Find common elements in three linked lists
Given three linked lists, find all common element among the three linked lists.
Examples:
Input : 10 15 20 25 12 10 12 13 15 10 12 15 24 25 26 Output : 10 12 15 Input : 1 2 3 4 5 1 2 3 4 6 9 8 1 2 4 5 10 Output : 1 2 4
Method 1 : (Simple)
Use three-pointers to iterate the given three linked lists and if any element common print that element.
Time complexity of the above solution will be O(N*N*N)
Method 2 : (Use Merge Sort)
In this method, we first sort the three lists and then we traverse the sorted lists to get the intersection.
Following are the steps to be followed to get intersection of three lists:
1) Sort the first Linked List using merge sort. This step takes O(mLogm) time. Refer this post for details of this step.
2) Sort the second Linked List using merge sort. This step takes O(nLogn) time. Refer this post for details of this step.
3) Sort the third Linked List using merge sort. This step takes O(pLogp) time. Refer this post for details of this step.
3) Linearly scan three sorted lists to get the intersection. This step takes O(m + n + p) time. This step can be implemented using the same algorithm as sorted arrays algorithm discussed here.
Time complexity of this method is O(mLogm + nLogn + plogp) which is better than method 1’s time complexity.
Method 3 : (Hashing)
Following are the steps to be followed to get intersection of three lists using hashing:
1) Create an empty hash table. Iterate through the first linked list and mark all the element frequency as 1 in the hash table. This step takes O(m) time.
2) Iterate through the second linked list and if current element frequency is 1 in hash table mark it as 2. This step takes O(n) time.
3) Iterate the third linked list and if the current element frequency is 2 in hash table mark it as 3. This step takes O(p) time.
4) Now iterate first linked list again to check the frequency of elements. if an element with frequency three exist in hash table, it will be present in the intersection of three linked lists. This step takes O(m) time.
Time complexity of this method is O(m + n + p) which is better than time complexity of method 1 and 2.
Below is the implementation of the above idea.
C++
// C++ program to find common element // in three unsorted linked list #include <bits/stdc++.h> #define max 1000000 using namespace std; /* Link list node */struct Node { int data; struct Node* next; }; /* A utility function to insert a node at the beginning of a linked list */void push(struct Node** head_ref, int new_data) { struct Node* new_node = (struct Node *)malloc(sizeof(struct Node)); new_node->data = new_data; new_node->next = (*head_ref); (*head_ref) = new_node; } /* print the common element in between given three linked list*/void Common(struct Node* head1, struct Node* head2, struct Node* head3) { // Creating empty hash table; unordered_map<int, int> hash; struct Node* p = head1; while (p != NULL) { // set frequency by 1 hash[p->data] = 1; p = p->next; } struct Node* q = head2; while (q != NULL) { // if the element is already exist in the // linked list set its frequency 2 if (hash.find(q->data) != hash.end()) hash[q->data] = 2; q = q->next; } struct Node* r = head3; while (r != NULL) { if (hash.find(r->data) != hash.end() && hash[r->data] == 2) // if the element frquancy is 2 it means // its present in both the first and second // linked list set its frquancy 3 hash[r->data] = 3; r = r->next; } for (auto x : hash) { // if current frequency is 3 its means // element is common in all the given // linked list if (x.second == 3) cout << x.first << " "; } } // Driver code int main() { // first list struct Node* head1 = NULL; push(&head1, 20); push(&head1, 5); push(&head1, 15); push(&head1, 10); // second list struct Node* head2 = NULL; push(&head2, 10); push(&head2, 20); push(&head2, 15); push(&head2, 8); // third list struct Node* head3 = NULL; push(&head3, 10); push(&head3, 2); push(&head3, 15); push(&head3, 20); Common(head1, head2, head3); return 0; } |
chevron_right
filter_none
Java
