Remove Duplicates from an Unsorted Linked List
Last Updated :
03 Feb, 2025
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Given an unsorted linked list containing n nodes, the task is to remove duplicate nodes while preserving the original order.
Examples:
Input: 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
Output: 12 -> 11 -> 21 -> 41 -> 43
Explanation: The second occurrence of 12 (the one after 11) and the second occurrence of 21 (the one at the end) are removed, resulting in a linked list that maintains the order of their first appearances.Input: 1 -> 2 -> 3 -> 2 -> 4
Output: 1 -> 2 -> 3 -> 4
Explanation: Similarly, the second occurrence of 2 is removed, ensuring that each number appears only once while maintaining the order of their first appearances.
Table of Content
[Naive Approach] Using Nested Loops – O(n^2) Time and O(1) Space
The idea is to use two loops to remove duplicates from an unsorted linked list. The first loop goes through each node one by one. For each node, the second loop checks all the nodes that come after it to see if there are any duplicates. If a duplicate is found, it removes it by changing the links. This way, we keep only the first occurrence of each number while maintaining their order.
// C++ program to remove duplicates from an
// unsorted linked list
#include <iostream>
using namespace std;
class Node {
public:
int data;
Node* next;
Node(int val) {
data = val;
next = nullptr;
}
};
// Function to remove duplicates using nested loops
Node* removeDuplicates(Node* head) {
Node* curr1 = head;
// Traverse each node in the list
while (curr1 != nullptr) {
Node* curr2 = curr1;
// Traverse the remaining nodes to find and
// remove duplicates
while (curr2->next != nullptr) {
// Check if the next node has the same
// data as the current node
if (curr2->next->data == curr1->data) {
// Duplicate found, remove it
Node* duplicate = curr2->next;
curr2->next = curr2->next->next;
// 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
curr2 = curr2->next;
}
}
// Move to the next node in the list
curr1 = curr1->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 singly linked list:
// 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
Node* head = new Node(12);
head->next = new Node(11);
head->next->next = new Node(12);
head->next->next->next = new Node(21);
head->next->next->next->next = new Node(41);
head->next->next->next->next->next = new Node(43);
head->next->next->next->next->next->next = new Node(21);
head = removeDuplicates(head);
printList(head);
return 0;
}
// C program to remove duplicates from an
// unsorted linked list
#include <stdio.h>
#include <stdlib.h>
struct Node {
int data;
struct Node* next;
};
// Function to remove duplicates using nested loops
struct Node* removeDuplicates(struct Node* head) {
struct Node* curr1 = head;
// Traverse each node in the list
while (curr1 != NULL) {
struct Node* curr2 = curr1;
// Traverse the remaining nodes to find and
// remove duplicates
while (curr2->next != NULL) {
// Check if the next node has the same
// data as the current node
if (curr2->next->data == curr1->data) {
// Duplicate found, remove it
struct Node* duplicate = curr2->next;
curr2->next = curr2->next->next;
// 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
curr2 = curr2->next;
}
}
// Move to the next node in the list
curr1 = curr1->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 data) {
struct Node* newNode =
(struct Node*)malloc(sizeof(struct Node));
newNode->data = data;
newNode->next = NULL;
return newNode;
}
int main() {
// Create a singly linked list:
// 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
struct Node* head = createNode(12);
head->next = createNode(11);
head->next->next = createNode(12);
head->next->next->next = createNode(21);
head->next->next->next->next = createNode(41);
head->next->next->next->next->next = createNode(43);
head->next->next->next->next->next->next = createNode(21);
head = removeDuplicates(head);
printList(head);
return 0;
}
// Java program to remove duplicates from an
// unsorted linked list
class Node {
int data;
Node next;
Node(int val) {
this.data = val;
this.next = null;
}
}
// Function to remove duplicates using nested loops
class GfG {
static Node removeDuplicates(Node head) {
Node curr1 = head;
// Traverse each node in the list
while (curr1 != null) {
Node curr2 = curr1;
// Traverse the remaining nodes to find and
// remove duplicates
while (curr2.next != null) {
// Check if the next node has the same
// data as the current node
if (curr2.next.data == curr1.data) {
// Duplicate found, remove it
curr2.next = curr2.next.next;
} else {
// If the next node has different data from
// the current node, move to the next node
curr2 = curr2.next;
}
}
// Move to the next node in the list
