# Longest common suffix of two linked lists

Given two singly linked lists, find the Longest common suffix of two linked lists. If there are no common characters which are suffixes, return the minimum length of the two linked lists.

Examples:

```Input : list1 = w -> a -> l -> k -> i -> n -> g
list2 = l -> i -> s -> t -> e -> n -> i -> n -> g
Output :i -> n -> g

Input : list1 = p -> a -> r -> t -> y
list2 = p -> a -> r -> t -> y -> i -> n -> g
Output :p -> a -> r -> t -> y
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to use auxiliary arrays to store linked lists. Then print longest commons suffix of two arrays.

The above solution requires extra space. We can save space by first doing reverse of both linked lists. After reversing, we can easily find length of longest common prefix. Reversing again to get the original lists back.
One important point here is, order of elements. We need to print nodes from n-th to end. We use the above found count and print nodes in required order using two pointer approach.

## C++

 `// C++ program to find the Longest Common ` `// suffix in linked lists ` `#include ` `using` `namespace` `std; ` ` `  `/* Linked list node */` `struct` `Node ` `{ ` `    ``char` `data; ` `    ``struct` `Node* next; ` `}; ` ` `  `/* Function to insert a node at the beginning of ` `   ``the linked list */` `void` `push(``struct` `Node **head_ref, ``int` `new_data) ` `{ ` `    ``struct` `Node* new_node = ``new` `Node; ` `    ``new_node->data = new_data; ` `    ``new_node->next = *head_ref; ` `    ``*head_ref = new_node; ` `} ` ` `  `/* Function to reverse the linked list */` `struct` `Node *reverseList(``struct` `Node *head_ref) ` `{ ` `    ``struct` `Node *current, *prev, *next; ` `    ``current = head_ref; ` `    ``prev = NULL; ` ` `  `    ``while` `(current != NULL) ` `    ``{ ` `        ``next = current->next; ` `        ``current->next = prev; ` `        ``prev = current; ` `        ``current = next; ` `    ``} ` `    ``return` `prev; ` `} ` ` `  ` `  `// Utility function to print last n nodes ` `void` `printLastNNode(``struct` `Node* head, ``int` `n) ` `{ ` `    ``// if n == 0 ` `    ``if` `(n <= 0) ` `        ``return``; ` ` `  `    ``// Move reference pointer n positions ahead ` `    ``struct` `Node* ref_ptr = head; ` `    ``while` `(ref_ptr != NULL && n--) ` `        ``ref_ptr = ref_ptr->next; ` ` `  `    ``// Now move main and reference pointers at ` `    ``// same speed. By the end of this loop, ` `    ``// reference pointer would point to end and ` `    ``// main pointer would point to n-th node ` `    ``// from end. ` `    ``Node *main_ptr = head; ` `    ``while` `(ref_ptr != NULL) { ` `        ``main_ptr = main_ptr->next; ` `        ``ref_ptr = ref_ptr->next; ` `    ``} ` ` `  `    ``// Print last n nodes. ` `    ``while` `(main_ptr != NULL) ` `    ``{ ` `       ``cout << main_ptr->data; ` `       ``main_ptr = main_ptr->next; ` `    ``} ` `} ` ` `  `// Prints the Longest Common suffix in  ` `// linked lists ` `void` `longestCommSuffix(Node *h1, Node *h2) ` `{ ` `    ``// Reverse Both Linked list ` `    ``h1 = reverseList(h1); ` `    ``h2 = reverseList(h2); ` ` `  `    ``// Now we print common nodes from head ` `    ``Node *temp1 = h1, *temp2 = h2; ` `    ``int` `count = 0; ` `    ``while` `(temp1!=NULL&&temp2!