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# Find count of common nodes in two Doubly Linked Lists

Given two doubly linked lists. The task is to find the total number of common nodes in both the doubly linked list.

Examples:

`Input : list 1 = 15 <=> 16 <=> 10 <=> 9 <=> 6 <=> 7 <=> 17 list 2 = 15 <=> 16 <=> 45 <=> 9 <=> 6Output : Number of common nodes: 4Input :list 1 = 18 <=> 30 <=> 92 <=> 46 <=> 72 <=> 1list 2 = 12 <=> 32 <=> 45 <=> 9 <=> 6 <=> 30Output : Number of common nodes: 1`

Approach 1: Traverse both lists till the end of the list using two nested loops. For every node in list 1 check if it matches with any node in list 2. If yes then increment the count of common nodes. Finally, print the count.

Algorithm:

•  Create a function called “push” that adds a new node to the doubly linked list’s beginning. The function requires two arguments: new data and head ref (a pointer to the list’s head) (the data value to be inserted). The updated head pointer is returned by the function.
• Create a function called “countCommonNodes” that accepts the parameters head ref (a pointer to the first doubly linked list’s head) and head (pointer to the head of the second doubly linked list). The number of shared nodes between the two lists is represented as an integer by the function’s output. s.
• Set up two pointers, ptr, and ptr1, to point to the respective heads of the two lists.
•  Set the count variable’s initial value to 0.
• Use ptr to navigate through the first list to the very end.
• Up to the list’s conclusion, traverse the second list using ptr1.
•  Increase the count and exit the inner loop if the data value of the current node in the first list equals the data value of the current node in the second list.
•   Reset ptr1 to the head of the second list.
• Move ptr to the next node in the first list.
• Repeat steps 6-10 until the end of the first list is reached.
• Return the count of common nodes.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to count``// common element in given two``// doubly linked list``#include ` `using` `namespace` `std;` `// Node of the doubly linked list``struct` `Node {``    ``int` `data;``    ``Node *prev, *next;``};` `// Function to insert a node at the beginning``// of the Doubly Linked List``void` `push(Node** head_ref, ``int` `new_data)``{``    ``// allocate node``    ``Node* new_node = (Node*)``malloc``(``sizeof``(``struct` `Node));` `    ``// put in the data``    ``new_node->data = new_data;` `    ``// since we are adding at the beginning,``    ``// prev is always NULL``    ``new_node->prev = NULL;` `    ``// link the old list of the new node``    ``new_node->next = (*head_ref);` `    ``// change prev of head node to new node``    ``if` `((*head_ref) != NULL)``        ``(*head_ref)->prev = new_node;` `    ``// move the head to point to the new node``    ``(*head_ref) = new_node;``}` `// Count common nodes in both list1 and list 2``int` `countCommonNodes(Node** head_ref, Node** head)``{``    ``// head for list 1``    ``Node* ptr = *head_ref;` `    ``// head for list 2``    ``Node* ptr1 = *head;` `    ``// initialize count = 0``    ``int` `count = 0;` `    ``// traverse list 1 till the end``    ``while` `(ptr != NULL) {``        ``// traverse list 2 till the end``        ``while` `(ptr1 != NULL) {``            ``// if node value is equal then``            ``// increment count``            ``if` `(ptr->data == ptr1->data) {``                ``count++;``                ``break``;``            ``}` `            ``// increment pointer list 2``            ``ptr1 = ptr1->next;``        ``}` `        ``// again list 2 start with starting point``        ``ptr1 = *head;` `        ``// increment pointer list 1``        ``ptr = ptr->next;``    ``}` `    ``// return count of common nodes``    ``return` `count;``}` `// Driver program``int` `main()``{``    ``// start with the empty list``    ``Node* head = NULL;``    ``Node* head1 = NULL;` `    ``// create the doubly linked list 1``    ``// 15 <-> 16 <-> 10 <-> 9 <-> 6 <-> 7 <-> 17``    ``push(&head, 17);``    ``push(&head, 7);``    ``push(&head, 6);``    ``push(&head, 9);``    ``push(&head, 10);``    ``push(&head, 16);``    ``push(&head, 15);` `    ``// create the doubly linked list 2``    ``// 15 <-> 16 <-> 45 <-> 9 <-> 6``    ``push(&head1, 6);``    ``push(&head1, 9);``    ``push(&head1, 45);``    ``push(&head1, 16);``    ``push(&head1, 15);` `    ``cout << ``"Number of common nodes:"``         ``<< countCommonNodes(&head, &head1);` `    ``return` `0;``}`

