Skip to content
geeksforgeeks
  • Courses
    • DSA to Development
    • Get IBM Certification
    • Newly Launched!
      • Master Django Framework
      • Become AWS Certified
    • For Working Professionals
      • Interview 101: DSA & System Design
      • Data Science Training Program
      • JAVA Backend Development (Live)
      • DevOps Engineering (LIVE)
      • Data Structures & Algorithms in Python
    • For Students
      • Placement Preparation Course
      • Data Science (Live)
      • Data Structure & Algorithm-Self Paced (C++/JAVA)
      • Master Competitive Programming (Live)
      • Full Stack Development with React & Node JS (Live)
    • Full Stack Development
    • Data Science Program
    • All Courses
  • Tutorials
    • Data Structures & Algorithms
    • ML & Data Science
    • Interview Corner
    • Programming Languages
    • Web Development
    • CS Subjects
    • DevOps And Linux
    • School Learning
  • Practice
    • Build your AI Agent
    • GfG 160
    • Problem of the Day
    • Practice Coding Problems
    • GfG SDE Sheet
  • Contests
    • Accenture Hackathon (Ending Soon!)
    • GfG Weekly [Rated Contest]
    • Job-A-Thon Hiring Challenge
    • All Contests and Events
  • DSA
  • Interview Problems on Linked List
  • Practice Linked List
  • MCQs on Linked List
  • Linked List Tutorial
  • Types of Linked List
  • Singly Linked List
  • Doubly Linked List
  • Circular Linked List
  • Circular Doubly Linked List
  • Linked List vs Array
  • Time & Space Complexity
  • Advantages & Disadvantages
Open In App

Difference of two Linked Lists using Merge sort

Last Updated : 04 Dec, 2023
Comments
Improve
Suggest changes
Like Article
Like
Report

Given two Linked List, the task is to create a Linked List to store the difference of Linked List 1 with Linked List 2, i.e. the elements present in List 1 but not in List 2.
Examples: 
 

Input: 
List1: 10 -> 15 -> 4 ->20, 
List2: 8 -> 4 -> 2 -> 10 
Output: 15 -> 20 
Explanation: 
In the given linked list elements 15 and 20 are present in the list 1 but not in list 2.
Input: 
List1: 2 -> 4 -> 8 -> 10, 
List2: 8 -> 10 
Output: 2 -> 4 
Explanation: 
In the given linked list 1 elements 2 and 4 are present in the list 1 but not in list 2. 
 

 

Approach: 
 

  • Sort both Linked Lists using merge sort.
  • Linearly scan both sorted lists to get the difference with two pointers on it p1 and p2 and compare the data of the nodes in the linked list and perform following in below three cases – 
    1. If p1.data == p2.data then, p1.data cannot be in the difference list, So move the pointers p1 and p2 ahead.
    2. If p1.data > p2.data then, p1.data may be present in list 2 in further nodes, So move the pointer p2 ahead.
    3. If p1.data > p2.data then, p1.data cannot be present in list 2 now, So add the data of p1 into difference list and move pointer p1 ahead.
  • If the end of list 2 is reached, insert all the remaining elements of list 1 into the difference list.

Below is the implementation of the above approach.
 

C++




//C++ program to implement above approach
 
#include <iostream>
 
class Node {
public:
    int data;
    Node* next;
 
    Node(int val) : data(val), next(nullptr) {}
};
 
class LinkedList {
public:
    Node* head;
 
    LinkedList() : head(nullptr) {}
 
    // Function to insert a node at the end of the linked list
    void append(int data) {
        Node* temp = new Node(data);
        if (head == nullptr) {
            head = temp;
        } else {
            Node* p = head;
            while (p->next != nullptr) {
                p = p->next;
            }
            p->next = temp;
        }
    }
 
    // Function to find the middle node of the linked list
    Node* get_mid(Node* head) {
        if (head == nullptr) {
            return head;
        }
        Node* slow = head;
        Node* fast = head;
        while (fast->next != nullptr && fast->next->next != nullptr) {
            slow = slow->next;
            fast = fast->next->next;
        }
        return slow;
    }
 
    // Recursive method to merge two halves after sorting
    Node* merge(Node* l, Node* r) {
        if (l == nullptr) return r;
        if (r == nullptr) return l;
 
        Node* result;
        if (l->data <= r->data) {
            result = l;
            result->next = merge(l->next, r);
        } else {
            result = r;
            result->next = merge(l, r->next);
        }
        return result;
    }
 