=NULL) ` `    ``{ ` `        ``// If a node is not common, break ` `        ``if` `(temp1 -> data != temp2 -> data) ` `            ``break``; ` ` `  `        ``// Keep printing while there are ` `        ``// common nodes. ` `        ``count++; ` `        ``temp1 = temp1 -> next; ` `        ``temp2 = temp2 -> next; ` `    ``} ` ` `  `    ``// Reversing linked lists to retain ` `    ``// original lists. ` `    ``h1 = reverseList(h1); ` `    ``h2 = reverseList(h2); ` ` `  `    ``printLastNNode(h1, count); ` `} ` ` `  `// Driver program to test above ` `int` `main() ` `{ ` `    ``struct` `Node *h1 = NULL, *h2 = NULL; ` ` `  `    ``// creating the 1 linked list ` `    ``push(&h1,``'g'``); ` `    ``push(&h1,``'n'``); ` `    ``push(&h1,``'i'``); ` `    ``push(&h1,``'k'``); ` `    ``push(&h1,``'l'``); ` `    ``push(&h1,``'a'``); ` `    ``push(&h1,``'w'``); ` ` `  `    ``// creating the 2 linked list ` `    ``push(&h2,``'g'``); ` `    ``push(&h2,``'n'``); ` `    ``push(&h2,``'i'``); ` `    ``push(&h2,``'n'``); ` `    ``push(&h2,``'e'``); ` `    ``push(&h2,``'t'``); ` `    ``push(&h2,``'s'``); ` `    ``push(&h2,``'i'``); ` `    ``push(&h2,``'l'``); ` ` `  `    ``longestCommSuffix(h1, h2); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find the Longest Common ` `// suffix in linked lists ` `import` `java.util.*; ` `class` `GFG ` `{ ` ` `  `/* Linked list node */` `static` `class` `Node  ` `{ ` `    ``int` `data; ` `    ``Node next; ` `}; ` `static` `Node h1,h2; ` ` `  `/* Function to insert a node at the beginning ` `of the 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; ` `} ` ` `  `/* Function to reverse the linked list */` `static` `Node reverseList(Node head_ref) ` `{ ` `    ``Node current, prev, next; ` `    ``current = head_ref; ` `    ``prev = ``null``; ` ` `  `    ``while` `(current != ``null``) ` `    ``{ ` `        ``next = current.next; ` `        ``current.next = prev; ` `        ``prev = current; ` `        ``current = next; ` `    ``} ` `    ``return` `prev; ` `} ` ` `  `// Utility function to print last n nodes ` `static` `void` `printLastNNode(Node head, ``int` `n) ` `{ ` `    ``// if n == 0 ` `    ``if` `(n <= ``0``) ` `        ``return``; ` ` `  `    ``// Move reference pointer n positions ahead ` `    ``Node ref_ptr = head; ` `    ``while` `(ref_ptr != ``null` `&& n-- > ``0``) ` `        ``ref_ptr = ref_ptr.next; ` ` `  `    ``// Now move main and reference pointers at ` `    ``// same speed. By the end of this loop, ` `    ``// reference pointer would point to end and ` `    ``// main pointer would point to n-th node ` `    ``// from end. ` `    ``Node main_ptr = head; ` `    ``while` `(ref_ptr != ``null``)  ` `    ``{ ` `        ``main_ptr = main_ptr.next; ` `        ``ref_ptr = ref_ptr.next; ` `    ``} ` ` `  `    ``// Print last n nodes. ` `    ``while` `(main_ptr != ``null``) ` `    ``{ ` `        ``System.out.print((``char``)(main_ptr.data)); ` `        ``main_ptr = main_ptr.next; ` `    ``} ` `} ` ` `  `// Prints the Longest Common suffix in  ` `// linked lists ` `static` `void` `longestCommSuffix() ` `{ ` `    ``// Reverse Both Linked list ` `    ``h1 = reverseList(h1); ` `    ``h2 = reverseList(h2); ` ` `  `    ``// Now we print common nodes from head ` `    ``Node temp1 = h1, temp2 = h2; ` `    ``int` `count = ``0``; ` `    ``while` `(temp1 != ``null` `&& temp2 != ``null``) ` `    ``{ ` `        ``// If a node is not common, break ` `        ``if` `(temp1 . data != temp2 . data) ` `            ``break``; ` ` `  `        ``// Keep printing while there are ` `        ``// common nodes. ` `        ``count++; ` `        ``temp1 = temp1 . next; ` `        ``temp2 = temp2 . next; ` `    ``} ` ` `  `    ``// Reversing linked lists to retain ` `    ``// original lists. ` `    ``h1 = reverseList(h1); ` `    ``h2 = reverseList(h2); ` ` `  `    ``printLastNNode(h1, count); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``h1 = ``null``; h2 = ``null``; ` ` `  `    ``// creating the 1 linked list ` `    ``h1 = push(h1, ``'g'``); ` `    ``h1 = push(h1, ``'n'``); ` `    ``h1 = push(h1, ``'i'``); ` `    ``h1 = push(h1, ``'k'``); ` `    ``h1 = push(h1, ``'l'``); ` `    ``h1 = push(h1, ``'a'``); ` `    ``h1 = push(h1, ``'w'``); ` ` `  `    ``// creating the 2 linked list ` `    ``h2 = push(h2, ``'g'``); ` `    ``h2 = push(h2, ``'n'``); ` `    ``h2 = push(h2, ``'i'``); ` `    ``h2 = push(h2, ``'n'``); ` `    ``h2 = push(h2, ``'e'``); ` `    ``h2 = push(h2, ``'t'``); ` `    ``h2 = push(h2, ``'s'``); ` `    ``h2 = push(h2, ``'i'``); ` `    ``h2 = push(h2, ``'l'``); ` ` `  `    ``longestCommSuffix(); ` `} ` `}  ` ` `  `// This code is contributed by Rajput-Ji `

