This article will help you to understand to find duplicate elements in linked list. Having knowledge about the linked lists will help you for clearing IT interviews. Having a good grasp of Linked Lists can be a huge plus point in a coding interview. Now, Let’s move to our topic to find duplicate in linked list.

### Problem Statement

In this question, we are given a singly linked list. We have to count the number of duplicate nodes in the linked list

### Problem Statement Understanding To Find Duplicate Elements In Linked List

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

Suppose the given list is:

We have to count the number of duplicate nodes in the list.

From the above given linked list, we can see that:

- Count of each 1 and 8 is 2 in the linked list.
- While the count of 7 is 1 in the linked list.

So, we can say that duplicates of 1 and 8 exist in the linked list, 1 duplicate each of 1 and 8 exists in the linked list. So, we will return the count of duplicate node in the linked list as (1+1) = 2.

If the given linked list is:

For the above-linked list, we can see that:

- Count of each 1, 2, 3 is 2 in the linked list.
- The count of 5 is 3.
- While the count of 4 is 1 in the linked list.

So, we can say that duplicates of 1, 2, 3, and 5 exist in the linked list, 1 duplicate each of 1, 2, and 3 exist and 2 duplicates of 5 exist. So, we will return the count of duplicate node in the linked list as (1+1+1+2) = 5.

Now, I think it is clear from the examples what we have to do with the problem. So let’s move to approach.

Before jumping to approach, just try to think how can you approach this problem?

It’s okay if your solution is brute force, we will try to optimize it together.

We will first make use of simple nested list traversal to find the duplicates, although it will be a brute-force approach, it is necessary to build the foundation.

Let us have a glance at brute force approaches.

### Approach and Algorithm (Brute Force) To Find Duplicate Elements In Linked List

The approach is going to be pretty simple.

- We will create a variable count and initialize it to 0.
- Now, we will traverse through the list, and for every node, we will traverse from its next node to the end of the list whenever we will find a match, we will increase the counter and will break out of the loop. Basically here what we are trying to find out is that is this particular node duplicate or not.

This method has a time complexity of O(n^{2}) as it is using nested traversal.

Can we do better?

Yes. We can do better. We can do it in O(n) with the help of hashing.

Let us see the efficient approach.

### Approach (Hashing) To Find Duplicate Elements In Linked List

In this approach, we will use hashing.

- Firstly, we will create an unordered set, and we will insert the data of the head in the set.
- We will also create a counter whose initial value will be 0.
- Now, we will traverse through the list, starting from the next node, and for every node, we will check if that node is already present in the set or not. If it is present, we will increment the counter. We will also keep inserting the nodes, that we are traversing, in the set.

### Algorithm To Find Duplicate Elements In Linked List

- Base Case – If the head is NULL, return 0.
- Create an unordered set, say s, and a variable count with an initial value of 0.
- Insert the head → data in the set.
- Traverse from the next node of the head till the end of the list.
- In every iteration, check if the node’s data is already present in the set or not. If present, increment the count.
- Insert the current node’s data in the set.
- In the end, return the count variable.

### Dry Run To Find Duplicate Elements In Linked List

### Code Implementation

#include<stdio.h> #include<stdlib.h> struct Node { int data; struct Node* next; }; // Function to insert a node at the beginning void insert(struct Node** head, int item) { struct Node* temp = (struct Node*)malloc(sizeof(struct Node)); temp->data = item; temp->next = *head; *head = temp; } // Function to count the number of // duplicate nodes in the linked list int countNode(struct Node* head) { int count = 0; while (head->next != NULL) { // Starting from the next node struct Node *ptr = head->next; while (ptr != NULL) { // If some duplicate node is found if (head->data == ptr->data) { count++; break; } ptr = ptr->next; } head = head->next; } // Return the count of duplicate nodes return count; } // Main function int main() { struct Node* head = NULL; insert(&head, 5); insert(&head, 7); insert(&head, 5); insert(&head, 4); insert(&head, 7); int ans = countNode(head); printf("%d", ans); return 0; }

#include <iostream> #include <unordered_set> using namespace std; struct Node { int data; Node* next; }; void insert(Node** head, int item) { Node* temp = new Node(); temp->data = item; temp->next = *head; *head = temp; } int countNode(Node* head) { if (head == NULL) return 0;; unordered_set<int> s; s.insert(head->data); int count = 0; for (Node *curr=head->next; curr != NULL; curr=curr->next) { if (s.find(curr->data) != s.end()) count++; s.insert(curr->data); } return count; } int main() { Node* head = NULL; insert(&head, 8); insert(&head, 8); insert(&head, 1); insert(&head, 7); insert(&head, 1); cout << countNode(head); return 0; }

import java.util.HashSet; public class PrepBytes { static class Node { int data; Node next; }; static Node head; static void insert(Node ref_head, int item) { Node temp = new Node(); temp.data = item; temp.next = ref_head; head = temp; } static int countNode(Node head) { if (head == null) return 0;; HashSet<integer>s = new HashSet<>(); s.add(head.data); int count = 0; for (Node curr=head.next; curr != null; curr=curr.next) { if (s.contains(curr.data)) count++; s.add(curr.data); } return count; } public static void main(String[] args) { insert(head, 8); insert(head, 8); insert(head, 1); insert(head, 7); insert(head, 1); System.out.println(countNode(head)); } }

class Node: def __init__(self, data = None, next = None): self.next = next self.data = data head = None def insert(ref_head, item): global head temp = Node() temp.data = item temp.next = ref_head head = temp def countNode(head): if (head == None): return 0 s = set() s.add(head.data) count = 0 curr = head.next while ( curr != None ) : if (curr.data in s): count = count + 1 s.add(curr.data) curr = curr.next return count insert(head, 8) insert(head, 8) insert(head, 1) insert(head, 7) insert(head, 1) print(countNode(head))

### Output

2

**Time Complexity To Find Duplicate Elements In Linked List:** O(n), as list traversal is needed.

This article will help you to understand how to find duplicate in linked list. There is a lot of stuff to prepare for the interviews but data structures and algorithms are the most important topics as these are the must needed skills for any company. This is an important question when it comes to coding interviews. If you want to solve more questions on Linked List, which is curated by our expert mentors at PrepBytes, you can follow this link Linked List.

## FAQ Related To Find Duplicate Elements In Linked List

**What is the time complexity for searching for an element on a Linked List?****How do you find duplicates in a linked list?****Is a linked list dynamic?**

Since there is no way of accessing any node without visiting its previous node, we have to traverse the entire Linked List every time we need to search for an element. Therefore, the worst-case time complexity is O(N) where N is the number of elements in a Linked List.

We traverse the whole linked list. For each node, we check in the remaining list whether the duplicate node exists or not. If it does then we increment the count.

Linked List is a dynamic structure, which means the list can grow or shrink depending on the data making it more powerful and flexible than Arrays. Unlike Arrays, Linked List is not stored in a contiguous memory location. Each element in the list is spread across the memory and are linked by the pointers in the Node.