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Count duplicates in a given linked list

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  • Last Updated : 21 Jun, 2021
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Given a linked list. The task is to count the number of duplicate nodes in the linked list. 
Examples: 
 

Input: 5 -> 7 -> 5 -> 1 -> 7 -> NULL 
Output: 2
Input: 5 -> 7 -> 8 -> 7 -> 1 -> NULL 
Output:
 

 

Simple Approach: 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.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <iostream>
#include <unordered_set>
using namespace std;
 
// Representation of node
struct Node {
    int data;
    Node* next;
};
 
 
// Function to insert a node at the beginning
void insert(Node** head, int item)
{
    Node* temp = new Node();
    temp->data = item;
    temp->next = *head;
    *head = temp;
}
 
// Function to count the number of
// duplicate nodes in the linked list
int countNode(Node* head)
{
    int count = 0;
 
    while (head->next != NULL) {
 
        // Starting from the next node
        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;
}
 
// Driver code
int main()
{
    Node* head = NULL;
    insert(&head, 5);
    insert(&head, 7);
    insert(&head, 5);
    insert(&head, 1);
    insert(&head, 7);
 
    cout << countNode(head);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
     
// Representation of node
static class Node
{
    int data;
    Node next;
};
 
 
// Function to insert a node at the beginning
static Node insert(Node head, int item)
{
    Node temp = new Node();
    temp.data = item;
    temp.next = head;
    head = temp;
    return head;
}
 
// Function to count the number of
// duplicate nodes in the linked list
static int countNode(Node head)
{
    int count = 0;
 
    while (head.next != null)
    {
 
        // Starting from the next node
        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;
}
 
// Driver code
public static void main(String args[])
{
    Node head = null;
    head = insert(head, 5);
    head = insert(head, 7);
    head = insert(head, 5);
    head = insert(head, 1);
    head = insert(head, 7);
 
    System.out.println( countNode(head));
}
}
 
// This code is contributed by Arnab Kundu

Python3




# Python3 implementation of the approach
import math
 
# Representation of node
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
 
# Function to push a node at the beginning
def push(head, item):
    temp = Node(item);
    temp.data = item;
    temp.next = head;
    head = temp;
    return head;
         
# Function to count the number of
# duplicate nodes in the linked list
def countNode(head):
    count = 0
    while (head.next != None):
         
        # print(1)
        # Starting from the next node
        ptr = head.next
        while (ptr != None):
             
            # print(2)
            # If some duplicate node is found
            if (head.data == ptr.data):
                count = count + 1
                break
            ptr = ptr.next
        head = head.next
         
    # Return the count of duplicate nodes
    return count
 
# Driver code
if __name__=='__main__':
 
    head = None;
    head = push(head, 5)
    head = push(head, 7)
    head = push(head, 5)
    head = push(head, 1)
    head = push(head, 7)
    print(countNode(head))
 
# This code is contributed by Srathore

C#




// C# implementation of the approach
using System;
     
class GFG
{
     
// Representation of node
public class Node
{
    public int data;
    public Node next;
};
 
 
// Function to insert a node at the beginning
static Node insert(Node head, int item)
{
    Node temp = new Node();
    temp.data = item;
    temp.next = head;
    head = temp;
    return head;
}
 
// Function to count the number of
// duplicate nodes in the linked list
static int countNode(Node head)
{
    int count = 0;
 
    while (head.next != null)
    {
 
        // Starting from the next node
        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;
}
 
// Driver code
public static void Main(String []args)
{
    Node head = null;
    head = insert(head, 5);
    head = insert(head, 7);
    head = insert(head, 5);
    head = insert(head, 1);
    head = insert(head, 7);
 
    Console.WriteLine( countNode(head));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Javascript implementation of the approach
 
// Representation of a node
class Node {
        constructor() {
                var data;
                var next;
             }
        }
         
// Function to insert a node at the beginning
function insert( head, item)
{
    var temp = new Node();
    temp.data = item;
    temp.next = head;
    head = temp;
    return head;
}
 
// Function to count the number of
// duplicate nodes in the linked list
function countNode( head)
{
    let count = 0;
 
    while (head.next != null)
    {
 
        // Starting from the next node
        let 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;
}
 
    // Driver Code
    var head = null;
    head = insert(head, 5);
    head = insert(head, 7);
    head = insert(head, 5);
    head = insert(head, 1);
    head = insert(head, 7);
 
    document.write( countNode(head));
 
// This code is contribute by jana_ayantan.
</script>

Output: 

2

 

Time Complexity: O(n*n)
Efficient Approach: The idea is to use hashing 
 

C++




// C++ implementation of the approach
#include <iostream>
#include <unordered_set>
using namespace std;
 
// Representation of node
struct Node {
    int data;
    Node* next;
};
 
// Function to insert a node at the beginning
void insert(Node** head, int item)
{
    Node* temp = new Node();
    temp->data = item;
    temp->next = *head;
    *head = temp;
}
 