## Java

 `// Java implementation to count``// common element in given two``// doubly linked list``class` `GFG``{` `// Node of the doubly linked list``static` `class` `Node``{``    ``int` `data;``    ``Node prev, next;``};` `// Function to insert a node at the beginning``// of the Doubly Linked List``static` `Node push(Node head_ref, ``int` `new_data)``{``    ``// allocate node``    ``Node new_node = ``new` `Node();` `    ``// put in the data``    ``new_node.data = new_data;` `    ``// since we are adding at the beginning,``    ``// prev is always null``    ``new_node.prev = ``null``;` `    ``// link the old list of the new node``    ``new_node.next = head_ref;` `    ``// change prev of head node to new node``    ``if` `(head_ref != ``null``)``        ``head_ref.prev = new_node;` `    ``// move the head to point to the new node``    ``head_ref = new_node;``    ``return` `head_ref;``}` `// Count common nodes in both list1 and list 2``static` `int` `countCommonNodes(Node head_ref,``                            ``Node head)``{``    ``// head for list 1``    ``Node ptr = head_ref;` `    ``// head for list 2``    ``Node ptr1 = head;` `    ``// initialize count = 0``    ``int` `count = ``0``;` `    ``// traverse list 1 till the end``    ``while` `(ptr != ``null``)``    ``{``        ` `        ``// traverse list 2 till the end``        ``while` `(ptr1 != ``null``)``        ``{``            ` `            ``// if node value is equal then``            ``// increment count``            ``if` `(ptr.data == ptr1.data)``            ``{``                ``count++;``                ``break``;``            ``}` `            ``// increment pointer list 2``            ``ptr1 = ptr1.next;``        ``}` `        ``// again list 2 start with starting point``        ``ptr1 = head;` `        ``// increment pointer list 1``        ``ptr = ptr.next;``    ``}` `    ``// return count of common nodes``    ``return` `count;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``// start with the empty list``    ``Node head = ``null``;``    ``Node head1 = ``null``;` `    ``// create the doubly linked list 1``    ``// 15 <. 16 <. 10 <. 9 <. 6 <. 7 <. 17``    ``head = push(head, ``17``);``    ``head = push(head, ``7``);``    ``head = push(head, ``6``);``    ``head = push(head, ``9``);``    ``head = push(head, ``10``);``    ``head = push(head, ``16``);``    ``head = push(head, ``15``);` `    ``// create the doubly linked list 2``    ``// 15 <. 16 <. 45 <. 9 <. 6``    ``head1 = push(head1, ``6``);``    ``head1 = push(head1, ``9``);``    ``head1 = push(head1, ``45``);``    ``head1 = push(head1, ``16``);``    ``head1 = push(head1, ``15``);` `    ``System.out.println(``"Number of common nodes: "` `+``                        ``countCommonNodes(head, head1));``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation to count``# common element in given two``# doubly linked list`` ` `# Node of the doubly linked list``class` `Node:   ``    ``def` `__init__(``self``, data):       ``        ``self``.data ``=` `data``        ``self``.prev ``=` `None``        ``self``.``next` `=` `None`` ` `# Function to insert a node at the beginning``# of the Doubly Linked List``def` `push(head_ref, new_data):``    ` `    ``# allocate node``    ``new_node ``=` `Node(new_data)`` ` `    ``# put in the data``    ``new_node.data ``=` `new_data;`` ` `    ``# since we are adding at the beginning,``    ``# prev is always None``    ``new_node.prev ``=` `None``;`` ` `    ``# link the old list of the new node``    ``new_node.``next` `=` `(head_ref);`` ` `    ``# change prev of head node to new node``    ``if` `((head_ref) !``=` `None``):``        ``(head_ref).prev ``=` `new_node;`` ` `    ``# move the head to point to the new node``    ``(head_ref) ``=` `new_node;   ``    ``return` `head_ref` `# Count common nodes in both list1 and list 2``def` `countCommonNodes(head_ref, head):` `    ``# head for list 1``    ``ptr ``=` `head_ref;`` ` `    ``# head for list 2``    ``ptr1 ``=` `head;`` ` `    ``# initialize count = 0``    ``count ``=` `0``;`` ` `    ``# traverse list 1 till the end``    ``while` `(ptr !``=` `None``):``      ` `        ``# traverse list 2 till the end``        ``while` `(ptr1 !``=` `None``):``          ` `            ``# if node value is equal then``            ``# increment count``            ``if` `(ptr.data ``=``=` `ptr1.data):``                ``count ``+``=` `1``                ``break``;``        ` `            ``# increment pointer list 2``            ``ptr1 ``=` `ptr1.``next``;`` ` `        ``# again list 2 start with starting point``        ``ptr1 ``=` `head;`` ` `        ``# increment pointer list 1``        ``ptr ``=` `ptr.``next``;`` ` `    ``# return count of common nodes``    ``return` `count;`` ` `# Driver program``if` `__name__``=``=``'__main__'``:``    ` `    ``# start with the empty list``    ``head ``=` `None``;``    ``head1 ``=` `None``;`` ` `    ``# create the doubly linked list 1``    ``# 15 <. 16 <. 10 <. 9 <. 6 <. 7 <. 17``    ``head ``=` `push(head, ``17``);``    ``head ``=` `push(head, ``7``);``    ``head ``=` `push(head, ``6``);``    ``head ``=` `push(head, ``9``);``    ``head ``=` `push(head, ``10``);``    ``head ``=` `push(head, ``16``);``    ``head ``=` `push( head, ``15``);`` ` `    ``# create the doubly linked list 2``    ``# 15 <. 16 <. 45 <. 9 <. 6``    ``head1 ``=` `push(head1, ``6``);``    ``head1 ``=` `push(head1, ``9``);``    ``head1 ``=` `push(head1, ``45``);``    ``head1 ``=` `push(head1, ``16``);``    ``head1 ``=` `push(head1, ``15``);`` ` `    ``print``(``"Number of common nodes: "` `+` `str``(countCommonNodes(head, head1)))``         ` `# This code is contributed by rutvik_56.`