    // Recursive method to divide the list into two halves until 1 node left
    Node* merge_sort(Node* head) {
        if (head == nullptr || head->next == nullptr) {
            return head;
        }
        Node* mid = get_mid(head);
        Node* next_to_mid = mid->next;
        mid->next = nullptr;
        Node* left = merge_sort(head);
        Node* right = merge_sort(next_to_mid);
        Node* sorted_merge = merge(left, right);
        return sorted_merge;
    }
 
    // Function to print the list elements
    void display() {
        Node* p = head;
        while (p != nullptr) {
            std::cout << p->data << " ";
            p = p->next;
        }
        std::cout << std::endl;
    }
};
 
// Function to get the difference list
LinkedList get_difference(Node* p1, Node* p2) {
    LinkedList difference_list;
    // Scan the lists
    while (p1 != nullptr && p2 != nullptr) {
        // Condition to check if the data of both pointers are the same, then move ahead
        if (p2->data == p1->data) {
            p1 = p1->next;
            p2 = p2->next;
        } else if (p2->data < p1->data) { // Condition to check if the data of the second pointer is smaller than the first, then move the second pointer ahead
            p2 = p2->next;
        } else { // Condition when the data of the first pointer is greater than the second, then append into the difference list and move
            difference_list.append(p1->data);
            p1 = p1->next;
        }
    }
 
    // If the end of list2 is reached, there may be some nodes in List 1 left to be scanned,
    // they all will be inserted in the difference list
    while (p1 != nullptr) {
        difference_list.append(p1->data);
        p1 = p1->next;
    }
 
    return difference_list;
}
 
int main() {
    // Linked List 1
    LinkedList list1;
    list1.append(2);
    list1.append(6);
    list1.append(8);
    list1.append(1);
 
    // Linked List 2
    LinkedList list2;
    list2.append(4);
    list2.append(1);
    list2.append(9);
 
    // Sort both linked lists
    list1.head = list1.merge_sort(list1.head);
    list2.head = list2.merge_sort(list2.head);
 
    // Get the difference list
    LinkedList result = get_difference(list1.head, list2.head);
 
    if (result.head) {
        result.display();
    } else {
        std::cout << "Lists are equal" << std::endl;
    }
 
    return 0;
}
 
 

Java




class Node {
    int data;
    Node next;
 
    Node(int data) {
        this.data = data;
        this.next = null;
    }
}
 
class LinkedList {
    Node head;
 
    LinkedList() {
        head = null;
    }
 
    // Function to insert a node at the end of Linked List
    void append(int data) {
        Node newNode = new Node(data);
        if (head == null) {
            head = newNode;
        } else {
            Node current = head;
            while (current.next != null) {
                current = current.next;
            }
            current.next = newNode;
        }
    }
 
    // Function to find the middle node of the Linked List
    Node getMid(Node head) {
        if (head == null) {
            return head;
        }
        Node slow = head;
        Node fast = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
 
    // Recursive method to merge two halves after sorting
    Node merge(Node left, Node right) {
        if (left == null) {
            return right;
        }
        if (right == null) {
            return left;
        }
 
        Node result;
        if (left.data <= right.data) {
            result = left;
            result.next = merge(left.next, right);
        } else {
            result = right;
            result.next = merge(left, right.next);
        }
        return result;
    }
 
    // Recursive method to divide the list into two halves until 1 node left
    Node mergeSort(Node head) {
        if (head == null || head.next == null) {
            return head;
        }
 
        Node mid = getMid(head);
        Node nextToMid = mid.next;
        mid.next = null;
        Node left = mergeSort(head);
        Node right = mergeSort(nextToMid);
        return merge(left, right);
    }
 
    // Function to print the list elements
    void display() {
        Node current = head;
        while (current != null) {
            System.out.print(current.data + " ");
            current = current.next;
        }
        System.out.println();
    }
 
    // Function to get the difference list
    static LinkedList getDifference(Node p1, Node p2) {
        LinkedList differenceList = new LinkedList();
 
        // Scan the lists
        while (p1 != null && p2 != null) {
 
            // Condition to check if the data of both pointers are the same, then move ahead
            if (p2.data == p1.data) {
                p1 = p1.next;
                p2 = p2.next;
            } else if (p2.data < p1.data) { // Condition to check if the data of the second pointer is smaller than the first, then move the second pointer ahead
                p2 = p2.next;
            } else { // Condition when the data of the first pointer is greater than the second, then append into the difference list and move
                differenceList.append(p1.data);
                p1 = p1.next;
            }
        }
 
