Count duplicates in a given linked list
Last Updated :
18 Jan, 2023
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: 1
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 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 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# 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
<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 contributed by jana_ayantan.
</script>
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 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
# 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# 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
<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>
Time Complexity : O(n)
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