curr1 = curr1.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 singly linked list:
// 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
Node head = new Node(12);
head.next = new Node(11);
head.next.next = new Node(12);
head.next.next.next = new Node(21);
head.next.next.next.next = new Node(41);
head.next.next.next.next.next = new Node(43);
head.next.next.next.next.next.next = new Node(21);
head = removeDuplicates(head);
printList(head);
}
}
# Python program to remove duplicates from an
# unsorted linked list
class Node:
def __init__(self, val):
self.data = val
self.next = None
# Function to remove duplicates using nested loops
def remove_duplicates(head):
curr1 = head
# Traverse each node in the list
while curr1:
curr2 = curr1
# Traverse the remaining nodes to find and
# remove duplicates
while curr2.next:
# Check if the next node has the same
# data as the current node
if curr2.next.data == curr1.data:
# Duplicate found, remove it
curr2.next = curr2.next.next
else:
# If the next node has different data from
# the current node, move to the next node
curr2 = curr2.next
# Move to the next node in the list
curr1 = curr1.next
return head
def print_list(head):
curr = head
while curr:
print(curr.data, end=" ")
curr = curr.next
print()
if __name__ == "__main__":
# Create a singly linked list:
# 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
head = Node(12)
head.next = Node(11)
head.next.next = Node(12)
head.next.next.next = Node(21)
head.next.next.next.next = Node(41)
head.next.next.next.next.next = Node(43)
head.next.next.next.next.next.next = Node(21)
head = remove_duplicates(head)
print_list(head)
// C# program to remove duplicates from an
// unsorted linked list
using System;
class Node {
public int Data;
public Node Next;
public Node(int val) {
this.Data = val;
this.Next = null;
}
}
// Function to remove duplicates using nested loops
class GfG {
static Node RemoveDuplicates(Node head) {
Node curr1 = head;
// Traverse each node in the list
while (curr1 != null) {
Node curr2 = curr1;
// Traverse the remaining nodes to find and
// remove duplicates
while (curr2.Next != null) {
// Check if the next node has the same
// data as the current node
if (curr2.Next.Data == curr1.Data) {
// Duplicate found, remove it
curr2.Next = curr2.Next.Next;
} else {
// If the next node has different data from
// the current node, move to the next node
curr2 = curr2.Next;
}
}
// Move to the next node in the list
curr1 = curr1.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 singly linked list:
// 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
Node head = new Node(12);
head.Next = new Node(11);
head.Next.Next = new Node(12);
head.Next.Next.Next = new Node(21);
head.Next.Next.Next.Next = new Node(41);
head.Next.Next.Next.Next.Next = new Node(43);
head.Next.Next.Next.Next.Next.Next = new Node(21);
head = RemoveDuplicates(head);
PrintList(head);
}
}
// JavaScript program to remove duplicates from an
// unsorted linked list
class Node {
constructor(val)
{
this.data = val;
this.next = null;
}
}
// Function to remove duplicates using nested loops
function removeDuplicates(head) {
let curr1 = head;
// Traverse each node in the list
while (curr1 != null) {
let curr2 = curr1;
// Traverse the remaining nodes to find and
// remove duplicates
while (curr2.next != null) {
// Check if the next node has the same
// data as the current node
if (curr2.next.data === curr1.data) {
// Duplicate found, remove it
curr2.next = curr2.next.next;
}
else {
// If the next node has different data from
// the current node, move to the next node
curr2 = curr2.next;
}
}
// Move to the next node in the list
curr1 = curr1.next;
}
return head;
}
function printList(head) {
let curr = head;
while (curr != null) {
console.log(curr.data + " ");
curr = curr.next;
}
console.log();
}
// Create a singly linked list:
// 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
let head = new Node(12);
head.next = new Node(11);
head.next.next = new Node(12);
head.next.next.next = new Node(21);
head.next.next.next.next = new Node(41);
head.next.next.next.next.next = new Node(43);
head.next.next.next.next.next.next = new Node(21);
head = removeDuplicates(head);
printList(head);
Output
12 11 21 41 43
Time Complexity: O(n^2), Due to two nested loops
Auxiliary Space: O(1)
[Expected Approach] Using HashSet – O(n) Time and O(n) Space
In this approach, we can use a hash set to keep track of the values (nodes) that have already been seen. As we traverse the linked list, for each node, we check if its value is already in the hash set. If the value is found, it means it’s a duplicate, so we remove that node by adjusting the pointers of the previous node to skip the current one. If the value is not found, we add it to the hash set and move to the next node.