        // If the end of list2 is reached, there may be some nodes in List 1 left to be scanned, they all will be inserted in the difference list
        while (p1 != null) {
            differenceList.append(p1.data);
            p1 = p1.next;
        }
 
        return differenceList;
    }
}
 
public class Main {
    public static void main(String[] args) {
        // Linked List 1
        LinkedList list1 = new LinkedList();
        list1.append(2);
        list1.append(6);
        list1.append(8);
        list1.append(1);
 
        // Linked List 2
        LinkedList list2 = new LinkedList();
        list2.append(4);
        list2.append(1);
        list2.append(9);
 
        // Sort both the linked lists
        list1.head = list1.mergeSort(list1.head);
        list2.head = list2.mergeSort(list2.head);
 
        // Get the difference list
        LinkedList result = LinkedList.getDifference(list1.head, list2.head);
 
        if (result.head != null) {
            result.display();
        } else {
            System.out.println("Lists are equal");
        }
    }
}
 
 

Python3




# Python implementation to create
# a difference Linked List of
# two Linked Lists
 
# Node of the Linked List
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
 
# Linked List
class linked_list:
    def __init__(self):
        self.head = None
     
    # Function to insert a node
    # at the end of Linked List
    def append(self, data):
        temp = Node(data)
        if self.head == None:
            self.head = temp
        else:
            p = self.head
            while p.next != None:
                p = p.next
            p.next = temp
 
    # Function to find the middle
    # node of the Linked List
    def get_mid(self, head):
        if head == None:
            return head
        slow = fast = head
        while fast.next != None \
           and fast.next.next != None:
            slow = slow.next
            fast = fast.next.next
        return slow
 
    # Recursive method to merge the
    # two half after sorting
    def merge(self, l, r):
        if l == None:return r
        if r == None:return l
 
        if l.data<= r.data:
            result = l
            result.next = \
                self.merge(l.next, r)
        else:
            result = r
            result.next = \
                self.merge(l, r.next)
        return result
 
    # Recursive method to divide the
    # list into two half until 1 node left
    def merge_sort(self, head):
        if head == None or head.next == None:
            return head
        mid = self.get_mid(head)
        next_to_mid = mid.next
        mid.next = None
        left = self.merge_sort(head)
        right = self.merge_sort(next_to_mid)
        sorted_merge = self.merge(left, right)
        return sorted_merge
 
    # Function to print the list elements
    def display(self):
        p = self.head
        while p != None:
            print(p.data, end =' ')
            p = p.next
        print()
 
# Function to get the difference list
def get_difference(p1, p2):
    difference_list = linked_list()
    # Scan the lists
    while p1 != None and p2 != None:
         
        # Condition to check if the
        # Data of the both pointer are
        # same then move ahead
        if p2.data == p1.data:
            p1 = p1.next
            p2 = p2.next
         
        # Condition to check if the
        # Data of the first pointer is
        # greater than second then
        # move second pointer ahead
        elif p2.data<p1.data:
            p2 = p2.next
         
        # Condition when first pointer
        # data is greater than the
        # second pointer then append
        # into the difference list and move
        else:
            difference_list.append(p1.data)
            p1 = p1.next
 
    # If end of list2 is reached,
    # there may be some nodes in
    # List 1 left to be scanned,
    # they all will be inserted
    # in the difference list
    if p2 == None:
        while p1:
            difference_list.append(p1.data)
            p1 = p1.next
 
    return difference_list
 
# Driver Code
if __name__ == '__main__':
    # Linked List 1
    list1 = linked_list()
    list1.append(2)
    list1.append(6)
    list1.append(8)
    list1.append(1)
     
    # Linked List 2
    list2 = linked_list()
    list2.append(4)
    list2.append(1)
    list2.append(9)
     
    # Sort both the linkedlists
    list1.head = list1.merge_sort(
                   list1.head
                 )
    list2.head = list2.merge_sort(
                   list2.head
                 )
     
    # Get difference list
    result = get_difference(
               list1.head, list2.head
             )
    if result.head:
        result.display()
     
    # if difference list is empty,
    # then lists are equal
    else:
        print('Lists are equal')
 
 

C#




using System;
class Node {
    public int data;
    public Node next;
 
    public Node(int data)
    {
        this.data = data;
        this.next = null;
    }
}
 
class LinkedList {
    public Node head;
 
    public LinkedList() { head = null; }
 
    // Function to insert a node at the end of Linked List
    public void Append(int data)
    {
        Node newNode = new Node(data);
        if (head == null) {
            head = newNode;
        }
        else {
            Node current = head;
            while (current.next != null) {
                current = current.next;
            }
            current.next = newNode;
        }
    }
 
