Union and Intersection of two Linked List using Hashing Last Updated : 10 Sep, 2024 Comments Improve Suggest changes Like Article Like Report Try it on GfG Practice 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 linked lists contains distinct node values.Note: The order of elements in output lists doesn’t matter. Examples:Input:head1 : 10 -> 15 -> 4 -> 20head2 : 8 -> 4 -> 2 -> 10Output:Intersection List: 4 -> 10Union List: 2 -> 8 -> 20 -> 4 -> 15 -> 10Explanation: In these two lists 4 and 10 nodes are common. The union lists contain all the nodes of both lists.Input:head1 : 1 -> 2 -> 3 -> 4head2 : 3 -> 4 -> 8 -> 10Output:Intersection List: 3 -> 4Union List: 1 -> 2 -> 3 -> 4 -> 8 -> 10Explanation: In these two lists 4 and 3 nodes are common. The union lists contain all the nodes of both lists.Approach:We have already discussed Method-1 and Method-2 of this question. In this post, its Method-3 (Using Hashing) is discussed with a Time Complexity of O(m + n) i.e. better than both methods discussed earlier.Intersection Function:Use a Hashset to keep track of elements from the first linked list.Traverse the second list and check if elements are present in the Hashset.if present , append the elements into the result list.Union Function:Use a Hashset to collect unique elements from both lists.Insert elements from both lists into the Hashset .Append the elements from Hashset into the result list.Below is the implementation of the above approach : C++ // C++ code for Union and Intersection // of two Linked List #include <bits/stdc++.h> using namespace std; class Node { public: int data; Node *next; Node(int x) { data = x; next = nullptr; } }; void printList(Node *head) { Node *curr = head; while (curr != nullptr) { cout << curr->data << " "; curr = curr->next; } cout << endl; } // Function to get the intersection of two lists Node *getIntersection(Node *head1, Node *head2) { unordered_set<int> set; Node *result = nullptr; // Traverse the first list and store // elements in a hash set Node *p = head1; while (p != nullptr) { set.insert(p->data); p = p->next; } // Traverse the second list and // check for common elements p = head2; while (p != nullptr) { if (set.find(p->data) != set.end()) { // Insert at the beginning of the result list Node *new_node = new Node(p->data); new_node->next = result; result = new_node; } p = p->next; } return result; } // Function to get the union of two lists Node *getUnion(Node *head1, Node *head2) { unordered_set<int> union_set; Node *p = head1; while (p != nullptr) { union_set.insert(p->data); p = p->next; } p = head2; while (p != nullptr) { union_set.insert(p->data); p = p->next; } Node *result = nullptr; Node *tail = nullptr; // Insert all unique values into // the result list for (int i : union_set) { Node *new_node = new Node(i); if (result == nullptr) { result = new_node; tail = new_node; } else { tail->next = new_node; tail = tail->next; } } return result; } // Function to print both union and intersection of two lists void printUnionIntersection(Node *head1, Node *head2) { Node *intersection_list = getIntersection(head1, head2); Node *union_list = getUnion(head1, head2); cout << "Intersection list is:" << endl; printList(intersection_list); cout << "Union list is:" << endl; printList(union_list); } int main() { //list1 : 1 -> 2 -> 3 -> 3 -> 4 -> 5 Node *head1 = new Node(1); head1->next = new Node(2); head1->next->next = new Node(3); head1->next->next->next = new Node(3); head1->next->next->next->next = new Node(4); head1->next->next->next->next->next = new Node(5); // list2 : 1 -> 5 -> 6 Node *head2 = new Node(1); head2->next = new Node(5); head2->next->next = new Node(6); printUnionIntersection(head1, head2); return 0; } Java // Java program to find union and intersection // of linked list import java.util.HashSet; import java.util.Set; class Node { int data; Node next; Node(int x) { data = x; next = null; } } class GfG { static void printList(Node head) { Node curr = head; while (curr != null) { System.out.print(curr.data + " "); curr = curr.next; } System.out.println(); } // Function to get the intersection of two lists static Node getIntersection(Node head1, Node head2) { Set<Integer> set = new HashSet<>(); Node result = null; // Traverse the first list and store elements in a // hash set Node p = head1; while (p != null) { set.add(p.data); p = p.next; } // Traverse the second list and check for common // elements p = head2; while (p != null) { if (set.contains(p.data)) { // Insert at the beginning of the result // list Node newNode = new Node(p.data); newNode.next = result; result = newNode; } p = p.next; } return result; } // Function to get the union of two lists static Node getUnion(Node head1, Node head2) { Set<Integer> unionSet = new HashSet<>(); Node p = head1; while (p != null) { unionSet.add(p.data); p = p.next; } p = head2; while (p != null) { unionSet.add(p.data); p = p.next; } Node result = null; Node tail = null; // Insert all unique values into the result list for (int i : unionSet) { Node newNode = new Node(i); if (result == null) { result = newNode; tail = newNode; } else { tail.next = newNode; tail = tail.next; } } return result; } // Function to print both union and intersection of two // lists static void printUnionIntersection(Node head1,Node head2) { Node intersectionList = getIntersection(head1, head2); Node unionList = getUnion(head1, head2); System.out.println("Intersection list is:"); printList(intersectionList); System.out.println("Union list is:"); printList(unionList); } public static void main(String[] args) { // List 1: 1 -> 2 -> 3 -> 3 -> 4 -> 5 Node head1 = new Node(1); head1.next = new Node(2); head1.next.next = new Node(3); head1.next.next.next = new Node(3); head1.next.next.next.next = new Node(4); head1.next.next.next.next.next = new Node(5); // List 2: 1 -> 5 -> 6 Node head2 = new Node(1); head2.next = new Node(5); head2.next.next = new Node(6); printUnionIntersection(head1, head2); } } Python # Python code for finding union and # intersection of linkedList class Node: def __init__(self, x): self.data = x self.next = None def print_list(head): curr = head while curr: print(curr.data, end=" ") curr = curr.next print() def get_intersection(head1, head2): seen = set() result = None # Traverse the first list and # store elements in a set p = head1 while p: seen.add(p.data) p = p.next # Traverse the second list and check # for common elements p = head2 while p: if p.data in seen: # Insert at the beginning of # the result list new_node = Node(p.data) new_node.next = result result = new_node p = p.next return result def get_union(head1, head2): union_set = set() p = head1 while p: union_set.add(p.data) p = p.next p = head2 while p: union_set.add(p.data) p = p.next result = None tail = None # Insert all unique values into # the result list for i in union_set: new_node = Node(i) if result is None: result = new_node tail = new_node else: tail.next = new_node tail = tail.next return result def print_union_intersection(head1, head2): intersection_list = get_intersection(head1, head2) union_list = get_union(head1, head2) print("Intersection list is:") print_list(intersection_list) print("Union list is:") print_list(union_list) if __name__ == "__main__": # List 1: 1 -> 2 -> 3 -> 3 -> 4 -> 5 head1 = Node(1) head1.next = Node(2) head1.next.next = Node(3) head1.next.next.next = Node(3) head1.next.next.next.next = Node(4) head1.next.next.next.next.next = Node(5) # List 2: 1 -> 5 -> 6 head2 = Node(1) head2.next = Node(5) head2.next.next = Node(6) print_union_intersection(head1, head2) C# // C# program to find union and intersection // of linked list 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 void PrintList(Node head) { Node curr = head; while (curr != null) { Console.Write(curr.Data + " "); curr = curr.Next; } Console.WriteLine(); } // Function to get the intersection of two lists static Node GetIntersection(Node head1, Node head2) { HashSet<int> set = new HashSet<int>(); Node result = null; // Traverse the first list and store elements in a // hash set Node p = head1; while (p != null) { set.Add(p.Data); p = p.Next; } // Traverse the second list and check for common // elements p = head2; while (p != null) { if (set.Contains(p.Data)) { // Insert at the beginning of the result // list Node newNode = new Node(p.Data); newNode.Next = result; result = newNode; } p = p.Next; } return