Open In App

# Check if two Linked Lists are permutations of each other

Given two singly Linked list of integer data. The task is to write a program that efficiently checks if two linked lists are permutations of each other.

Examples:

`Input: 1 -> 2 -> 3 -> 4 -> 5        2 -> 1 -> 3 -> 5 -> 4Output: YesInput: 10 -> 20 -> 30 -> 40        20 -> 50 -> 60 -> 70Output: No`

Approach: Do the following for both linked lists:

1. Take a temporary node pointing to the head of the linked list.
2. Start traversing through the linked list, and keep sum and multiplications of data of nodes.

Note: After having sum and multiplication of both linked list, check if sum and multiplication of both linked lists are equal. If they are equal, it means linked lists are permutations of each other, else not.

Below is the implementation of the above approach:

## C++

 `// C++ program to check if linked lists``// are permutations of each other``#include ` `using` `namespace` `std;` `// A linked list node``struct` `Node {``    ``int` `data;``    ``struct` `Node* next;``};` `/*Function to check if two linked lists``* are permutations of each other``* first : reference to head of first linked list``* second : reference to head of second linked list``*/``bool` `isPermutation(``struct` `Node* first, ``struct` `Node* second)``{` `    ``// Variables to keep track of sum and multiplication``    ``int` `sum1 = 0, sum2 = 0, mul1 = 1, mul2 = 1;` `    ``struct` `Node* temp1 = first;` `    ``// Traversing through linked list``    ``// and calculating sum and multiply``    ``while` `(temp1 != NULL) {``        ``sum1 += temp1->data;``        ``mul1 *= temp1->data;``        ``temp1 = temp1->next;``    ``}` `    ``struct` `Node* temp2 = second;` `    ``// Traversing through linked list``    ``// and calculating sum and multiply``    ``while` `(temp2 != NULL) {``        ``sum2 += temp2->data;``        ``mul2 *= temp2->data;``        ``temp2 = temp2->next;``    ``}` `    ``return` `((sum1 == sum2) && (mul1 == mul2));``}` `// Function to add a node at the``// beginning of Linked List``void` `push(``struct` `Node** head_ref, ``int` `new_data)``{``    ``/* allocate node */``    ``struct` `Node* new_node = (``struct` `Node*)``malloc``(``sizeof``(``struct` `Node));` `    ``/* put in the data */``    ``new_node->data = new_data;` `    ``/* link the old list of the new node */``    ``new_node->next = (*head_ref);` `    ``/* move the head to point to the new node */``    ``(*head_ref) = new_node;``}` `// Driver program to test above function``int` `main()``{``    ``struct` `Node* first = NULL;` `    ``/* First constructed linked list is:``    ``12 -> 35 -> 1 -> 10 -> 34 -> 1 */``    ``push(&first, 1);``    ``push(&first, 34);``    ``push(&first, 10);``    ``push(&first, 1);``    ``push(&first, 35);``    ``push(&first, 12);` `    ``struct` `Node* second = NULL;``    ``/* Second constructed linked list is:``    ``35 -> 1 -> 12 -> 1 -> 10 -> 34 */``    ``push(&second, 35);``    ``push(&second, 1);``    ``push(&second, 12);``    ``push(&second, 1);``    ``push(&second, 10);``    ``push(&second, 34);` `    ``if` `(isPermutation(first, second)) {``        ``cout << ``"Yes"` `<< endl;``    ``}``    ``else` `{``        ``cout << ``"No"` `<< endl;``    ``}` `    ``return` `0;``}`