    // Function to find the middle node of the Linked List
    public Node GetMid(Node head)
    {
        if (head == null) {
            return head;
        }
        Node slow = head;
        Node fast = head;
        while (fast.next != null
               && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
 
    // Recursive method to merge two halves after sorting
    public Node Merge(Node left, Node right)
    {
        if (left == null) {
            return right;
        }
        if (right == null) {
            return left;
        }
 
        Node result;
        if (left.data <= right.data) {
            result = left;
            result.next = Merge(left.next, right);
        }
        else {
            result = right;
            result.next = Merge(left, right.next);
        }
        return result;
    }
 
    // Recursive method to divide the list into two halves
    // until 1 node left
    public Node MergeSort(Node head)
    {
        if (head == null || head.next == null) {
            return head;
        }
 
        Node mid = GetMid(head);
        Node nextToMid = mid.next;
        mid.next = null;
        Node left = MergeSort(head);
        Node right = MergeSort(nextToMid);
        return Merge(left, right);
    }
 
    // Function to print the list elements
    public void Display()
    {
        Node current = head;
        while (current != null) {
            Console.Write(current.data + " ");
            current = current.next;
        }
        Console.WriteLine();
    }
 
    // Function to get the difference list
    public static LinkedList GetDifference(Node p1, Node p2)
    {
        LinkedList differenceList = new LinkedList();
 
        // Scan the lists
        while (p1 != null && p2 != null) {
            // Condition to check if the data of both
            // pointers are the same, then move ahead
            if (p2.data == p1.data) {
                p1 = p1.next;
                p2 = p2.next;
            }
            else if (p2.data
                     < p1.data) // Condition to check if the
                                // data of the second
                                // pointer is smaller than
                                // the first, then move the
                                // second pointer ahead
            {
                p2 = p2.next;
            }
            else { // Condition when the data of the first
                   // pointer is greater than the second,
                   // then append into the difference list
                   // and move
                differenceList.Append(p1.data);
                p1 = p1.next;
            }
        }
 
        // If the end of list2 is reached, there may be some
        // nodes in List 1 left to be scanned, they all will
        // be inserted in the difference list
        while (p1 != null) {
            differenceList.Append(p1.data);
            p1 = p1.next;
        }
 
        return differenceList;
    }
}
 
class Program {
    public static void Main(string[] args)
    {
        // Linked List 1
        LinkedList list1 = new LinkedList();
        list1.Append(2);
        list1.Append(6);
        list1.Append(8);
        list1.Append(1);
 
        // Linked List 2
        LinkedList list2 = new LinkedList();
        list2.Append(4);
        list2.Append(1);
        list2.Append(9);
 
        // Sort both the linked lists
        list1.head = list1.MergeSort(list1.head);
        list2.head = list2.MergeSort(list2.head);
 
        // Get the difference list
        LinkedList result = LinkedList.GetDifference(
            list1.head, list2.head);
 
        if (result.head != null) {
            result.Display();
        }
        else {
            Console.WriteLine("Lists are equal");
        }
    }
}
 
 

Javascript




class Node {
  constructor(data) {
    this.data = data;
    this.next = null;
  }
}
 
class LinkedList {
  constructor() {
    this.head = null;
  }
 
  // Function to insert a node at the end of Linked List
  append(data) {
    const newNode = new Node(data);
    if (this.head === null) {
      this.head = newNode;
    } else {
      let current = this.head;
      while (current.next !== null) {
        current = current.next;
      }
      current.next = newNode;
    }
  }
 
  // Function to find the middle node of the Linked List
  getMid(head) {
    if (head === null) {
      return head;
    }
    let slow = head;
    let fast = head;
    while (fast.next !== null && fast.next.next !== null) {
      slow = slow.next;
      fast = fast.next.next;
    }
    return slow;
  }
 
  // Recursive method to merge two halves after sorting
  merge(left, right) {
    if (left === null) {
      return right;
    }
    if (right === null) {
      return left;
    }
 
    let result;
    if (left.data <= right.data) {
      result = left;
      result.next = this.merge(left.next, right);
    } else {
      result = right;
      result.next = this.merge(left, right.next);
    }
    return result;
  }
 
  // Recursive method to divide the list into two halves until 1 node left
  mergeSort(head) {
    if (head === null || head.next === null) {
      return head;
    }
 
    const mid = this.getMid(head);
    const nextToMid = mid.next;
    mid.next = null;
    const left = this.mergeSort(head);
    const right = this.mergeSort(nextToMid);
    return this.merge(left, right);
  }
 
