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.
#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!