Remove duplicates from an unsorted doubly linked list
Last Updated :
28 Aug, 2024
Given an unsorted doubly linked list containing n nodes, the task is to remove duplicate nodes while preserving the original order.
Examples:
Input: Doubly Linked List = 1 <-> 2 <-> 3 <-> 2 <-> 4
Output: Doubly Linked List = 1 <-> 2 <-> 3 <-> 4
Removal of duplicate node 2 from Doubly Linked ListInput: Doubly Linked List = 8 <-> 4 <-> 4 <-> 6 <-> 4 <-> 8 <-> 4 <-> 10 <-> 12 <-> 12
Output: Doubly Linked List = 8 <-> 4 <-> 6 <-> 10 <-> 12
[Naive Approach] Using Nested Loops - O(n^2) Time and O(1) Space
The idea is to use two nested loops. The outer loop is used to pick the elements one by one and the inner loop compares the picked element with the rest of the elements in the doubly linked list.
Below is the implementation of the above approach:
C++
// C++ program to remove duplicates from an
// unsorted doubly linked list
#include <iostream>
using namespace std;
class Node {
public:
int data;
Node* next;
Node* prev;
Node(int val) {
data = val;
next = nullptr;
prev = nullptr;
}
};
// Function to remove duplicates using nested loops
Node* removeDuplicates(Node* head) {
Node* curr = head;
// Traverse each node in the list
while (curr != nullptr) {
Node* temp = curr;
// Traverse the remaining nodes to find and
// remove duplicates
while (temp->next != nullptr) {
// Check if the next node has the same
// data as the current node
if (temp->next->data == curr->data) {
// Duplicate found, remove it
Node* duplicate = temp->next;
temp->next = temp->next->next;
// Update the previous pointer of the
// next node, if it exists
if (temp->next != nullptr)
temp->next->prev = temp;
// Free the memory of the duplicate node
delete duplicate;
}
else {
// If the next node has different data from
// the current node, move to the next node
temp = temp->next;
}
}
// Move to the next node in the list
curr = curr->next;
}
return head;
}
void printList(Node* head) {
Node* curr = head;
while (curr != nullptr) {
cout << curr->data << " ";
curr = curr->next;
}
cout << endl;
}
int main() {
// Create a doubly linked list:
// 1 <-> 2 <-> 3 <-> 2 <-> 4
Node* head = new Node(1);
head->next = new Node(2);
head->next->prev = head;
head->next->next = new Node(3);
head->next->next->prev = head->next;
head->next->next->next = new Node(2);
head->next->next->next->prev = head->next->next;
head->next->next->next->next = new Node(4);
head->next->next->next->next->prev = head->next->next->next;
head = removeDuplicates(head);
printList(head);
return 0;
}
// C program to remove duplicates
// from an unsorted doubly linked list
#include <stdio.h>
struct Node {
int data;
struct Node *next;
struct Node *prev;
};
// Function to remove duplicates using nested loops
struct Node *removeDuplicates(struct Node *head) {
struct Node *curr = head;
// Traverse each node in the list
while (curr != NULL) {
struct Node *temp = curr;
// Traverse the remaining nodes to
// find and remove duplicates
while (temp->next != NULL) {
// Check if the next node has the same
// data as the current node
if (temp->next->data == curr->data) {
// Duplicate found, remove it
struct Node *duplicate = temp->next;
temp->next = temp->next->next;
// Update the previous pointer of the next
// node, if it exists
if (temp->next != NULL)
temp->next->prev = temp;
// Free the memory of the duplicate node
free(duplicate);
}
else {
// If the next node has different data from
// the current node, move to the next node
temp = temp->next;
}
}
// Move to the next node in the list
curr = curr->next;
}
return head;
}
void printList(struct Node *head) {
struct Node *curr = head;
while (curr != NULL) {
printf("%d ", curr->data);
curr = curr->next;
}
printf("\n");
}
struct Node *createNode(int val) {
struct Node *newNode = (struct Node *)malloc(sizeof(struct Node));
newNode->data = val;
newNode->next = NULL;
newNode->prev = NULL;
return newNode;
}
int main() {
// Create a doubly linked list:
// 1 <-> 2 <-> 3 <-> 2 <-> 4
struct Node *head = createNode(1);
head->next = createNode(2);
head->next->prev = head;
head->next->next = createNode(3);
head->next->next->prev = head->next;
head->next->next->next = createNode(2);
head->next->next->next->prev = head->next->next;
head->next->next->next->next = createNode(4);
head->next->next->next->next->prev = head->next->next->next;
head = removeDuplicates(head);
printList(head);
return 0;
}
// Java program to remove duplicates from
// an unsorted doubly linked list
class Node {
int data;
Node next;
Node prev;
Node(int val) {
data = val;
next = null;
prev = null;
}
}
class GfG {
static Node removeDuplicates(Node head) {
Node curr = head;
// Traverse each node in the list
while (curr != null) {
Node temp = curr;
// Traverse the remaining nodes to
// find and remove duplicates
while (temp.next != null) {
// Check if the next node has the same data
// as the current node
if (temp.next.data == curr.data) {
// Duplicate found, remove it
Node duplicate = temp.next;
temp.next = temp.next.next;
// Update the previous pointer of the
// next node, if it exists
if (temp.next != null)
temp.next.prev = temp;
// Free the memory of the duplicate node
duplicate = null;
}
else {
// If the next node has different data from