// Java program to find common element // in three unsorted linked list import java.util.*; class GFG { static int max = 1000000; /* Link list node */static class Node { int data; Node next; }; /* A utility function to insert a node at the beginning of a linked list */static Node push(Node head_ref, int new_data) { Node new_node = new Node(); new_node.data = new_data; new_node.next = head_ref; head_ref = new_node; return head_ref; } /* print the common element in between given three linked list*/static void Common(Node head1, Node head2, Node head3) { // Creating empty hash table; HashMap<Integer, Integer> hash = new HashMap<Integer, Integer>(); Node p = head1; while (p != null) { // set frequency by 1 hash. put(p.data, 1); p = p.next; } Node q = head2; while (q != null) { // if the element is already exist in the // linked list set its frequency 2 if (hash.containsKey(q.data)) hash. put(q.data, 2); q = q.next; } Node r = head3; while (r != null) { if (hash.containsKey(r.data)&& hash.get(r.data) == 2) // if the element frquancy is 2 it means // its present in both the first and second // linked list set its frquancy 3 hash. put(r.data, 3); r = r.next; } for (Map.Entry<Integer, Integer> x : hash.entrySet()) { // if current frequency is 3 its means // element is common in all the given // linked list if (x.getValue() == 3) System.out.println(x.getKey() + " "); } } // Driver code public static void main(String[] args) { // first list Node head1 = null; head1 = push(head1, 20); head1 = push(head1, 5); head1 = push(head1, 15); head1 = push(head1, 10); // second list Node head2 = null; head2 = push(head2, 10); head2 = push(head2, 20); head2 = push(head2, 15); head2 = push(head2, 8); // third list Node head3 = null; head3 = push(head3, 10); head3 = push(head3, 2); head3 = push(head3, 15); head3 = push(head3, 20); Common(head1, head2, head3); } } // This code is contributed by 29AjayKumar |
chevron_right
filter_none
C#
// C# program to find common element // in three unsorted linked list using System; using System.Collections.Generic; class GFG { static int max = 1000000; /* Link list node */public class Node { public int data; public Node next; }; /* A utility function to insert a node at the beginning of a linked list */static Node push(Node head_ref, int new_data) { Node new_node = new Node(); new_node.data = new_data; new_node.next = head_ref; head_ref = new_node; return head_ref; } /* print the common element in between given three linked list*/static void Common(Node head1, Node head2, Node head3) { // Creating empty hash table; Dictionary<int, int> hash = new Dictionary<int, int>(); Node p = head1; while (p != null) { // set frequency by 1 hash.Add(p.data, 1); p = p.next; } Node q = head2; while (q != null) { // if the element is already exist in the // linked list set its frequency 2 if (hash.ContainsKey(q.data)) hash[q.data] = 2; q = q.next; } Node r = head3; while (r != null) { if (hash.ContainsKey(r.data)&& hash[r.data] == 2) // if the element frquancy is 2 it means // its present in both the first and // second linked list set its frquancy 3 hash[r.data] = 3; r = r.next; } foreach(KeyValuePair<int, int> x in hash) { // if current frequency is 3 its means // element is common in all the given // linked list if (x.Value == 3) Console.Write(x.Key + " "); } } // Driver code public static void Main(String[] args) { // first list Node head1 = null; head1 = push(head1, 20); head1 = push(head1, 5); head1 = push(head1, 15); head1 = push(head1, 10); // second list Node head2 = null; head2 = push(head2, 10); head2 = push(head2, 20); head2 = push(head2, 15); head2 = push(head2, 8); // third list Node head3 = null; head3 = push(head3, 10); head3 = push(head3, 2); head3 = push(head3, 15); head3 = push(head3, 20); Common(head1, head2, head3); } } // This code is contributed by Princi Singh |
chevron_right
filter_none
Output:
10 15 20
Time Complexity : O(m + n + p)
Recommended Posts:
- Find count of common nodes in two Doubly Linked Lists
- Construct a Maximum Sum Linked List out of two Sorted Linked Lists having some Common nodes
- First common element in two linked lists
- Longest common suffix of two linked lists
- Minimum Index Sum for Common Elements of Two Lists
- Find the common nodes in two singly linked list
- Find a triplet from three linked lists with sum equal to a given number
- Find distinct elements common to all rows of a matrix
- Find smallest range containing elements from k lists
- Create a linked list from two linked lists by choosing max element at each position
- Count items common to both the lists but with different prices
- Find unique elements in linked list
- Find smallest and largest elements in singly linked list
- Find minimum and maximum elements in singly Circular Linked List
- Identical Linked Lists
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