## C#

 `// C# implementation to count``// common element in given two``// doubly linked list``using` `System;``    ` `class` `GFG``{` `// Node of the doubly linked list``public` `class` `Node``{``    ``public` `int` `data;``    ``public` `Node prev, next;``};` `// Function to insert a node at the beginning``// of the Doubly Linked List``static` `Node push(Node head_ref, ``int` `new_data)``{``    ``// allocate node``    ``Node new_node = ``new` `Node();` `    ``// put in the data``    ``new_node.data = new_data;` `    ``// since we are adding at the beginning,``    ``// prev is always null``    ``new_node.prev = ``null``;` `    ``// link the old list of the new node``    ``new_node.next = head_ref;` `    ``// change prev of head node to new node``    ``if` `(head_ref != ``null``)``        ``head_ref.prev = new_node;` `    ``// move the head to point to the new node``    ``head_ref = new_node;``    ``return` `head_ref;``}` `// Count common nodes in both list1 and list 2``static` `int` `countCommonNodes(Node head_ref,``                            ``Node head)``{``    ``// head for list 1``    ``Node ptr = head_ref;` `    ``// head for list 2``    ``Node ptr1 = head;` `    ``// initialize count = 0``    ``int` `count = 0;` `    ``// traverse list 1 till the end``    ``while` `(ptr != ``null``)``    ``{``        ` `        ``// traverse list 2 till the end``        ``while` `(ptr1 != ``null``)``        ``{``            ` `            ``// if node value is equal then``            ``// increment count``            ``if` `(ptr.data == ptr1.data)``            ``{``                ``count++;``                ``break``;``            ``}` `            ``// increment pointer list 2``            ``ptr1 = ptr1.next;``        ``}` `        ``// again list 2 start with starting point``        ``ptr1 = head;` `        ``// increment pointer list 1``        ``ptr = ptr.next;``    ``}` `    ``// return count of common nodes``    ``return` `count;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``// start with the empty list``    ``Node head = ``null``;``    ``Node head1 = ``null``;` `    ``// create the doubly linked list 1``    ``// 15 <. 16 <. 10 <. 9 <. 6 <. 7 <. 17``    ``head = push(head, 17);``    ``head = push(head, 7);``    ``head = push(head, 6);``    ``head = push(head, 9);``    ``head = push(head, 10);``    ``head = push(head, 16);``    ``head = push(head, 15);` `    ``// create the doubly linked list 2``    ``// 15 <. 16 <. 45 <. 9 <. 6``    ``head1 = push(head1, 6);``    ``head1 = push(head1, 9);``    ``head1 = push(head1, 45);``    ``head1 = push(head1, 16);``    ``head1 = push(head1, 15);` `    ``Console.WriteLine(``"Number of common nodes: "` `+``                   ``countCommonNodes(head, head1));``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output

```Number of common nodes:4

```

Complexity Analysis:

• Time Complexity: O(n1*n2) where n1 is the length of the first linked list and n2 is the length of the second linked list
• Auxiliary Space: O(1)

Approach 2:  Use hash tables or sets to keep track of the values in one list and then traverse the other list to count the common nodes.

