Given two linked lists, the task is to check whether the first list is present in 2nd list or not.
Examples:
Input: list1 = 10->20 list2 = 5->10->20 Output : LIST FOUND
Input: list1 = 1->2->3->4 list2 = 1->2->1->2->3->4 Output: LIST FOUND
Input: list1 = 1->2->3->4 list2 = 1->2->2->1->2->3 Output: LIST NOT FOUND
Approach 1:
- Take first node of second list.
- Start matching the first list from this first node.
- If whole lists match return true.
- Else break and take first list to the first node again.
- And take second list to its second node.
- Repeat these steps until any of linked lists becomes empty.
- If first list becomes empty then list found else not.
Implementation:
// C++ program to find a list in second list #include <bits/stdc++.h> using namespace std;
// A Linked List node struct Node
{ int data;
Node* next;
}; // Returns true if first list is present in second // list bool findList(Node* first, Node* second)
{ Node* ptr1 = first, *ptr2 = second;
// If both linked lists are empty, return true
if (first == NULL && second == NULL)
return true ;
// Else If one is empty and other is not return
// false
if ( first == NULL ||
(first != NULL && second == NULL))
return false ;
// Traverse the second list by picking nodes
// one by one
while (second != NULL)
{
// Initialize ptr2 with current node of second
ptr2 = second;
// Start matching first list with second list
while (ptr1 != NULL)
{
// If second list becomes empty and first
// not then return false
if (ptr2 == NULL)
return false ;
// If data part is same, go to next
// of both lists
else if (ptr1->data == ptr2->data)
{
ptr1 = ptr1->next;
ptr2 = ptr2->next;
}
// If not equal then break the loop
else break ;
}
// Return true if first list gets traversed
// completely that means it is matched.
if (ptr1 == NULL)
return true ;
// Initialize ptr1 with first again
ptr1 = first;
// And go to next node of second list
second = second->next;
}
return false ;
} /* Function to print nodes in a given linked list */ void printList(Node* node)
{ while (node != NULL)
{
printf ( "%d " , node->data);
node = node->next;
}
} // Function to add new node to linked lists Node *newNode( int key)
{ Node *temp = new Node;
temp-> data= key;
temp->next = NULL;
return temp;
} /* Driver program to test above functions*/ int main()
{ /* Let us create two linked lists to test
the above functions. Created lists shall be
a: 1->2->3->4
b: 1->2->1->2->3->4*/
Node *a = newNode(1);
a->next = newNode(2);
a->next->next = newNode(3);
a->next->next->next = newNode(4);
Node *b = newNode(1);
b->next = newNode(2);
b->next->next = newNode(1);
b->next->next->next = newNode(2);
b->next->next->next->next = newNode(3);
b->next->next->next->next->next = newNode(4);
findList(a,b) ? cout << "LIST FOUND" :
cout << "LIST NOT FOUND" ;
return 0;
} |
// Java program to find a list in second list import java.util.*;
public class GFG
{ // A Linked List node static class Node
{ int data;
Node next;
}; // Returns true if first list is // present in second list static boolean findList(Node first,
Node second)
{ Node ptr1 = first, ptr2 = second;
// If both linked lists are empty,
// return true
if (first == null && second == null )
return true ;
// Else If one is empty and
// other is not, return false
if (first == null ||
(first != null && second == null ))
return false ;
// Traverse the second list by
// picking nodes one by one
while (second != null )
{
// Initialize ptr2 with
// current node of second
ptr2 = second;
// Start matching first list
// with second list
while (ptr1 != null )
{
// If second list becomes empty and
// first not then return false
if (ptr2 == null )
return false ;
// If data part is same, go to next
// of both lists
else if (ptr1.data == ptr2.data)
{
ptr1 = ptr1.next;
ptr2 = ptr2.next;
}
// If not equal then break the loop
else break ;
}
// Return true if first list gets traversed
// completely that means it is matched.