// C++ implementation to remove duplicates from
// an unsorted singly linked list using hashing
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node *next;
Node(int x) {
data = x;
next = nullptr;
}
};
Node *removeDuplicates(Node *head) {
unordered_set<int> hashSet;
Node *curr = head;
Node *prev = nullptr;
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
prev->next = curr->next;
// 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);
prev = curr;
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 singly linked list:
// 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
Node* head = new Node(12);
head->next = new Node(11);
head->next->next = new Node(12);
head->next->next->next = new Node(21);
head->next->next->next->next = new Node(41);
head->next->next->next->next->next = new Node(43);
head->next->next->next->next->next->next = new Node(21);
head = removeDuplicates(head);
printList(head);
return 0;
}
// Java implementation to remove duplicates from
// an unsorted singly linked list using hashing
import java.util.HashSet;
class Node {
int data;
Node next;
Node(int x) {
data = x;
next = null;
}
}
class GfG {
static Node removeDuplicates(Node head) {
HashSet<Integer> hashSet = new HashSet<>();
Node curr = head;
Node prev = null;
while (curr != null) {
// Check if the element is already in the hash table
if (hashSet.contains(curr.data)) {
// Element is present, remove it
prev.next = curr.next;
} else {
// Element is not present, add it to hash table
hashSet.add(curr.data);
prev = curr;
}
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 singly linked list:
// 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
Node head = new Node(12);
head.next = new Node(11);
head.next.next = new Node(12);
head.next.next.next = new Node(21);
head.next.next.next.next = new Node(41);
head.next.next.next.next.next = new Node(43);
head.next.next.next.next.next.next = new Node(21);
head = removeDuplicates(head);
printList(head);
}
}
# Python implementation to remove duplicates from
# an unsorted singly linked list using hashing
class Node:
def __init__(self, x):
self.data = x
self.next = None
def remove_duplicates(head):
hash_set = set()
curr = head
prev = None
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
prev.next = curr.next
curr = curr.next
else:
# Element is not present, add it to hash table
hash_set.add(curr.data)
prev = curr
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 singly linked list:
# 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
head = Node(12)
head.next = Node(11)
head.next.next = Node(12)
head.next.next.next = Node(21)
head.next.next.next.next = Node(41)
head.next.next.next.next.next = Node(43)
head.next.next.next.next.next.next = Node(21)
head = remove_duplicates(head)
print_list(head)
// C# implementation to remove duplicates from
// an unsorted singly linked list using hashing
using System;
using System.Collections.Generic;
class Node {
public int Data;
public Node Next;
public Node(int x) {
Data = x;
Next = null;
}
}
class GfG {
static Node RemoveDuplicates(Node head) {
HashSet<int> hashSet = new HashSet<int>();
Node curr = head;
Node prev = null;
while (curr != null) {
// Check if the element is already in the hash table
if (hashSet.Contains(curr.Data)) {
// Element is present, remove it
prev.Next = curr.Next;
curr = curr.Next;
} else {
// Element is not present, add it to hash table
hashSet.Add(curr.Data);
prev = curr;
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 singly linked list:
// 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
Node head = new Node(12);
head.Next = new Node(11);
head.Next.Next = new Node(12);
head.Next.Next.Next = new Node(21);
head.Next.Next.Next.Next = new Node(41);
head.Next.Next.Next.Next.Next = new Node(43);
head.Next.Next.Next.Next.Next.Next = new Node(21);
head = RemoveDuplicates(head);
PrintList(head);
}
}
// JavaScript implementation to remove duplicates from
// an unsorted singly linked list using hashing
class Node {
constructor(x) {
this.data = x;
this.next = null;
}
}
function removeDuplicates(head) {
const hashSet = new Set();
let curr = head;
let prev = null;
while (curr !== null) {
// Check if the element is already in the hash table
if (hashSet.has(curr.data)) {
// Element is present, remove it
prev.next = curr.next;
curr = curr.next;
} else {
// Element is not present, add it to hash table
hashSet.add(curr.data);
prev = curr;
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 singly linked list:
// 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
let head = new Node(12);
head.next = new Node(11);
head.next.next = new Node(12);
head.next.next.next = new Node(21);
head.next.next.next.next = new Node(41);
head.next.next.next.next.next = new Node(43);
head.next.next.next.next.next.next = new Node(21);
head = removeDuplicates(head);
printList(head);
Output
12 11 21 41 43
Time Complexity: O(n), where n are the number of nodes in the linked list.
Auxiliary Space: O(n)