Algorithm:

• Define a struct Node which consists of an integer data value, a pointer to the previous node, and a pointer to the next node.
• Create a push function to insert a new node at the beginning of the doubly linked list. This function takes the address of the head pointer and the new data as inputs.
• Create a countCommonNodes function to count the number of common nodes in two doubly linked lists. This function takes the addresses of the head pointers of both lists as inputs and returns an integer value.
• Inside the countCommonNodes function, initialize two node pointers ptr and ptr1 to the head of the first and second linked lists, respectively.
• Initialize a count variable to 0 to keep track of the number of common nodes.
• Traverse the first linked list using ptr and for each node of the first list, traverse the second linked list using ptr1.
• If the data value of the current node in the first list matches the data value of any node in the second list, increment the count variable and break the inner loop.
• After the inner loop completes, reset ptr1 to the head of the second list.
Move ptr to the next node of the first list and repeat steps 6-8 until ptr reaches the end of the first list.
• Return the count of common nodes.
• In the main function, create two doubly linked lists using push function and call the countCommonNodes function passing the head pointers of both the lists as inputs.
• Print the returned count value.

Below is the implementation of the above approach:

## C++

 `//C++ code for above approach``#include ` `using` `namespace` `std;` `// Node of the doubly linked list``struct` `Node {``    ``int` `data;``    ``Node *prev, *next;``};` `// Function to insert a node at the beginning``// of the Doubly Linked List``void` `push(Node** head_ref, ``int` `new_data)``{``    ``// allocate node``    ``Node* new_node = (Node*)``malloc``(``sizeof``(``struct` `Node));` `    ``// put in the data``    ``new_node->data = new_data;` `    ``// since we are adding at the beginning,``    ``// prev is always NULL``    ``new_node->prev = NULL;` `    ``// link the old list of the new node``    ``new_node->next = (*head_ref);` `    ``// change prev of head node to new node``    ``if` `((*head_ref) != NULL)``        ``(*head_ref)->prev = new_node;` `    ``// move the head to point to the new node``    ``(*head_ref) = new_node;``}` `// Count common nodes in both list1 and list 2``int` `countCommonNodes(Node** head_ref, Node** head)``{``    ``// head for list 1``    ``Node* ptr = *head_ref;` `    ``// head for list 2``    ``Node* ptr1 = *head;` `    ``// initialize count = 0``    ``int` `count = 0;` `    ``// set to store values of list 1``    ``unordered_set<``int``> values;` `    ``// insert values of list 1 in the set``    ``while` `(ptr != NULL) {``        ``values.insert(ptr->data);``        ``ptr = ptr->next;``    ``}` `    ``// traverse list 2 and count common nodes``    ``while` `(ptr1 != NULL) {``        ``if` `(values.count(ptr1->data) > 0) {``            ``count++;``        ``}``        ``ptr1 = ptr1->next;``    ``}` `    ``// return count of common nodes``    ``return` `count;``}` `// Driver program``int` `main()``{``    ``// start with the empty list``    ``Node* head = NULL;``    ``Node* head1 = NULL;` `    ``// create the doubly linked list 1``    ``// 15 <-> 16 <-> 10 <-> 9 <-> 6 <-> 7 <-> 17``    ``push(&head, 17);``    ``push(&head, 7);``    ``push(&head, 6);``    ``push(&head, 9);``    ``push(&head, 10);``    ``push(&head, 16);``    ``push(&head, 15);` `    ``// create the doubly linked list 2``    ``// 15 <-> 16 <-> 45 <-> 9 <-> 6``    ``push(&head1, 6);``    ``push(&head1, 9);``    ``push(&head1, 45);``    ``push(&head1, 16);``    ``push(&head1, 15);` `    ``cout << ``"Number of common nodes:"``         ``<< countCommonNodes(&head, &head1);` `    ``return` `0;``}`