if (ptr1 == null )
return true ;
// Initialize ptr1 with first again
ptr1 = first;
// And go to next node of second list
second = second.next;
}
return false ;
} /* Function to print nodes in a given linked list */ static void printList(Node node)
{ while (node != null )
{
System.out.printf( "%d " , node.data);
node = node.next;
}
} // Function to add new node to linked lists static Node newNode( int key)
{ Node temp = new Node();
temp.data= key;
temp.next = null ;
return temp;
} // Driver Code public static void main(String[] args)
{ /* Let us create two linked lists to test
the above functions. Created lists shall be
a: 1->2->3->4
b: 1->2->1->2->3->4*/
Node a = newNode( 1 );
a.next = newNode( 2 );
a.next.next = newNode( 3 );
a.next.next.next = newNode( 4 );
Node b = newNode( 1 );
b.next = newNode( 2 );
b.next.next = newNode( 1 );
b.next.next.next = newNode( 2 );
b.next.next.next.next = newNode( 3 );
b.next.next.next.next.next = newNode( 4 );
if (findList(a, b) == true )
System.out.println( "LIST FOUND" );
else
System.out.println( "LIST NOT FOUND" );
} } // This code is contributed by Princi Singh |
# Python3 program to find a list in second list class Node:
def __init__( self , value = 0 ):
self .value = value
self . next = None
# Returns true if first list is # present in second list def findList(first, second):
# If both linked lists are empty/None,
# return True
if not first and not second:
return True
# If ONLY one of them is empty,
# return False
if not first or not second:
return False
ptr1 = first
ptr2 = second
# Traverse the second LL by
# picking nodes one by one
while ptr2:
# Initialize 'ptr2' with current
# node of 'second'
ptr2 = second
# Start matching first LL
# with second LL
while ptr1:
# If second LL become empty and
# first not, return False,
# since first LL has not been
# traversed completely
if not ptr2:
return False
# If value of both nodes from both
# LLs are equal, increment pointers
# for both LLs so that next value
# can be matched
elif ptr1.value = = ptr2.value:
ptr1 = ptr1. next
ptr2 = ptr2. next
# If a single mismatch is found
# OR ptr1 is None/empty,break out
# of the while loop and do some checks
else :
break
# check 1 :
# If 'ptr1' is None/empty,that means
# the 'first LL' has been completely
# traversed and matched so return True
if not ptr1:
return True
# If check 1 fails, that means, some
# items for 'first' LL are still yet
# to be matched, so start again by
# bringing back the 'ptr1' to point
# to 1st node of 'first' LL
ptr1 = first
# And increment second node element to next
second = second. next
return False
# Driver Code # Let us create two linked lists to # test the above functions. # Created lists would be # node_a: 1->2->3->4 # node_b: 1->2->1->2->3->4 node_a = Node( 1 )
node_a. next = Node( 2 )
node_a. next . next = Node( 3 )
node_a. next . next . next = Node( 4 )
node_b = Node( 1 )
node_b. next = Node( 2 )
node_b. next . next = Node( 1 )
node_b. next . next . next = Node( 2 )
node_b. next . next . next . next = Node( 3 )
node_b. next . next . next . next . next = Node( 4 )
if findList(node_a, node_b):
print ( "LIST FOUND" )
else :
print ( "LIST NOT FOUND" )
# This code is contributed by GauriShankarBadola |
// C# program to find a list in second list using System;
class GFG
{ // A Linked List node class Node
{ public int data;
public Node next;
}; // Returns true if first list is // present in second list static Boolean findList(Node first,
Node second)
{ Node ptr1 = first, ptr2 = second;
// If both linked lists are empty,
// return true
if (first == null && second == null )
return true ;
// Else If one is empty and
// other is not, return false
if (first == null ||
(first != null && second == null ))
return false ;
// Traverse the second list by
// picking nodes one by one
while (second != null )
{
// Initialize ptr2 with
// current node of second
ptr2 = second;
// Start matching first list
// with second list
while (ptr1 != null )
{
// If second list becomes empty and
// first not then return false
if (ptr2 == null )
return false ;
// If data part is same, go to next
// of both lists
else if (ptr1.data == ptr2.data)
{
ptr1 = ptr1.next;
ptr2 = ptr2.next;
}
// If not equal then break the loop
else break ;
}
// Return true if first list gets traversed
// completely that means it is matched.