  // Function to print the list elements
  display() {
    let current = this.head;
    let result = '';
    while (current !== null) {
      result += current.data + ' ';
      current = current.next;
    }
    console.log(result.trim());
  }
 
  // Function to get the difference list
  static getDifference(p1, p2) {
    const differenceList = new LinkedList();
 
    // Scan the lists
    while (p1 !== null && p2 !== null) {
      // Condition to check if the data of both pointers are the same, then move ahead
      if (p2.data === p1.data) {
        p1 = p1.next;
        p2 = p2.next;
      } else if (p2.data < p1.data) {
        // Condition to check if the data of the second pointer is smaller than the first, then move the second pointer ahead
        p2 = p2.next;
      } else {
        // Condition when the data of the first pointer is greater than the second, then append into the difference list and move
        differenceList.append(p1.data);
        p1 = p1.next;
      }
    }
 
    // If the end of list2 is reached, there may be some nodes in List 1 left to be scanned, they all will be inserted in the difference list
    while (p1 !== null) {
      differenceList.append(p1.data);
      p1 = p1.next;
    }
 
    return differenceList;
  }
}
 
// Linked List 1
const list1 = new LinkedList();
list1.append(2);
list1.append(6);
list1.append(8);
list1.append(1);
 
// Linked List 2
const list2 = new LinkedList();
list2.append(4);
list2.append(1);
list2.append(9);
 
// Sort both the linked lists
list1.head = list1.mergeSort(list1.head);
list2.head = list2.mergeSort(list2.head);
 
// Get the difference list
const result = LinkedList.getDifference(list1.head, list2.head);
 
if (result.head !== null) {
  result.display();
} else {
  console.log("Lists are equal");
}
 
 
Output
2 6 8        

Time complexity: O(M Log M + N Log N).
 



N

naina024
Improve
Article Tags :
  • Data Structures
  • DSA
  • Linked List
Practice Tags :
  • Data Structures
  • Linked List

Similar Reads

  • Merge Sort - Data Structure and Algorithms Tutorials
    Merge sort is a popular sorting algorithm known for its efficiency and stability. It follows the divide-and-conquer approach. It works by recursively dividing the input array into two halves, recursively sorting the two halves and finally merging them back together to obtain the sorted array. How do
    14 min read
  • Merge sort in different languages

    • C Program for Merge Sort
      Merge Sort is a comparison-based sorting algorithm that works by dividing the input array into two halves, then calling itself for these two halves, and finally it merges the two sorted halves. In this article, we will learn how to implement merge sort in C language. What is Merge Sort Algorithm?Mer
      3 min read

    • C++ Program For Merge Sort
      Merge Sort is a comparison-based sorting algorithm that uses divide and conquer paradigm to sort the given dataset. It divides the dataset into two halves, calls itself for these two halves, and then it merges the two sorted halves. In this article, we will learn how to implement merge sort in a C++
      4 min read

    • Java Program for Merge Sort
      Merge Sort is a divide-and-conquer algorithm. It divides the input array into two halves, calls itself the two halves, and then merges the two sorted halves. The merge() function is used for merging two halves. The merge(arr, l, m, r) is a key process that assumes that arr[l..m] and arr[m+1..r] are
      3 min read

    • Merge Sort in Python
      Merge Sort is a Divide and Conquer algorithm. It divides input array in two halves, calls itself for the two halves and then merges the two sorted halves. The merge() function is used for merging two halves. The merge(arr, l, m, r) is key process that assumes that arr[l..m] and arr[m+1..r] are sorte
      4 min read

    • Merge Sort using Multi-threading
      Merge Sort is a popular sorting technique which divides an array or list into two halves and then start merging them when sufficient depth is reached. Time complexity of merge sort is O(nlogn).Threads are lightweight processes and threads shares with other threads their code section, data section an
      14 min read

    Variations of Merge Sort

    • 3-way Merge Sort
      Merge Sort is a divide-and-conquer algorithm that recursively splits an array into two halves, sorts each half, and then merges them. A variation of this is 3-way Merge Sort, where instead of splitting the array into two parts, we divide it into three equal parts. In traditional Merge Sort, the arra
      14 min read

    • Iterative Merge Sort
      Given an array of size n, the task is to sort the given array using iterative merge sort. Examples: Input: arr[] = [4, 1, 3, 9, 7]Output: [1, 3, 4, 7, 9]Explanation: The output array is sorted. Input: arr[] = [1, 3 , 2]Output: [1, 2, 3]Explanation: The output array is sorted. You can refer to Merge
      9 min read