// the current node, move to the next node
temp = temp.next;
}
}
// Move to the next node in the list
curr = curr.next;
}
return head;
}
static void printList(Node head) {
Node curr = head;
while (curr != null) {
System.out.print(curr.data + " ");
curr = curr.next;
}
System.out.println();
}
public static void main(String[] args) {
// Create a doubly linked list:
// 1 <-> 2 <-> 3 <-> 2 <-> 4
Node head = new Node(1);
head.next = new Node(2);
head.next.prev = head;
head.next.next = new Node(3);
head.next.next.prev = head.next;
head.next.next.next = new Node(2);
head.next.next.next.prev = head.next.next;
head.next.next.next.next = new Node(4);
head.next.next.next.next.prev = head.next.next.next;
head = removeDuplicates(head);
printList(head);
}
}
# Python program to remove duplicates
# from an unsorted doubly linked list
class Node:
def __init__(self, val):
self.data = val
self.next = None
self.prev = None
def print_list(head):
curr = head
while curr is not None:
print(curr.data, end=" ")
curr = curr.next
print()
def remove_duplicates(head):
curr = head
# Traverse each node in the list
while curr is not None:
temp = curr
# Traverse the remaining nodes to
# find and remove duplicates
while temp.next is not None:
# Check if the next node has the same
# data as the curr node
if temp.next.data == curr.data:
# Duplicate found, remove it
duplicate = temp.next
temp.next = temp.next.next
# Update the previous pointer of
# the next node, if it exists
if temp.next is not None:
temp.next.prev = temp
# Free the memory of the duplicate node
del duplicate
else:
# If the next node has different data from
# the current node, move to the next node
temp = temp.next
# Move to the next node in the list
curr = curr.next
return head
if __name__ == "__main__":
# Create a doubly linked list:
# 1 <-> 2 <-> 3 <-> 2 <-> 4
head = Node(1)
head.next = Node(2)
head.next.prev = head
head.next.next = Node(3)
head.next.next.prev = head.next
head.next.next.next = Node(2)
head.next.next.next.prev = head.next.next
head.next.next.next.next = Node(4)
head.next.next.next.next.prev = head.next.next.next
head = remove_duplicates(head)
print_list(head)
// C# program to remove duplicates from an unsorted
// doubly linked list
using System;
class Node {
public int Data;
public Node Next;
public Node Prev;
public Node(int val) {
Data = val;
Next = null;
Prev = null;
}
}
class GfG {
static void PrintList(Node head) {
Node curr = head;
while (curr != null) {
Console.Write(curr.Data + " ");
curr = curr.Next;
}
Console.WriteLine();
}
static Node RemoveDuplicates(Node head) {
Node curr = head;
// Traverse each node in the list
while (curr != null) {
Node temp = curr;
// Traverse the remaining nodes to
// find and remove duplicates
while (temp.Next != null) {
// Check if the next node has the same data
// as the curr node
if (temp.Next.Data == curr.Data) {
// Duplicate found, remove it
temp.Next = temp.Next.Next;
// Update the previous pointer of the
// next node, if it exists
if (temp.Next != null) {
temp.Next.Prev = temp;
}
}
else {
// If the next node has different data from
// the current node, move to the next node
temp = temp.Next;
}
}
// Move to the next node in the list
curr = curr.Next;
}
return head;
}
static void Main() {
// Create a doubly linked list:
// 1 <-> 2 <-> 3 <-> 2 <-> 4
Node head = new Node(1);
head.Next = new Node(2);
head.Next.Prev = head;
head.Next.Next = new Node(3);
head.Next.Next.Prev = head.Next;
head.Next.Next.Next = new Node(2);
head.Next.Next.Next.Prev = head.Next.Next;
head.Next.Next.Next.Next = new Node(4);
head.Next.Next.Next.Next.Prev = head.Next.Next.Next;
head = RemoveDuplicates(head);
PrintList(head);
}
}
// Javascript program to remove duplicates from
// an unsorted doubly linked list
class Node {
constructor(val) {
this.data = val;
this.next = null;
this.prev = null;
}
}
function printList(head) {
let curr = head;
while (curr !== null) {
console.log(curr.data + " ");
curr = curr.next;
}
console.log();
}
function removeDuplicates(head) {
let curr = head;
// Traverse each node in the list
while (curr !== null) {
let temp = curr;
// Traverse the remaining nodes to find and remove
// duplicates
while (temp.next !== null) {
// Check if the next node has the same data as
// the curr node
if (temp.next.data === curr.data) {
// Duplicate found, remove it
let duplicate = temp.next;
temp.next = temp.next.next;
// Update the previous pointer of the next
// node, if it exists
if (temp.next !== null) {
temp.next.prev = temp;
}
// Free the memory of the duplicate node
delete duplicate;
}
else {
// If the next node has different data from
// the current node, move to the next node
temp = temp.next;
}
}
// Move to the next node in the list
curr = curr.next;
}
return head;
}
// Create a doubly linked list:
// 1 <-> 2 <-> 3 <-> 2 <-> 4
let head = new Node(1);
head.next = new Node(2);
head.next.prev = head;
head.next.next = new Node(3);
head.next.next.prev = head.next;
head.next.next.next = new Node(2);
head.next.next.next.prev = head.next.next;
head.next.next.next.next = new Node(4);
head.next.next.next.next.prev = head.next.next.next;
head = removeDuplicates(head);
printList(head);
Time Complexity: O(n^2), Two nested loops are used to find the duplicates.