if (ptr1 == null )
return true ;
// Initialize ptr1 with first again
ptr1 = first;
// And go to next node of second list
second = second.next;
}
return false ;
} /* Function to print nodes in a given linked list */ static void printList(Node node)
{ while (node != null )
{
Console.Write( "{0} " , node.data);
node = node.next;
}
} // Function to add new node to linked lists static Node newNode( int key)
{ Node temp = new Node();
temp.data= key;
temp.next = null ;
return temp;
} // Driver Code public static void Main(String[] args)
{ /* Let us create two linked lists to test
the above functions. Created lists shall be
a: 1->2->3->4
b: 1->2->1->2->3->4*/
Node a = newNode(1);
a.next = newNode(2);
a.next.next = newNode(3);
a.next.next.next = newNode(4);
Node b = newNode(1);
b.next = newNode(2);
b.next.next = newNode(1);
b.next.next.next = newNode(2);
b.next.next.next.next = newNode(3);
b.next.next.next.next.next = newNode(4);
if (findList(a, b) == true )
Console.Write( "LIST FOUND" );
else
Console.Write( "LIST NOT FOUND" );
} } // This code is contributed by Rajput-Ji |
<script> // JavaScript program to find a // list in second list // A Linked List node
class Node {
constructor() {
this .data = 0;
this .next = null ;
}
} // Returns true if first list is
// present in second list
function findList(first, second) {
var ptr1 = first, ptr2 = second;
// If both linked lists are empty,
// return true
if (first == null && second == null )
return true ;
// Else If one is empty and
// other is not, return false
if (first == null || (first != null &&
second == null ))
return false ;
// Traverse the second list by
// picking nodes one by one
while (second != null ) {
// Initialize ptr2 with
// current node of second
ptr2 = second;
// Start matching first list
// with second list
while (ptr1 != null ) {
// If second list becomes empty and
// first not then return false
if (ptr2 == null )
return false ;
// If data part is same, go to next
// of both lists
else if (ptr1.data == ptr2.data) {
ptr1 = ptr1.next;
ptr2 = ptr2.next;
}
// If not equal then break the loop
else
break ;
}
// Return true if first list gets traversed
// completely that means it is matched.
if (ptr1 == null )
return true ;
// Initialize ptr1 with first again
ptr1 = first;
// And go to next node of second list
second = second.next;
}
return false ;
}
/* Function to print nodes in a given linked list */
function printList(node) {
while (node != null ) {
document.write( "%d " , node.data);
node = node.next;
}
}
// Function to add new node to linked lists
function newNode(key) {
var temp = new Node();
temp.data = key;
temp.next = null ;
return temp;
}
// Driver Code
/*
Let us create two linked lists to test
the above functions. Created lists
shall be a: 1->2->3->4 b: 1->2->1->2->3->4
*/
var a = newNode(1);
a.next = newNode(2);
a.next.next = newNode(3);
a.next.next.next = newNode(4);
var b = newNode(1);
b.next = newNode(2);
b.next.next = newNode(1);
b.next.next.next = newNode(2);
b.next.next.next.next = newNode(3);
b.next.next.next.next.next = newNode(4);
if (findList(a, b) == true )
document.write( "LIST FOUND" );
else
document.write( "LIST NOT FOUND" );
// This code contributed by gauravrajput1 </script> |
LIST FOUND
Time Complexity: O(m*n) where m is the number of nodes in the second list and n in the first.
Auxiliary Space: O(1)
Optimization:
Above code can be optimized by using extra space i.e. storing the list into two strings and applying KMP algorithm.
Approach 2: Using recursion
Time complexity :- O(n*m)
The time complexity of this implementation is O(n*m), where n is the length of the main list and m is the length of the sublist. This is because in the worst case, we may have to iterate through the entire main list for each node in the sublist.
Space complexity :- O(m)
The space complexity of this implementation is O(m), where m is the length of the sublist. This is because we are only keeping track of one node at a time from the sublist, so the space required is proportional to the length of the sublist.
- Define a Node class with an int data member and a Node* pointer to the next node in the linked list.
- Define a recursive isSublist function that takes two Node* pointers as parameters: list (the main list) and sublist (the sublist to search for in the main list).
- In the isSublist function, handle the base cases:
a. If sublist is nullptr, i.e., the end of the sublist has been reached, return true because the sublist has been found.
b. If list is nullptr, i.e., the end of the main list has been reached but the sublist has not been found, return false. - In the recursive case of the isSublist function:
a. If the data member of the current list node is equal to the data member of the current sublist node, recursively call isSublist with the next nodes in both lists (list->next and sublist->next).
b. If the data member of the current list node is not equal to the data member of the current sublist node, recursively call isSublist with the next node in the main list (list->next) and the entire sublist (sublist). - In the main function:
a. Create a main linked list called list with 6 nodes, each with an int data member.
b. Create a sublist called sublist with 4 nodes, each with an int data member.
c. Call the isSublist function with list and sublist as arguments.
d. If the isSublist function returns true, print “LIST FOUND”. Otherwise, print “LIST NOT FOUND”.
Return 0 to exit the program.