    • In-Place Merge Sort
      Implement Merge Sort i.e. standard implementation keeping the sorting algorithm as in-place. In-place means it does not occupy extra memory for merge operation as in the standard case. Examples: Input: arr[] = {2, 3, 4, 1} Output: 1 2 3 4 Input: arr[] = {56, 2, 45} Output: 2 45 56 Approach 1: Mainta
      15+ min read

    • In-Place Merge Sort | Set 2
      Given an array A[] of size N, the task is to sort the array in increasing order using In-Place Merge Sort. Examples: Input: A = {5, 6, 3, 2, 1, 6, 7}Output: {1, 2, 3, 5, 6, 6, 7} Input: A = {2, 3, 4, 1}Output: {1, 2, 3, 4} Approach: The idea is to use the inplace_merge() function to merge the sorted
      7 min read

    • Merge Sort with O(1) extra space merge and O(n log n) time [Unsigned Integers Only]
      We have discussed Merge sort. How to modify the algorithm so that merge works in O(1) extra space and algorithm still works in O(n Log n) time. We may assume that the input values are integers only. Examples: Input : 5 4 3 2 1 Output : 1 2 3 4 5 Input : 999 612 589 856 56 945 243 Output : 56 243 589
      10 min read

    Merge Sort in Linked List

    • Merge Sort for Linked Lists
      Given a singly linked list, The task is to sort the linked list in non-decreasing order using merge sort. Examples: Input: 40 -> 20 -> 60 -> 10 -> 50 -> 30 -> NULLOutput: 10 -> 20 -> 30 -> 40 -> 50 -> 60 -> NULL Input: 9 -> 5 -> 2 -> 8 -> NULLOutput: 2
      12 min read

    • Merge Sort for Doubly Linked List
      Given a doubly linked list, The task is to sort the doubly linked list in non-decreasing order using merge sort.Examples: Input: 10 <-> 8 <-> 4 <-> 2Output: 2 <-> 4 <-> 8 <-> 10Input: 5 <-> 3 <-> 2Output: 2 <-> 3 <-> 5 Note: Merge sort for
      13 min read

    • Iterative Merge Sort for Linked List
      Given a singly linked list of integers, the task is to sort it using iterative merge sort. Examples: Input: 40 -> 20 -> 60 -> 10 -> 50 -> 30 -> NULLOutput: 10 -> 20 -> 30 -> 40 -> 50 -> 60 -> NULL Input: 9 -> 5 -> 2 -> 8 -> NULLOutput: 2 -> 5 ->
      14 min read

    • Merge two sorted lists (in-place)
      Given two sorted linked lists consisting of n and m nodes respectively. The task is to merge both of the lists and return the head of the merged list. Example: Input: Output: Input: Output: Approach: The idea is to iteratively merge two sorted linked lists using a dummy node to simplify the process.
      9 min read

    • Merge K sorted Doubly Linked List in Sorted Order
      Given K sorted doubly linked list. The task is to merge all sorted doubly linked list in single sorted doubly linked list means final list must be sorted.Examples: Input: List 1 : 2 <-> 7 <-> 8 <-> 12 <-> 15 <-> NULL List 2 : 4 <-> 9 <-> 10 <-> NULL Li
      15+ min read

    • Merge a linked list into another linked list at alternate positions
      Given two singly linked lists, The task is to insert nodes of the second list into the first list at alternate positions of the first list and leave the remaining nodes of the second list if it is longer. Example: Input: head1: 1->2->3 , head2: 4->5->6->7->8Output: head1: 1->4-
      8 min read

  • Find a permutation that causes worst case of Merge Sort
    Given a set of elements, find which permutation of these elements would result in worst case of Merge Sort.Asymptotically, merge sort always takes O(n Log n) time, but the cases that require more comparisons generally take more time in practice. We basically need to find a permutation of input eleme
    12 min read
  • How to make Mergesort to perform O(n) comparisons in best case?
    As we know, Mergesort is a divide and conquer algorithm that splits the array to halves recursively until it reaches an array of the size of 1, and after that it merges sorted subarrays until the original array is fully sorted. Typical implementation of merge sort works in O(n Log n) time in all thr
    3 min read
  • Concurrent Merge Sort in Shared Memory
    Given a number 'n' and a n numbers, sort the numbers using Concurrent Merge Sort. (Hint: Try to use shmget, shmat system calls).Part1: The algorithm (HOW?) Recursively make two child processes, one for the left half, one of the right half. If the number of elements in the array for a process is less
    10 min read
  • Visualization of Merge Sort

