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

• Difficulty Level : Easy
• Last Updated : 10 Jan, 2023

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.
Auxiliary Space: O(1)

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

This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

My Personal Notes arrow_drop_up
Related Articles