Linked list

Sublist Search (Search a linked list in another list)

PrepBytes


The linked list is one of the most important concepts and data structures to learn while preparing for interviews. Having a good grasp of Linked Lists can be a huge plus point in a coding interview.

Problem Statement

Given two linked lists, our task is to check whether the first list is present in the second list or not.

Problem Statement Understanding

Let’s try to understand the problem statement with the help of examples.

According to the problem statement, we will be given two linked lists list1 and list2, and we need to check whether list1 is present in list2 or not.

  • If list1 is present in list2, we will output Found.
  • Otherwise, we will output Not Found.

If the given linked lists are list1: 1→2→4 and list2: 1→2→1→2→4→3.

  • As we can see, in list2 starting from the 3rd index and up to 5th index (1→2→4) (considering 1 based indexing), list2 contains all the elements of list 1 in the same order as they are present in list1. So we can say that list2 contains list1.
  • We will output Found, as we found our list1 in list2.

Say, if the list1: 1→2→4 and list2: 1→2→1→2→3→4.

  • As we can see that list2 does not contain all the elements of list1 in the same order as they were present in list1, so we will output Not Found.
Some more examples

Sample Input 1: list1 = 3→5, list2 =5→3→5.
Sample Output 1: Found

Sample Input 2: list1 = 1→3→4, list2 = 1→2→1→3→5.
Sample Output 2: Not Found

Remember: Here we are doing a sublist search, so if all the elements of list1 are present in list2 in the same order as they were present in list1 and these elements are consecutive in list2, then only we can say that we found list1 in list2.

Now I think from the above examples, the problem statement is clear. So let's see how we will approach it. Any Ideas?

  • If not, it's okay. We will see in the next section thoroughly how we can approach this problem.

Let’s move to the next section.

Approach and Algorithm

  • Start traversing through both the list.
  • Match every node of the 2nd list (list2) with the first node of the 1st list (list1).
  • If the first node of the 1st list matches with the current node of the 2nd list.
    • Then, we have to check whether the remaining nodes of 1st List matches the nodes of 2nd list or not.
    • If all nodes of 1st list (list1) are matched with 2nd list (list2), then return true.
  • If all nodes of list1 didn’t match with list2 nodes, we will move forward in list2 and repeat the above process from step 2.
  • Until any of the list1 or list2 becomes empty, we will repeat the above process.
  • If our list1 got empty, then we can say that list1 found in list2, else not.

Dry Run

Code Implementation

#include 
using namespace std;
 
//  Linked List node structure
struct Node
{
    int data;
    Node* next;
};
 
//This searchList function will return true if list1 is present in list2
bool searchList(Node* list1, Node* list2)
{
    Node* p1 = list1, *p2 = list2;
 
    if (list1 == NULL && list2 == NULL)
        return true;
 
    if ( list1 == NULL || (list1 != NULL && list2 == NULL))
        return false;
 
    while (list2 != NULL)
    {
        p2 = list2;
 
        while (p1 != NULL)
        {
            if (p2 == NULL)
                return false;
 
            
            else if (p1->data == p2->data)
            {
                p1 = p1->next;
                p2 = p2->next;
            }
            else break;
        }
 
        if (p1 == NULL)
            return true;
 
        p1 = list1;
 
        list2 = list2->next;
    }
 
    return false;
}
 
/* This function is used to print the nodes of a given linked list */
void printList(Node* node)
{
    while (node != NULL)
    {
        printf("%d ", node->data);
        node = node->next;
    }
}
 
// This function is used to add a new node to a linked list
Node *newNode(int key)
{
    Node *temp = new Node;
    temp-> data= key;
    temp->next = NULL;
    return temp;
}
 
int main()
{
    Node *list1 = newNode(1);
    list1->next = newNode(2);
    list1->next->next = newNode(4);
 
    Node *list2 = newNode(1);
    list2->next = newNode(2);
    list2->next->next = newNode(1);
    list2->next->next->next = newNode(2);
    list2->next->next->next->next = newNode(4);
    list2->next->next->next->next->next = newNode(3);
 
    searchList(list1,list2) ? cout << "Found" :
                    cout << "Not Found";
 
    return 0;
}

#include
#include
#include

struct Node
{
    int data;
    struct Node* next;
};
 
// Returns true if first list is present in second
// list
bool findList(struct Node* first,struct Node* second)
{
    struct 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(struct Node* node)
{
    while (node != NULL)
    {
        printf("%d ", node->data);
        node = node->next;
    }
}
 
// Function to add new node to linked lists
struct Node *newNode(int key)
{
    struct Node* temp = (struct Node*)malloc(sizeof(struct 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*/
    struct Node *a = newNode(1);
    a->next = newNode(2);
    a->next->next = newNode(3);
    a->next->next->next = newNode(4);
 
    struct 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) ? printf("LIST FOUND") :
                    printf("LIST NOT FOUND");
 
    return 0;
}

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 not first and not second:
		return True

	if not first or not second:
		return False

	ptr1 = first
	ptr2 = second

	while ptr2:

		ptr2 = second
		while ptr1:

			if not ptr2:
				return False
			elif ptr1.value == ptr2.value:
				ptr1 = ptr1.next
				ptr2 = ptr2.next

			else:
				break

		if not ptr1:
			return True

		ptr1 = first
		second = second.next
		
	return False

node_a = Node(1)
node_a.next = Node(2)
node_a.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(4)
node_b.next.next.next.next.next = Node(3)

if findList(node_a, node_b):
	print("LIST FOUND")
else:
	print("LIST NOT FOUND")

Output

Found

Time Complexity: O(M * N), M is the size of lsit1 and N is the size of list2.
[forminator_quiz id="4878"]

So, In this blog, we have learned How to Search a linked list in another list. If you want to solve more questions on Linked List, which are curated by our expert mentors at PrepBytes, you can follow this link Linked List.

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