Remove Duplicates from an Unsorted Linked List (original) (raw)

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
• [Expected Approach] Using HashSet - O(n) Time and O(n) Space

[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++ `

// C++ program to remove duplicates from an // unsorted linked list

#include 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

// 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

// 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

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#

// 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

// 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);

`

**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++ `

// 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 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

// 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 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

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#

// 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 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;
            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

// 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);

`

**Time Complexity: O(n), where n are the number of nodes in the linked list.
**Auxiliary Space: O(n)