Auxiliary Space: O(1)
[Expected Approach] Using HashSet - O(n) Time and O(n) Space
The idea is to traverse the doubly linked list from head to end. For every newly encountered element
- If the element is already present in the HashSet, remove it from the list.
- If the element is not in the HashSet, add it to the HashSet and move ahead in the list.
Step-by-step implementation:
- Create a hash
set for keeping track of data nodes. - Start traversing from the head of the list.
- For each node check if node’s data is in the set:
- If present, update pointers to remove the node and delete the node.
- If not present then add the data into the set.
Below is the implementation of the above approach:
C++
// C++ implementation to remove duplicates from
// an unsorted doubly linked list using hashing
#include <iostream>
#include <unordered_set>
using namespace std;
class Node {
public:
int data;
Node *next;
Node *prev;
Node(int x) {
data = x;
prev = nullptr;
next = nullptr;
}
};
Node *removeDuplicates(Node *head) {
unordered_set<int> hashSet;
Node *curr = head;
while (curr != nullptr) {
// Check if the element is already in the hash table
if (hashSet.find(curr->data) != hashSet.end()) {
// Element is present, remove it
// Adjust previous node's next pointer
if (curr->prev)
curr->prev->next = curr->next;
// Adjust next node's prev pointer
if (curr->next)
curr->next->prev = curr->prev;
// Delete the curr node
Node *temp = curr;
curr = curr->next;
delete temp;
}
else {
// Element is not present, add it to hash table
hashSet.insert(curr->data);
curr = curr->next;
}
}
return head;
}
void printList(Node *head) {
Node *curr = head;
while (curr != nullptr) {
cout << curr->data << " ";
curr = curr->next;
}
cout << endl;
}
int main() {
// Create a doubly linked list:
// 1 <-> 2 <-> 3 <-> 2 <-> 4
Node *head = new Node(1);
head->next = new Node(2);
head->next->prev = head;
head->next->next = new Node(3);
head->next->next->prev = head->next;
head->next->next->next = new Node(2);
head->next->next->next->prev = head->next->next;
head->next->next->next->next = new Node(4);
head->next->next->next->next->prev = head->next->next->next;
head = removeDuplicates(head);
printList(head);
return 0;
}
// Java implementation to remove duplicates from
// an unsorted doubly linked list using hashing
import java.util.HashSet;
class Node {
int data;
Node next;
Node prev;
Node(int x) {
data = x;
next = null;
prev = null;
}
}
class GfG {
static Node removeDuplicates(Node head) {
HashSet<Integer> hashSet = new HashSet<>();
Node curr = head;
while (curr != null) {
// Check if the element is already in the hash
// table
if (hashSet.contains(curr.data)) {
// Element is present, remove it
// Adjust previous node's next pointer
if (curr.prev != null)
curr.prev.next = curr.next;
// Adjust next node's prev pointer
if (curr.next != null)
curr.next.prev = curr.prev;
// Delete the curr node
Node temp = curr;
curr = curr.next;
temp = null;
}
else {
// Element is not present, add it to hash
// table
hashSet.add(curr.data);
curr = curr.next;
}
}
return head;
}
static void printList(Node head) {
Node curr = head;
while (curr != null) {
System.out.print(curr.data + " ");
curr = curr.next;
}
System.out.println();
}
public static void main(String[] args) {
// Create a doubly linked list:
// 1 <-> 2 <-> 3 <-> 2 <-> 4
Node head = new Node(1);
head.next = new Node(2);
head.next.prev = head;
head.next.next = new Node(3);
head.next.next.prev = head.next;
head.next.next.next = new Node(2);
head.next.next.next.prev = head.next.next;
head.next.next.next.next = new Node(4);
head.next.next.next.next.prev = head.next.next.next;
head = removeDuplicates(head);
printList(head);
}
}
# Python implementation to remove duplicates
# from an unsorted doubly linked list using hashing
class Node:
def __init__(self, x):
self.data = x
self.next = None
self.prev = None
def remove_duplicates(head):
hash_set = set()
