (a != None) :
b = headB
c = headC
# For every node of list a, prick two nodes
# from lists b abd c
while (b != None and c != None) :
# If this a triplet with given sum, print
# it and return true
sum = a.data + b.data + c.data
if (sum == givenNumber) :
print "Triplet Found: " , a.data , " " , b.data , " " , c.data,
return True
# If sum of this triplet is smaller, look for
# greater values in b
elif (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
print("No such triplet")
return False
# Driver code
# Start with the empty list
headA = None
headB = None
headC = None
# create a linked list 'a' 10.15.5.20
headA = push (headA, 20)
headA = push (headA, 4)
headA = push (headA, 15)
headA = push (headA, 10)
# create a sorted linked list 'b' 2.4.9.10
headB = push (headB, 10)
headB = push (headB, 9)
headB = push (headB, 4)
headB = push (headB, 2)
# create another sorted
# linked list 'c' 8.4.2.1
headC = push (headC, 1)
headC = push (headC, 2)
headC = push (headC, 4)
headC = push (headC, 8)
givenNumber = 25
isSumSorted (headA, headB, headC, givenNumber)
# This code is contributed by Arnab Kundu
// 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
<script>
// Javascript program to find a triplet from three linked lists with
// sum equal to a given number
/* Linked list Node*/
class Node
{
constructor(d)
{
this.data = d;
this.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. */
function isSumSorted(la,lb,lc,givenNumber)
{
let a = la;
// Traverse all nodes of la
while (a != null)
{
let b = lb;
let c = lc;
// for every node in la pick 2 nodes from lb and lc
while (b != null && c!=null)
{
let sum = a.data + b.data + c.data;
if (sum == givenNumber)
{
document.write("Triplet found " + a.data +
" " + b.data + " " + c.data+"<br>");
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;
}
document.write("No Triplet found<br>");
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. */
function push(head_ref,new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
let new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = (head_ref);
(head_ref) = new_node;
return head_ref;
}
let headA =null;
headA = push (headA, 20)
headA = push (headA, 4)
headA = push (headA, 15)
headA = push (headA, 10)
// create a sorted linked list 'b' 2.4.9.10
let headB = null;
headB = push (headB, 10)
headB = push (headB, 9)
headB = push (headB, 4)
headB = push (headB, 2)
// create another sorted
// linked list 'c' 8.4.2.1
let headC = null;
headC = push (headC, 1)
headC = push (headC, 2)
headC = push (headC, 4)
headC = push (headC, 8)
let givenNumber = 25
isSumSorted (headA, headB, headC, givenNumber)
// This code is contributed by avanitrachhadiya2155
</script>
Output:
Triplet Found: 15 2 8
Time complexity: O(n2)
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.
Auxiliary Space: O(1)
Here constant extra space is used, but if we wish to maintain the original order of lists b and c our time complexity will become O(n) as extra space will be used to store the sorted list elements.
This article is compiled by Abhinav Priyadarshi and reviewed by GeeksforGeeks team.
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