Program For Finding A Triplet From Three Linked Lists With Sum Equal To A Given Number
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;
class Node {
public :
int data;
Node* next;
Node( int d)
{
this ->data = d;
this ->next = NULL;
}
};
Node* head;
bool isSumSorted(Node*& la, Node* lb, Node* lc,
int givenNumber)
{
Node* a = la;
while (a != NULL) {
Node* b = lb;
Node* c = 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 ;
}
else if (sum < givenNumber) {
b = b->next;
}
else {
c = c->next;
}
}
a = a->next;
}
cout << "No Triplet found" << endl;
return false ;
}
void push(Node*& list, int new_data)
{
Node* new_node = new Node(new_data);
new_node->next = list;
list = new_node;
}
int main()
{
Node* list1 = new Node(20);
Node* list2 = new Node(10);
Node* list3 = new Node(1);
push(list1, 5);
push(list1, 15);
push(list1, 100);
push(list2, 9);
push(list2, 4);
push(list2, 2);
push(list3, 2);
push(list3, 4);
push(list3, 8);
int givenNumber = 25;
isSumSorted(list1, list2, list3, givenNumber);
return 0;
}
|
Java
public class LinkedList {
public class Node {
int data;
Node next;
public Node( int d)
{
this .data = d;
next = null ;
}
}
public Node head;
public boolean isSumSorted(LinkedList la, LinkedList lb,
LinkedList lc,
int givenNumber)
{
Node a = la.head;
while (a != null ) {
Node b = lb.head;
Node c = lc.head;
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 ;
}
else if (sum < givenNumber) {
b = b.next;
}
else {
c = c.next;
}
}
a = a.next;
}
System.out.println( "No Triplet found" );
return false ;
}
public void push( int new_data)
{
Node new_node = new Node(new_data);
new_node.next = head;
head = new_node;
}
public static void main(String[] args)
{
LinkedList list1 = new LinkedList();
LinkedList list2 = new LinkedList();
LinkedList list3 = new LinkedList();
list1.push( 20 );
list1.push( 5 );
list1.push( 15 );
list1.push( 100 );
list2.push( 10 );
list2.push( 9 );
list2.push( 4 );
list2.push( 2 );
list3.push( 1 );
list3.push( 2 );
list3.push( 4 );
list3.push( 8 );
int givenNumber = 25 ;
list1.isSumSorted(list1, list2, list3, givenNumber);
}
}
|
Python3
class Node:
def __init__( self , data):
self .data = data
self . next = None
class LinkedList:
def __init__( self ):
self .head = None
def push( self , new_data):
new_node = Node(new_data)
new_node. next = self .head
self .head = new_node
def isSumSorted( self , la, lb, lc, givenNumber):
a = la.head
while a is not None :
b = lb.head
c = lc.head
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
elif sum < givenNumber:
b = b. next
else :
c = c. next
a = a. next
print ( "No Triplet found" )
return False
if __name__ = = '__main__' :
llist1 = LinkedList()
llist2 = LinkedList()
llist3 = LinkedList()
llist1.push( 20 )
llist1.push( 5 )
llist1.push( 15 )
llist1.push( 100 )
llist2.push( 10 )
llist2.push( 9 )
llist2.push( 4 )
llist2.push( 2 )
llist3.push( 1 )
llist3.push( 2 )
llist3.push( 4 )
llist3.push( 8 )
givenNumber = 25
llist1.isSumSorted(llist1, llist2, llist3, givenNumber)
|
C#
using System;
public class LinkedList
{
public Node head;
public class Node
{
public int data;
public Node next;
public Node( int d)
{
data = d; next = null ;
}
}
bool isSumSorted(LinkedList la, LinkedList lb,
LinkedList lc, int givenNumber)
{
Node a = la.head;
while (a != null )
{
Node b = lb.head;
Node c = lc.head;
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 ;
}
else if (sum < givenNumber)
b = b.next;
else
c = c.next;
}
a = a.next;
}
Console.WriteLine( "No Triplet found" );
return false ;
}
void push( int new_data)
{
Node new_node = new Node(new_data);
new_node.next = head;
head = new_node;
}
public static void Main(String []args)
{
LinkedList llist1 = new LinkedList();
LinkedList llist2 = new LinkedList();
LinkedList llist3 = new LinkedList();
llist1.push(20);
llist1.push(5);
llist1.push(15);
llist1.push(100);
llist2.push(10);
llist2.push(9);
llist2.push(4);
llist2.push(2);
llist3.push(1);
llist3.push(2);
llist3.push(4);
llist3.push(8);
int givenNumber = 25;
llist1.isSumSorted(llist1,llist2,llist3,givenNumber);
}
}
|
Javascript
class Node {
constructor(data) {
this .data = data;
this .next = null ;
}
}
class LinkedList {
constructor() {
this .head = null ;
}
push(newData) {
const newNode = new Node(newData);
newNode.next = this .head;
this .head = newNode;
}
isSumSorted(la, lb, lc, givenNumber) {
let a = la.head;
while (a !== null ) {
let b = lb.head;
let c = lc.head;
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 ;
} else if (sum < givenNumber) {
b = b.next;
} else {
c = c.next;
}
}
a = a.next;
}
console.log( "No Triplet found" );
return false ;
}
}
if ( typeof exports !== "undefined" ) {
exports.LinkedList = LinkedList;
}
const llist1 = new LinkedList();
const llist2 = new LinkedList();
const llist3 = new LinkedList();
llist1.push(20);
llist1.push(5);
llist1.push(15);
llist1.push(100);
llist2.push(10);
llist2.push(9);
llist2.push(4);
llist2.push(2);
llist3.push(1);
llist3.push(2);
llist3.push(4);
llist3.push(8);
const givenNumber = 25;
llist1.isSumSorted(llist1, llist2, llist3, givenNumber);
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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!
Last Updated :
18 Jan, 2023
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