    • Sorting Algorithm Visualization : Merge Sort
      The human brain can easily process visuals instead of long codes to understand the algorithms. In this article, a program that program visualizes the Merge sort Algorithm has been implemented. The GUI(Graphical User Interface) is implemented using pygame package in python. Approach: An array of rand
      3 min read

    • Merge Sort Visualization in JavaScript
      GUI(Graphical User Interface) helps users with better understanding programs. In this article, we will visualize Merge Sort using JavaScript. We will see how the arrays are divided and merged after sorting to get the final sorted array.  Refer: Merge SortCanvas in HTMLAsynchronous Function in JavaSc
      4 min read

    • Visualize Merge sort Using Tkinter in Python
      Prerequisites: Python GUI – tkinter In this article, we will create a GUI application that will help us to visualize the algorithm of merge sort using Tkinter in Python. Merge Sort is a popular sorting algorithm. It has a time complexity of N(logN) which is faster than other sorting algorithms like
      5 min read

    • Visualization of Merge sort using Matplotlib
      Prerequisites: Introduction to Matplotlib, Merge Sort Visualizing algorithms makes it easier to understand them by analyzing and comparing the number of operations that took place to compare and swap the elements. For this we will use matplotlib, to plot bar graphs to represent the elements of the a
      3 min read

    • 3D Visualisation of Merge Sort using Matplotlib
      Visualizing algorithms makes it easier to understand them by analyzing and comparing the number of operations that took place to compare and swap the elements. 3D visualization of algorithms is less common, for this we will use matplotlib to plot bar graphs and animate them to represent the elements
      3 min read

    Some problems on Merge Sort

    • Count Inversions of an Array
      Given an integer array arr[] of size n, find the inversion count in the array. Two array elements arr[i] and arr[j] form an inversion if arr[i] > arr[j] and i < j. Note: Inversion Count for an array indicates that how far (or close) the array is from being sorted. If the array is already sorte
      15+ min read

    • Count of smaller elements on right side of each element in an Array using Merge sort
      Given an array arr[] of N integers, the task is to count the number of smaller elements on the right side for each of the element in the array Examples: Input: arr[] = {6, 3, 7, 2} Output: 2, 1, 1, 0 Explanation: Smaller elements after 6 = 2 [3, 2] Smaller elements after 3 = 1 [2] Smaller elements a
      12 min read

    • Sort a nearly sorted (or K sorted) array
      Given an array arr[] and a number k . The array is sorted in a way that every element is at max k distance away from it sorted position. It means if we completely sort the array, then the index of the element can go from i - k to i + k where i is index in the given array. Our task is to completely s
      6 min read

    • Median of two Sorted Arrays of Different Sizes
      Given two sorted arrays, a[] and b[], the task is to find the median of these sorted arrays. Assume that the two sorted arrays are merged and then median is selected from the combined array. This is an extension of Median of two sorted arrays of equal size problem. Here we handle arrays of unequal s
      15+ min read

    • Merge k Sorted Arrays
      Given K sorted arrays, merge them and print the sorted output. Examples: Input: K = 3, arr = { {1, 3, 5, 7}, {2, 4, 6, 8}, {0, 9, 10, 11}}Output: 0 1 2 3 4 5 6 7 8 9 10 11 Input: k = 4, arr = { {1}, {2, 4}, {3, 7, 9, 11}, {13} }Output: 1 2 3 4 7 9 11 13 Table of Content Naive - Concatenate all and S
      15+ min read

    • Merge K sorted arrays of different sizes | ( Divide and Conquer Approach )
      Given k sorted arrays of different length, merge them into a single array such that the merged array is also sorted.Examples: Input : {{3, 13}, {8, 10, 11} {9, 15}} Output : {3, 8, 9, 10, 11, 13, 15} Input : {{1, 5}, {2, 3, 4}} Output : {1, 2, 3, 4, 5} Let S be the total number of elements in all th
      8 min read

    • Merge K sorted linked lists
      Given k sorted linked lists of different sizes, the task is to merge them all maintaining their sorted order. Examples: Input: Output: Merged lists in a sorted order where every element is greater than the previous element. Input: Output: Merged lists in a sorted order where every element is greater
      15+ min read

    • Union and Intersection of two Linked List using Merge Sort
      Given two singly Linked Lists, create union and intersection lists that contain the union and intersection of the elements present in the given lists. Each of the two lists contains distinct node values. Note: The order of elements in output lists doesn't matter. Examples: Input: head1: 10 -> 15
      15+ min read