## C#

 `// C# program to find the Longest Common ` `// suffix in linked lists ` `using` `System; ` `     `  `class` `GFG ` `{ ` ` `  `/* Linked list node */` `public` `class` `Node  ` `{ ` `    ``public` `int` `data; ` `    ``public` `Node next; ` `}; ` `static` `Node h1, h2; ` ` `  `/* Function to insert a node  ` `at the beginning of the 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; ` `} ` ` `  `/* Function to reverse the linked list */` `static` `Node reverseList(Node head_ref) ` `{ ` `    ``Node current, prev, next; ` `    ``current = head_ref; ` `    ``prev = ``null``; ` ` `  `    ``while` `(current != ``null``) ` `    ``{ ` `        ``next = current.next; ` `        ``current.next = prev; ` `        ``prev = current; ` `        ``current = next; ` `    ``} ` `    ``return` `prev; ` `} ` ` `  `// Utility function to print last n nodes ` `static` `void` `printLastNNode(Node head, ``int` `n) ` `{ ` `    ``// if n == 0 ` `    ``if` `(n <= 0) ` `        ``return``; ` ` `  `    ``// Move reference pointer n positions ahead ` `    ``Node ref_ptr = head; ` `    ``while` `(ref_ptr != ``null` `&& n-- > 0) ` `        ``ref_ptr = ref_ptr.next; ` ` `  `    ``// Now move main and reference pointers at ` `    ``// same speed. By the end of this loop, ` `    ``// reference pointer would point to end and ` `    ``// main pointer would point to n-th node ` `    ``// from end. ` `    ``Node main_ptr = head; ` `    ``while` `(ref_ptr != ``null``)  ` `    ``{ ` `        ``main_ptr = main_ptr.next; ` `        ``ref_ptr = ref_ptr.next; ` `    ``} ` ` `  `    ``// Print last n nodes. ` `    ``while` `(main_ptr != ``null``) ` `    ``{ ` `        ``Console.Write((``char``)(main_ptr.data)); ` `        ``main_ptr = main_ptr.next; ` `    ``} ` `} ` ` `  `// Prints the Longest Common suffix in  ` `// linked lists ` `static` `void` `longestCommSuffix() ` `{ ` `    ``// Reverse Both Linked list ` `    ``h1 = reverseList(h1); ` `    ``h2 = reverseList(h2); ` ` `  `    ``// Now we print common nodes from head ` `    ``Node temp1 = h1, temp2 = h2; ` `    ``int` `count = 0; ` `    ``while` `(temp1 != ``null` `&& temp2 != ``null``) ` `    ``{ ` `        ``// If a node is not common, break ` `        ``if` `(temp1 . data != temp2 . data) ` `            ``break``; ` ` `  `        ``// Keep printing while there are ` `        ``// common nodes. ` `        ``count++; ` `        ``temp1 = temp1 . next; ` `        ``temp2 = temp2 . next; ` `    ``} ` ` `  `    ``// Reversing linked lists to retain ` `    ``// original lists. ` `    ``h1 = reverseList(h1); ` `    ``h2 = reverseList(h2); ` ` `  `    ``printLastNNode(h1, count); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``h1 = ``null``; h2 = ``null``; ` ` `  `    ``// creating the 1 linked list ` `    ``h1 = push(h1, ``'g'``); ` `    ``h1 = push(h1, ``'n'``); ` `    ``h1 = push(h1, ``'i'``); ` `    ``h1 = push(h1, ``'k'``); ` `    ``h1 = push(h1, ``'l'``); ` `    ``h1 = push(h1, ``'a'``); ` `    ``h1 = push(h1, ``'w'``); ` ` `  `    ``// creating the 2 linked list ` `    ``h2 = push(h2, ``'g'``); ` `    ``h2 = push(h2, ``'n'``); ` `    ``h2 = push(h2, ``'i'``); ` `    ``h2 = push(h2, ``'n'``); ` `    ``h2 = push(h2, ``'e'``); ` `    ``h2 = push(h2, ``'t'``); ` `    ``h2 = push(h2, ``'s'``); ` `    ``h2 = push(h2, ``'i'``); ` `    ``h2 = push(h2, ``'l'``); ` ` `  `    ``longestCommSuffix(); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```ing
```

Time Complexity : O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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.

Improved By : Rajput-Ji, princiraj1992