curr = head
while curr is not None:
# Check if the element is already in the hash table
if curr.data in hash_set:
# Element is present, remove it
# Adjust previous node's next pointer
if curr.prev:
curr.prev.next = curr.next
# Adjust next node's prev pointer
if curr.next:
curr.next.prev = curr.prev
# Delete the curr node
temp = curr
curr = curr.next
del temp
else:
# Element is not present, add it to hash table
hash_set.add(curr.data)
curr = curr.next
return head
def print_list(head):
curr = head
while curr is not None:
print(curr.data, end=" ")
curr = curr.next
print()
if __name__ == "__main__":
# Create a doubly linked list:
# 1 <-> 2 <-> 3 <-> 2 <-> 4
head = Node(1)
head.next = Node(2)
head.next.prev = head
head.next.next = Node(3)
head.next.next.prev = head.next
head.next.next.next = Node(2)
head.next.next.next.prev = head.next.next
head.next.next.next.next = Node(4)
head.next.next.next.next.prev = head.next.next.next
head = remove_duplicates(head)
print_list(head)
// C# implementation to remove duplicates from
// an unsorted doubly linked list using hashing
using System;
using System.Collections.Generic;
class Node {
public int Data;
public Node Next;
public Node Prev;
public Node(int x) {
Data = x;
Next = null;
Prev = null;
}
}
class GfG {
static Node RemoveDuplicates(Node head) {
HashSet<int> hashSet = new HashSet<int>();
Node curr = head;
while (curr != null) {
// Check if the element is already in the hash table
if (hashSet.Contains(curr.Data)) {
// Element is present, remove it
// Adjust previous node's next pointer
if (curr.Prev != null)
curr.Prev.Next = curr.Next;
// Adjust next node's prev pointer
if (curr.Next != null)
curr.Next.Prev = curr.Prev;
// Delete the current node
curr = curr.Next;
}
else {
// Element is not present, add it to hash table
hashSet.Add(curr.Data);
curr = curr.Next;
}
}
return head;
}
static void PrintList(Node head) {
Node curr = head;
while (curr != null) {
Console.Write(curr.Data + " ");
curr = curr.Next;
}
Console.WriteLine();
}
static void Main() {
// Create a doubly linked list:
// 1 <-> 2 <-> 3 <-> 2 <-> 4
Node head = new Node(1);
head.Next = new Node(2);
head.Next.Prev = head;
head.Next.Next = new Node(3);
head.Next.Next.Prev = head.Next;
head.Next.Next.Next = new Node(2);
head.Next.Next.Next.Prev = head.Next.Next;
head.Next.Next.Next.Next = new Node(4);
head.Next.Next.Next.Next.Prev = head.Next.Next.Next;
head = RemoveDuplicates(head);
PrintList(head);
}
}
// Javascript implementation to remove duplicates
// from an unsorted doubly linked list using hashing
class Node {
constructor(x) {
this.data = x;
this.next = null;
this.prev = null;
}
}
function removeDuplicates(head) {
const hashSet = new Set();
let curr = head;
while (curr !== null) {
// Check if the element is already in the hash table
if (hashSet.has(curr.data)) {
// Element is present, remove it
// Adjust previous node's next pointer
if (curr.prev)
curr.prev.next = curr.next;
// Adjust next node's prev pointer
if (curr.next)
curr.next.prev = curr.prev;
// Delete the curr node
let temp = curr;
curr = curr.next;
temp = null;
}
else {
// Element is not present, add it to hash table
hashSet.add(curr.data);
curr = curr.next;
}
}
return head;
}
function printList(head) {
let curr = head;
while (curr !== null) {
console.log(curr.data + " ");
curr = curr.next;
}
console.log();
}
// Create a doubly linked list:
// 1 <-> 2 <-> 3 <-> 2 <-> 4
let head = new Node(1);
head.next = new Node(2);
head.next.prev = head;
head.next.next = new Node(3);
head.next.next.prev = head.next;
head.next.next.next = new Node(2);
head.next.next.next.prev = head.next.next;
head.next.next.next.next = new Node(4);
head.next.next.next.next.prev = head.next.next.next;
head = removeDuplicates(head);
printList(head);
Time Complexity: O(n) , where n are the number of nodes in the doubly linked list.
Auxiliary Space: O(n)
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