result; } // Function to get the union of two lists static Node GetUnion(Node head1, Node head2) { HashSet<int> unionSet = new HashSet<int>(); Node p = head1; while (p != null) { unionSet.Add(p.Data); p = p.Next; } p = head2; while (p != null) { unionSet.Add(p.Data); p = p.Next; } Node result = null; Node tail = null; // Insert all unique values into the result list foreach(int i in unionSet) { Node newNode = new Node(i); if (result == null) { result = newNode; tail = newNode; } else { tail.Next = newNode; tail = tail.Next; } } return result; } // Function to print both union and intersection of two // lists static void PrintUnionIntersection(Node head1,Node head2) { Node intersectionList = GetIntersection(head1, head2); Node unionList = GetUnion(head1, head2); Console.WriteLine("Intersection list is:"); PrintList(intersectionList); Console.WriteLine("Union list is:"); PrintList(unionList); } static void Main() { // List 1: 1 -> 2 -> 3 -> 3 -> 4 -> 5 Node head1 = new Node(1); head1.Next = new Node(2); head1.Next.Next = new Node(3); head1.Next.Next.Next = new Node(3); head1.Next.Next.Next.Next = new Node(4); head1.Next.Next.Next.Next.Next = new Node(5); // List 2: 1 -> 5 -> 6 Node head2 = new Node(1); head2.Next = new Node(5); head2.Next.Next = new Node(6); PrintUnionIntersection(head1, head2); } } JavaScript // JavaScript program to find union and // intersection of linked list class Node { constructor(x) { this.data = x; this.next = null; } } function printList(head) { let curr = head; while (curr !== null) { console.log(curr.data + " "); curr = curr.next; } console.log(); } // Function to get the intersection of two lists function getIntersection(head1, head2) { const set = new Set(); let result = null; // Traverse the first list and store // elements in a set let p = head1; while (p !== null) { set.add(p.data); p = p.next; } // Traverse the second list and check for common // elements p = head2; while (p !== null) { if (set.has(p.data)) { // Insert at the beginning of the result list const newNode = new Node(p.data); newNode.next = result; result = newNode; } p = p.next; } return result; } // Function to get the union of two lists function getUnion(head1, head2) { const unionSet = new Set(); let p = head1; while (p !== null) { unionSet.add(p.data); p = p.next; } p = head2; while (p !== null) { unionSet.add(p.data); p = p.next; } let result = null; let tail = null; // Insert all unique values into the result list for (let i of unionSet) { const newNode = new Node(i); if (result === null) { result = newNode; tail = newNode; } else { tail.next = newNode; tail = tail.next; } } return result; } // Function to print both union and intersection of two // lists function printUnionIntersection(head1, head2) { const intersectionList = getIntersection(head1, head2); const unionList = getUnion(head1, head2); console.log("Intersection list is:"); printList(intersectionList); console.log("Union list is:"); printList(unionList); } // List 1: 1 -> 2 -> 3 -> 3 -> 4 -> 5 const head1 = new Node(1); head1.next = new Node(2); head1.next.next = new Node(3); head1.next.next.next = new Node(3); head1.next.next.next.next = new Node(4); head1.next.next.next.next.next = new Node(5); // List 2: 1 -> 5 -> 6 const head2 = new Node(1); head2.next = new Node(5); head2.next.next = new Node(6); printUnionIntersection(head1, head2); OutputIntersection list is: 5 1 Union list is: 6 5 1 2 3 4 Time Complexity: O(m+n) , where m and n are number of elements present in first and second lists respectively.Auxiliary Space: O(m+n) Comment More infoAdvertise with us Next Article Two Sum - Pair with given Sum S Sahil Chhabra Improve Article Tags : Linked List Sorting Hash DSA Microsoft Amazon Flipkart Walmart VMWare Komli Media 24*7 Innovation Labs Taxi4Sure +8 More Practice Tags : 24*7 Innovation LabsAmazonFlipkartKomli MediaMicrosoftTaxi4SureVMWareWalmartHashLinked ListSorting +7 More Similar Reads Hashing in Data Structure Hashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. 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If no such subarray exists, print -1 and If multiple subarrays meet the criteria, return the one with the smalle 14 min read Like