// Function to count the number of
// duplicate nodes in the linked list
int countNode(Node* head)
{
    if (head == NULL)
       return 0;;
 
    // Create a hash table insert head
    unordered_set<int> s;
    s.insert(head->data);
 
    // Traverse through remaining nodes   
    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 the count of duplicate nodes
    return count;
}
 
// Driver code
int main()
{
    Node* head = NULL;
    insert(&head, 5);
    insert(&head, 7);
    insert(&head, 5);
    insert(&head, 1);
    insert(&head, 7);
 
    cout << countNode(head);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.HashSet;
 
class GFG
{
 
// Representation of node
static class Node
{
    int data;
    Node next;
};
static Node head;
 
// Function to insert a node at the beginning
static void insert(Node ref_head, int item)
{
    Node temp = new Node();
    temp.data = item;
    temp.next = ref_head;
    head = temp;
     
}
 
// Function to count the number of
// duplicate nodes in the linked list
static int countNode(Node head)
{
    if (head == null)
    return 0;;
 
    // Create a hash table insert head
    HashSet<Integer>s = new HashSet<>();
    s.add(head.data);
 
    // Traverse through remaining nodes
    int count = 0;
    for (Node curr=head.next; curr != null; curr=curr.next)
    {
        if (s.contains(curr.data))
            count++;
 
        s.add(curr.data);
    }
 
    // Return the count of duplicate nodes
    return count;
}
 
// Driver code
public static void main(String[] args)
{
 
    insert(head, 5);
    insert(head, 7);
    insert(head, 5);
    insert(head, 1);
    insert(head, 7);
 
    System.out.println(countNode(head));
}
}
 
// This code is contributed by Princi Singh

Python




# Python3 implementation of the approach
 
# Node of a linked list
class Node:
    def __init__(self, data = None, next = None):
        self.next = next
        self.data = data
 
head = None
 
# Function to insert a node at the beginning
def insert(ref_head, item):
    global head
    temp = Node()
    temp.data = item
    temp.next = ref_head
    head = temp
     
# Function to count the number of
# duplicate nodes in the linked list
def countNode(head):
 
    if (head == None):
        return 0
 
    # Create a hash table insert head
    s = set()
    s.add(head.data)
 
    # Traverse through remaining nodes
    count = 0
    curr = head.next
    while ( curr != None ) :
        if (curr.data in s):
            count = count + 1
 
        s.add(curr.data)
        curr = curr.next
 
    # Return the count of duplicate nodes
    return count
 
# Driver code
insert(head, 5)
insert(head, 7)
insert(head, 5)
insert(head, 1)
insert(head, 7)
 
print(countNode(head))
 
# This code is contributed by Arnab Kundu

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
     
class GFG
{
 
// Representation of node
public class Node
{
    public int data;
    public Node next;
};
static Node head;
 
// Function to insert a node at the beginning
static void insert(Node ref_head, int item)
{
    Node temp = new Node();
    temp.data = item;
    temp.next = ref_head;
    head = temp;
     
}
 
// Function to count the number of
// duplicate nodes in the linked list
static int countNode(Node head)
{
    if (head == null)
    return 0;;
 
    // Create a hash table insert head
    HashSet<int>s = new HashSet<int>();
    s.Add(head.data);
 
    // Traverse through remaining nodes
    int count = 0;
    for (Node curr=head.next; curr != null; curr=curr.next)
    {
        if (s.Contains(curr.data))
            count++;
 
        s.Add(curr.data);
    }
 
    // Return the count of duplicate nodes
    return count;
}
 
// Driver code
public static void Main(String[] args)
{
 
    insert(head, 5);
    insert(head, 7);
    insert(head, 5);
    insert(head, 1);
    insert(head, 7);
 
    Console.WriteLine(countNode(head));
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Representation of node
class Node {
 
    constructor()
    {
        this.data = 0;
        this.next = null;
    }
};
 
// Function to insert a node at the beginning
function insert(head, item)
{
    var temp = new Node();
    temp.data = item;
    temp.next = head;
    head = temp;
    return head;
}
 
// Function to count the number of
// duplicate nodes in the linked list
function countNode(head)
{
    if (head == null)
       return 0;;
 
    // Create a hash table insert head
    var s = new Set();
    s.add(head.data);
 
    // Traverse through remaining nodes   
    var count = 0;
    for (var curr=head.next; curr != null; curr=curr.next) {
        if (s.has(curr.data))
             count++;
 
        s.add(curr.data);
    }
 
    // Return the count of duplicate nodes
    return count;
}
 
// Driver code
var head = null;
head = insert(head, 5);
head = insert(head, 7);
head = insert(head, 5);
head = insert(head, 1);
head = insert(head, 7);
document.write( countNode(head));
 
 
</script>

Output: 

2

 

Time Complexity : O(n)
 


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