Given an unsorted Linked List, the task is to remove duplicates from the list.
Examples:
Input: linked_list = 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
Output: 12 -> 11 -> 21 -> 41 -> 43
Explanation: Second occurrence of 12 and 21 are removed.Input: linked_list = 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
Output: 12 -> 11 -> 21 -> 41 -> 43
Naive Approach to Remove Duplicates from an Unsorted Linked List:
The most simple approach to solve this, is to check each node for duplicate in the Linked List one by one.
Below is the Implementation of the above approach:
Javascript
// Javascript program to remove duplicates from unsorted // linked list var head; class Node { constructor(val) { this.data = val; this.next = null; } } /* * Function to remove duplicates from an unsorted linked list */ function remove_duplicates() { var ptr1 = null, ptr2 = null, dup = null; ptr1 = head; /* Pick elements one by one */ while (ptr1 != null && ptr1.next != null) { ptr2 = ptr1; /* * Compare the picked element with rest of the elements */ while (ptr2.next != null) { /* If duplicate then delete it */ if (ptr1.data == ptr2.next.data) { /* sequence of steps is important here */ dup = ptr2.next; ptr2.next = ptr2.next.next; } else /* This is tricky */ { ptr2 = ptr2.next; } } ptr1 = ptr1.next; } } function printList( node) { while (node != null) { console.log(node.data + " "); node = node.next; } } head = new Node(10); head.next = new Node(12); head.next.next = new Node(11); head.next.next.next = new Node(11); head.next.next.next.next = new Node(12); head.next.next.next.next.next = new Node(11); head.next.next.next.next.next.next = new Node(10); console.log("Linked List before removing duplicates : <br/> "); printList(head); remove_duplicates(); console.log("<br/>"); console.log("Linked List after removing duplicates : <br/> "); printList(head);// This code contributed by aashish1995 |
Output
Linked List before removing duplicates : <br/> 10 12 11 11 12 11 10 <br/> Linked List after removing duplicates : <br/> 10 12 11
Time Complexity: O(N2)
Auxiliary Space: O(1)
Remove duplicates from an Unsorted Linked List using Sorting:
Follow the below steps to Implement the idea:
- Sort the elements using Merge Sort for Linked Lists.
- Remove duplicates in linear time using the algorithm for removing duplicates in sorted Linked List.
Below is the implementation for above approach:
Javascript
// structure of a node in the linked listclass Node {constructor(data) {this.data = data;this.next = null;}}// function to insert a node in the linked listfunction push(head_ref, new_data) {const new_node = new Node(new_data);new_node.next = head_ref;head_ref = new_node;return head_ref;}// function to sort the linked listfunction sortList(head) {// pointer to traverse the linked listlet current = head;let index = null;// loop to traverse the linked listwhile (current !== null) { // loop to compare current node with all other nodes index = current.next;while (index !== null) { // checking for duplicate values if (current.data === index.data) { // deleting the duplicate node current.next = index.next; index = current.next; } else { index = index.next; }}current = current.next;}}// function to display the linked listfunction printList(node) {let str = "";while (node !== null) {str += node.data + " ";node = node.next;}console.log(str);}// main functionlet head = null;head = push(head, 20);head = push(head, 13);head = push(head, 13);head = push(head, 11);head = push(head, 11);head = push(head, 11);console.log("Linked List before removing duplicates : ");printList(head);sortList(head);console.log("Linked List after removing duplicates : ");printList(head); |
Output
Linked List before removing duplicates : 11 11 11 13 13 20 Linked List after removing duplicates : 11 13 20
Time Complexity: O(N log N)
Auxiliary Space: O(1)
Remove duplicates from an Unsorted Linked List using Hashing:
The idea for this approach is based on the following observations:
- Traverse the link list from head to end.
- For every newly encountered element, check whether if it is in the hash table:
- if yes, remove it
- otherwise put it in the hash table.
- At the end, the Hash table will contain only the unique elements.
Below is the implementation of the above approach:
Javascript
// JavaScript program to remove duplicates// from unsorted linkedlist class node { constructor(val) { this.val = val; this.next = null; } } /* Function to remove duplicates from a unsorted linked list */ function removeDuplicate( head) { // Hash to store seen values var hs = new Set(); /* Pick elements one by one */ var current = head; var prev = null; while (current != null) { var curval = current.val; // If current value is seen before if (hs.has(curval)) { prev.next = current.next; } else { hs.add(curval); prev = current; } current = current.next; } } /* Function to print nodes in a given linked list */ function printList( head) { while (head != null) { console.log(head.val + " "); head = head.next; } } /* * The constructed linked list is: 10->12->11->11->12->11->10 */ start = new node(10); start.next = new node(12); start.next.next = new node(11); start.next.next.next = new node(11); start.next.next.next.next = new node(12); start.next.next.next.next.next = new node(11); start.next.next.next.next.next.next = new node(10); console.log( "Linked list before removing duplicates :<br/>" ); printList(start); removeDuplicate(start); console.log( "<br/>Linked list after removing duplicates :<br/>" ); printList(start);// This code is contributed by todaysgaurav |
Output
Linked list before removing duplicates :<br/> 10 12 11 11 12 11 10 <br/>Linked list after removing duplicates :<br/> 10 12 11
Time Complexity: O(N), on average (assuming that hash table access time is O(1) on average).
Auxiliary Space: O(N), As extra space is used to store the elements in the stack.
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