# Sublist Search (Search a linked list in another list)

• Difficulty Level : Easy
• Last Updated : 26 Jul, 2022

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

Algorithm:

1. Take first node of second list.
2. Start matching the first list from this first node.
3. If whole lists match return true.
4. Else break and take first list to the first node again.
5. And take second list to its second node.
6. Repeat these steps until any of linked lists becomes empty.
7. If first list becomes empty then list found else not.

Below is the implementation.

## C++

 `// C++ program to find a list in second list``#include ``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

 `// Java program to find a list in second list``import` `java.util.*;``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

 `# 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``            ``else` `if` `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 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#

 `// 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`

## Javascript

 ``

Output

`LIST FOUND` Time Complexity: O(m*n) where m is the number of nodes in second list and n in first.

Optimization :

Above code can be optimized by using extra space i.e. stores the list into two strings and apply KMP algorithm. Refer https://ide.geeksforgeeks.org/3fXUaV for implementation provided by Nishant Singh

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