## Java

 `import` `java.util.HashSet;` `// Node of the doubly linked list``class` `Node {``    ``int` `data;``    ``Node prev, next;``}` `public` `class` `CommonNodeCount {``    ``// Function to insert a node at the beginning``    ``// of the Doubly Linked List``    ``static` `void` `push(Node[] headRef, ``int` `newData)``    ``{``        ``// Allocate node``        ``Node newNode = ``new` `Node();` `        ``// Put in the data``        ``newNode.data = newData;` `        ``// Since we are adding at the beginning,``        ``// prev is always null``        ``newNode.prev = ``null``;` `        ``// Link the old list to the new node``        ``newNode.next = headRef[``0``];` `        ``// Change prev of the head node to new node``        ``if` `(headRef[``0``] != ``null``) {``            ``headRef[``0``].prev = newNode;``        ``}` `        ``// Move the head to point to the new node``        ``headRef[``0``] = newNode;``    ``}` `    ``// Count common nodes in both list1 and list2``    ``static` `int` `countCommonNodes(Node[] headRef, Node[] head)``    ``{``        ``// Head for list1``        ``Node ptr = headRef[``0``];` `        ``// Head for list2``        ``Node ptr1 = head[``0``];` `        ``// Initialize count = 0``        ``int` `count = ``0``;` `        ``// Set to store values of list1``        ``HashSet values = ``new` `HashSet<>();` `        ``// Insert values of list1 in the set``        ``while` `(ptr != ``null``) {``            ``values.add(ptr.data);``            ``ptr = ptr.next;``        ``}` `        ``// Traverse list2 and count common nodes``        ``while` `(ptr1 != ``null``) {``            ``if` `(values.contains(ptr1.data)) {``                ``count++;``            ``}``            ``ptr1 = ptr1.next;``        ``}` `        ``// Return count of common nodes``        ``return` `count;``    ``}` `    ``// Driver program``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// Start with the empty list``        ``Node[] head = { ``null` `};``        ``Node[] head1 = { ``null` `};` `        ``// Create the doubly linked list 1``        ``// 15 <-> 16 <-> 10 <-> 9 <-> 6 <-> 7 <-> 17``        ``push(head, ``17``);``        ``push(head, ``7``);``        ``push(head, ``6``);``        ``push(head, ``9``);``        ``push(head, ``10``);``        ``push(head, ``16``);``        ``push(head, ``15``);` `        ``// Create the doubly linked list 2``        ``// 15 <-> 16 <-> 45 <-> 9 <-> 6``        ``push(head1, ``6``);``        ``push(head1, ``9``);``        ``push(head1, ``45``);``        ``push(head1, ``16``);``        ``push(head1, ``15``);` `        ``System.out.println(``"Number of common nodes: "``                           ``+ countCommonNodes(head, head1));``    ``}``}` `// This code is contributed by Samim Hossain Mondal.`

## Python3

 `#Python3 approach for the above code``# Node of the doubly linked list``class` `Node:``    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.prev ``=` `None``        ``self``.``next` `=` `None` `# Function to insert a node at the beginning``# of the Doubly Linked List``def` `push(head_ref, new_data):``    ``# allocate node``    ``new_node ``=` `Node(new_data)` `    ``# link the old list of the new node``    ``new_node.``next` `=` `head_ref` `    ``# change prev of head node to new node``    ``if` `head_ref !``=` `None``:``        ``head_ref.prev ``=` `new_node` `    ``# move the head to point to the new node``    ``head_ref ``=` `new_node``    ``return` `head_ref` `# Count common nodes in both list1 and list 2``def` `countCommonNodes(head_ref, head):``    ``# head for list 1``    ``ptr ``=` `head_ref` `    ``# head for list 2``    ``ptr1 ``=` `head` `    ``# initialize count = 0``    ``count ``=` `0` `    ``# set to store values of list 1``    ``values ``=` `set``()` `    ``# insert values of list 1 in the set``    ``while` `ptr !``=` `None``:``        ``values.add(ptr.data)``        ``ptr ``=` `ptr.``next` `    ``# traverse list 2 and count common nodes``    ``while` `ptr1 !``=` `None``:``        ``if` `ptr1.data ``in` `values:``            ``count ``+``=` `1``        ``ptr1 ``=` `ptr1.``next` `    ``# return count of common nodes``    ``return` `count` `# Driver program``if` `__name__ ``=``=` `'__main__'``:``    ``# start with the empty list``    ``head ``=` `None``    ``head1 ``=` `None` `    ``# create the doubly linked list 1``    ``# 15 <-> 16 <-> 10 <-> 9 <-> 6 <-> 7 <-> 17``    ``head ``=` `push(head, ``17``)``    ``head ``=` `push(head, ``7``)``    ``head ``=` `push(head, ``6``)``    ``head ``=` `push(head, ``9``)``    ``head ``=` `push(head, ``10``)``    ``head ``=` `push(head, ``16``)``    ``head ``=` `push(head, ``15``)` `    ``# create the doubly linked list 2``    ``# 15 <-> 16 <-> 45 <-> 9 <-> 6``    ``head1 ``=` `push(head1, ``6``)``    ``head1 ``=` `push(head1, ``9``)``    ``head1 ``=` `push(head1, ``45``)``    ``head1 ``=` `push(head1, ``16``)``    ``head1 ``=` `push(head1, ``15``)` `    ``print``(``"Number of common nodes:"``,``          ``countCommonNodes(head, head1))`