    • Sorting by combining Insertion Sort and Merge Sort algorithms
      Insertion sort: The array is virtually split into a sorted and an unsorted part. Values from the unsorted part are picked and placed at the correct position in the sorted part.Advantages: Following are the advantages of insertion sort: If the size of the list to be sorted is small, insertion sort ru
      2 min read

    • Find array with k number of merge sort calls
      Given two numbers n and k, find an array containing values in [1, n] and requires exactly k calls of recursive merge sort function. Examples: Input : n = 3 k = 3 Output : a[] = {2, 1, 3} Explanation: Here, a[] = {2, 1, 3} First of all, mergesort(0, 3) will be called, which then sets mid = 1 and call
      6 min read

    • Difference of two Linked Lists using Merge sort
      Given two Linked List, the task is to create a Linked List to store the difference of Linked List 1 with Linked List 2, i.e. the elements present in List 1 but not in List 2.Examples: Input: List1: 10 -> 15 -> 4 ->20, List2: 8 -> 4 -> 2 -> 10 Output: 15 -> 20 Explanation: In the
      14 min read

geeksforgeeks-footer-logo
Corporate & Communications Address:
A-143, 7th Floor, Sovereign Corporate Tower, Sector- 136, Noida, Uttar Pradesh (201305)
Registered Address:
K 061, Tower K, Gulshan Vivante Apartment, Sector 137, Noida, Gautam Buddh Nagar, Uttar Pradesh, 201305
GFG App on Play Store GFG App on App Store
Advertise with us
  • Company
  • About Us
  • Legal
  • Privacy Policy
  • In Media
  • Contact Us
  • Advertise with us
  • GFG Corporate Solution
  • Placement Training Program
  • Languages
  • Python
  • Java
  • C++
  • PHP
  • GoLang
  • SQL
  • R Language
  • Android Tutorial
  • Tutorials Archive
  • DSA
  • Data Structures
  • Algorithms
  • DSA for Beginners
  • Basic DSA Problems
  • DSA Roadmap
  • Top 100 DSA Interview Problems
  • DSA Roadmap by Sandeep Jain
  • All Cheat Sheets
  • Data Science & ML
  • Data Science With Python
  • Data Science For Beginner
  • Machine Learning
  • ML Maths
  • Data Visualisation
  • Pandas
  • NumPy
  • NLP
  • Deep Learning
  • Web Technologies
  • HTML
  • CSS
  • JavaScript
  • TypeScript
  • ReactJS
  • NextJS
  • Bootstrap
  • Web Design
  • Python Tutorial
  • Python Programming Examples
  • Python Projects
  • Python Tkinter
  • Python Web Scraping
  • OpenCV Tutorial
  • Python Interview Question
  • Django
  • Computer Science
  • Operating Systems
  • Computer Network
  • Database Management System
  • Software Engineering
  • Digital Logic Design
  • Engineering Maths
  • Software Development
  • Software Testing
  • DevOps
  • Git
  • Linux
  • AWS
  • Docker
  • Kubernetes
  • Azure
  • GCP
  • DevOps Roadmap
  • System Design
  • High Level Design
  • Low Level Design
  • UML Diagrams
  • Interview Guide
  • Design Patterns
  • OOAD
  • System Design Bootcamp
  • Interview Questions
  • Inteview Preparation
  • Competitive Programming
  • Top DS or Algo for CP
  • Company-Wise Recruitment Process
  • Company-Wise Preparation
  • Aptitude Preparation
  • Puzzles
  • School Subjects
  • Mathematics
  • Physics
  • Chemistry
  • Biology
  • Social Science
  • English Grammar
  • Commerce
  • World GK
  • GeeksforGeeks Videos
  • DSA
  • Python
  • Java
  • C++
  • Web Development
  • Data Science
  • CS Subjects
@GeeksforGeeks, Sanchhaya Education Private Limited, All rights reserved
We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy
Lightbox
Improvement
Suggest Changes
Help us improve. Share your suggestions to enhance the article. Contribute your expertise and make a difference in the GeeksforGeeks portal.
geeksforgeeks-suggest-icon
Create Improvement
Enhance the article with your expertise. Contribute to the GeeksforGeeks community and help create better learning resources for all.
geeksforgeeks-improvement-icon
Suggest Changes
min 4 words, max Words Limit:1000

Thank You!

Your suggestions are valuable to us.

What kind of Experience do you want to share?

Interview Experiences
Admission Experiences
Career Journeys
Work Experiences
Campus Experiences
Competitive Exam Experiences