#include <iostream> using namespace std;
class Node {
public :
int data;
Node* next;
Node( int data)
{
this ->data = data;
next = nullptr;
}
}; bool isSublist(Node* list, Node* sublist)
{ // Base cases
if (sublist == nullptr) {
return true ;
}
if (list == nullptr) {
return false ;
}
// Recursive case
if (list->data == sublist->data) {
return isSublist(list->next, sublist->next);
}
else {
return isSublist(list->next, sublist);
}
} int main()
{ // Create the main list
Node* list = new Node(1);
list->next = new Node(2);
list->next->next = new Node(1);
list->next->next->next = new Node(2);
list->next->next->next->next = new Node(3);
list->next->next->next->next->next = new Node(4);
// Create the sublist
Node* sublist = new Node(1);
sublist->next = new Node(2);
sublist->next->next = new Node(3);
sublist->next->next->next = new Node(4);
// Check if sublist is present in list
if (isSublist(list, sublist)) {
cout << "LIST FOUND" << endl;
}
else {
cout << "LIST NOT FOUND" << endl;
}
return 0;
} |
class Node {
public int data;
public Node next;
public Node( int data)
{
this .data = data;
next = null ;
}
} public class Main {
public static boolean isSublist(Node list, Node sublist)
{
// Base cases
if (sublist == null ) {
return true ;
}
if (list == null ) {
return false ;
}
// Recursive case
if (list.data == sublist.data) {
return isSublist(list.next, sublist.next);
}
else {
return isSublist(list.next, sublist);
}
}
public static void main(String[] args)
{
// Create the main list
Node list = new Node( 1 );
list.next = new Node( 2 );
list.next.next = new Node( 1 );
list.next.next.next = new Node( 2 );
list.next.next.next.next = new Node( 3 );
list.next.next.next.next.next = new Node( 4 );
// Create the sublist
Node sublist = new Node( 1 );
sublist.next = new Node( 2 );
sublist.next.next = new Node( 3 );
sublist.next.next.next = new Node( 4 );
// Check if sublist is present in list
if (isSublist(list, sublist)) {
System.out.println( "LIST FOUND" );
}
else {
System.out.println( "LIST NOT FOUND" );
}
}
} |
class Node:
def __init__( self , data):
self .data = data
self . next = None
def isSublist( list , sublist):
# Base cases
if sublist is None :
return True
if list is None :
return False
# Recursive case
if list .data = = sublist.data:
return isSublist( list . next , sublist. next )
else :
return isSublist( list . next , sublist)
# Create the main list list = Node( 1 )
list . next = Node( 2 )
list . next . next = Node( 1 )
list . next . next . next = Node( 2 )
list . next . next . next . next = Node( 3 )
list . next . next . next . next . next = Node( 4 )
# Create the sublist sublist = Node( 1 )
sublist. next = Node( 2 )
sublist. next . next = Node( 3 )
sublist. next . next . next = Node( 4 )
# Check if sublist is present in list if isSublist( list , sublist):
print ( "LIST FOUND" )
else :
print ( "LIST NOT FOUND" )
|
using System;
class Node {
public int data;
public Node next;
public Node( int data) {
this .data = data;
this .next = null ;
}
} class Sublist {
public static bool IsSublist(Node list, Node sublist) {
// Base cases
if (sublist == null ) {
return true ;
}
if (list == null ) {
return false ;
}
// Recursive case
if (list.data == sublist.data) {
return IsSublist(list.next, sublist.next);
} else {
return IsSublist(list.next, sublist);
}
}
} class Program {
static void Main( string [] args) {
// Create the main list
Node list = new Node(1);
list.next = new Node(2);
list.next.next = new Node(1);
list.next.next.next = new Node(2);
list.next.next.next.next = new Node(3);
list.next.next.next.next.next = new Node(4);
// Create the sublist
Node sublist = new Node(1);
sublist.next = new Node(2);
sublist.next.next = new Node(3);
sublist.next.next.next = new Node(4);
// Check if sublist is present in list
if (Sublist.IsSublist(list, sublist)) {
Console.WriteLine( "LIST FOUND" );
} else {
Console.WriteLine( "LIST NOT FOUND" );
}
}
} |
class Node { constructor(data) {
this .data = data;
this .next = null ;
}
} class Sublist { static isSublist(list, sublist) {
// Base cases
if (sublist == null ) {
return true ;
}
if (list == null ) {
return false ;
}
// Recursive case
if (list.data == sublist.data) {
return Sublist.isSublist(list.next, sublist.next);
} else {
return Sublist.isSublist(list.next, sublist);
}
}
} // Create the main list let list = new Node(1);
list.next = new Node(2);
list.next.next = new Node(1);
list.next.next.next = new Node(2);
list.next.next.next.next = new Node(3);
list.next.next.next.next.next = new Node(4);
// Create the sublist let sublist = new Node(1);
sublist.next = new Node(2);
sublist.next.next = new Node(3);
sublist.next.next.next = new Node(4);
// Check if sublist is present in list if (Sublist.isSublist(list, sublist)) {
console.log( "LIST FOUND" );
} else {
console.log( "LIST NOT FOUND" );
} |
LIST FOUND
Time Complexity: O(m*n)
Auxiliary Space: O(m)