## C#

 `using` `System;``using` `System.Collections.Generic;` `// Node of the doubly linked list``public` `class` `Node {``    ``public` `int` `data;``    ``public` `Node prev, next;``}` `class` `Program {``    ``// Function to insert a node at the beginning``    ``// of the Doubly Linked List``    ``static` `void` `Push(``ref` `Node head_ref, ``int` `new_data)``    ``{``        ``// allocate node``        ``Node new_node = ``new` `Node();` `        ``// put in the data``        ``new_node.data = new_data;` `        ``// since we are adding at the beginning,``        ``// prev is always null``        ``new_node.prev = ``null``;` `        ``// link the old list to the new node``        ``new_node.next = head_ref;` `        ``// change prev of head node to new node``        ``if` `(head_ref != ``null``)``            ``head_ref.prev = new_node;` `        ``// move the head to point to the new node``        ``head_ref = new_node;``    ``}` `    ``// Count common nodes in both list1 and list2``    ``static` `int` `CountCommonNodes(``ref` `Node head_ref,``                                ``ref` `Node head)``    ``{``        ``// head for list 1``        ``Node ptr = head_ref;` `        ``// head for list 2``        ``Node ptr1 = head;` `        ``// initialize count = 0``        ``int` `count = 0;` `        ``// set to store values of list 1``        ``HashSet<``int``> values = ``new` `HashSet<``int``>();` `        ``// insert values of list 1 in the set``        ``while` `(ptr != ``null``) {``            ``values.Add(ptr.data);``            ``ptr = ptr.next;``        ``}` `        ``// traverse list 2 and count common nodes``        ``while` `(ptr1 != ``null``) {``            ``if` `(values.Contains(ptr1.data)) {``                ``count++;``            ``}``            ``ptr1 = ptr1.next;``        ``}` `        ``// return count of common nodes``        ``return` `count;``    ``}` `    ``// Driver program``    ``static` `void` `Main()``    ``{``        ``// start with the empty list``        ``Node head = ``null``;``        ``Node head1 = ``null``;` `        ``// create the doubly linked list 1``        ``// 15 <-> 16 <-> 10 <-> 9 <-> 6 <-> 7 <-> 17``        ``Push(``ref` `head, 17);``        ``Push(``ref` `head, 7);``        ``Push(``ref` `head, 6);``        ``Push(``ref` `head, 9);``        ``Push(``ref` `head, 10);``        ``Push(``ref` `head, 16);``        ``Push(``ref` `head, 15);` `        ``// create the doubly linked list 2``        ``// 15 <-> 16 <-> 45 <-> 9 <-> 6``        ``Push(``ref` `head1, 6);``        ``Push(``ref` `head1, 9);``        ``Push(``ref` `head1, 45);``        ``Push(``ref` `head1, 16);``        ``Push(``ref` `head1, 15);` `        ``Console.WriteLine(``            ``"Number of common nodes: "``            ``+ CountCommonNodes(``ref` `head, ``ref` `head1));``    ``}``}`

Output:

`Number of common nodes:4`

Time complexity: O(m*n)
The time complexity of the code is O(mn), where m and n are the sizes of the two input linked lists. This is because we have to traverse both linked lists completely to find common nodes. The nested while loop inside the function iterates over each node of the second list for each node of the first list. This results in a time complexity of O(mn).

Auxiliary space: O(1)
The auxiliary space complexity of the code is O(1) because we are not using any extra space for storing information. We are just using pointers to traverse the linked lists and an integer variable to store the count of common nodes. Therefore, the space complexity remains constant, i.e., O(1).