## Java

 `// Java program to check if linked lists``// are permutations of each other``import` `java.util.*;` `class` `GFG``{``static` `class` `Node``{``    ``int` `data;``    ``Node next;``};` `/*Function to check if two linked lists``* are permutations of each other``* first : reference to head of first linked list``* second : reference to head of second linked list``*/``static` `boolean` `isPermutation(Node first,``                             ``Node second)``{` `    ``// Variables to keep track of``    ``// sum and multiplication``    ``int` `sum1 = ``0``, sum2 = ``0``,``        ``mul1 = ``1``, mul2 = ``1``;` `    ``Node temp1 = first;` `    ``// Traversing through linked list``    ``// and calculating sum and multiply``    ``while` `(temp1 != ``null``)``    ``{``        ``sum1 += temp1.data;``        ``mul1 *= temp1.data;``        ``temp1 = temp1.next;``    ``}` `    ``Node temp2 = second;` `    ``// Traversing through linked list``    ``// and calculating sum and multiply``    ``while` `(temp2 != ``null``)``    ``{``        ``sum2 += temp2.data;``        ``mul2 *= temp2.data;``        ``temp2 = temp2.next;``    ``}` `    ``return` `((sum1 == sum2) &&``            ``(mul1 == mul2));``}` `// Function to add a node at the``// beginning of 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;` `    ``/* link the old list of the new node */``    ``new_node.next = head_ref;` `    ``/* move the head to point to the new node */``    ``head_ref = new_node;``    ``return` `head_ref;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``Node first = ``null``;` `    ``/* First constructed linked list is:``    ``12 . 35 . 1 . 10 . 34 . 1 */``    ``first = push(first, ``1``);``    ``first = push(first, ``34``);``    ``first = push(first, ``10``);``    ``first = push(first, ``1``);``    ``first = push(first, ``35``);``    ``first = push(first, ``12``);` `    ``Node second = ``null``;``    ` `    ``/* Second constructed linked list is:``    ``35 . 1 . 12 . 1 . 10 . 34 */``    ``second = push(second, ``35``);``    ``second = push(second, ``1``);``    ``second = push(second, ``12``);``    ``second = push(second, ``1``);``    ``second = push(second, ``10``);``    ``second = push(second, ``34``);` `    ``if` `(isPermutation(first, second))``    ``{``        ``System.out.print(``"Yes"``);``    ``}``    ``else``    ``{``        ``System.out.print(``"No"``);``    ``}``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to check if linked lists``# are permutations of each other``class` `Node:``    ` `    ``def` `__init__(``self``):``        ` `        ``self``.data ``=` `0``        ``self``.``next` `=` `None` `# Function to check if two linked lists``# are permutations of each other``# first : reference to head of first linked list``# second : reference to head of second linked list``def` `isPermutation(first, second):``    ` `    ``# Variables to keep track of``    ``# sum and multiplication``    ``sum1 ``=` `0``    ``sum2 ``=` `0``    ``mul1 ``=` `1``    ``mul2 ``=` `1`` ` `    ``temp1 ``=` `first`` ` `    ``# Traversing through linked list``    ``# and calculating sum and multiply``    ``while` `(temp1 !``=` `None``):``        ``sum1 ``+``=` `temp1.data``        ``mul1 ``*``=` `temp1.data``        ``temp1 ``=` `temp1.``next` `    ``temp2 ``=` `second``    ` `    ``# Traversing through linked list``    ``# and calculating sum and multiply``    ``while` `(temp2 !``=` `None``):``        ``sum2 ``+``=` `temp2.data``        ``mul2 ``*``=` `temp2.data``        ``temp2 ``=` `temp2.``next``    ` `    ``return` `((sum1 ``=``=` `sum2) ``and` `(mul1 ``=``=` `mul2))` `# Function to add a node at the``# beginning of Linked List``def` `push(head_ref, new_data):``    ` `    ``# Allocate node``    ``new_node ``=` `Node()``    ` `    ``# Put in the data``    ``new_node.data ``=` `new_data`` ` `    ``# Link the old list of the new node``    ``new_node.``next` `=` `head_ref`` ` `    ``# Move the head to point to the new node``    ``head_ref ``=` `new_node``    ``return` `head_ref` `# Driver Code``if` `__name__``=``=``'__main__'``:``    ` `    ``first ``=` `None`` ` `    ``# First constructed linked list is:``    ``# 12 . 35 . 1 . 10 . 34 . 1``    ``first ``=` `push(first, ``1``)``    ``first ``=` `push(first, ``34``)``    ``first ``=` `push(first, ``10``)``    ``first ``=` `push(first, ``1``)``    ``first ``=` `push(first, ``35``)``    ``first ``=` `push(first, ``12``)`` ` `    ``second ``=` `None``     ` `    ``# Second constructed linked list is:``    ``# 35 . 1 . 12 . 1 . 10 . 34``    ``second ``=` `push(second, ``35``)``    ``second ``=` `push(second, ``1``)``    ``second ``=` `push(second, ``12``)``    ``second ``=` `push(second, ``1``)``    ``second ``=` `push(second, ``10``)``    ``second ``=` `push(second, ``34``)`` ` `    ``if` `(isPermutation(first, second)):``        ``print``(``"Yes"``)``    ``else``:``        ``print``(``"No"``)``    ` `# This code is contributed by pratham76`

## C#

 `// C# program to check if linked lists``// are permutations of each other``using` `System;` `class` `GFG``{``public` `class` `Node``{``    ``public` `int` `data;``    ``public` `Node next;``};` `/*Function to check if two linked lists``* are permutations of each other``* first : reference to head of first linked list``* second : reference to head of second linked list``*/``static` `bool` `isPermutation(Node first,``                          ``Node second)``{` `    ``// Variables to keep track of``    ``// sum and multiplication``    ``int` `sum1 = 0, sum2 = 0,``        ``mul1 = 1, mul2 = 1;` `    ``Node temp1 = first;` `    ``// Traversing through linked list``    ``// and calculating sum and multiply``    ``while` `(temp1 != ``null``)``    ``{``        ``sum1 += temp1.data;``        ``mul1 *= temp1.data;``        ``temp1 = temp1.next;``    ``}` `    ``Node temp2 = second;` `    ``// Traversing through linked list``    ``// and calculating sum and multiply``    ``while` `(temp2 != ``null``)``    ``{``        ``sum2 += temp2.data;``        ``mul2 *= temp2.data;``        ``temp2 = temp2.next;``    ``}` `    ``return` `((sum1 == sum2) &&``            ``(mul1 == mul2));``}` `// Function to add a node at the``// beginning of 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;` `    ``/* link the old list of the new node */``    ``new_node.next = head_ref;` `    ``/* move the head to point to the new node */``    ``head_ref = new_node;``    ``return` `head_ref;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``Node first = ``null``;` `    ``/* First constructed linked list is:``    ``12 . 35 . 1 . 10 . 34 . 1 */``    ``first = push(first, 1);``    ``first = push(first, 34);``    ``first = push(first, 10);``    ``first = push(first, 1);``    ``first = push(first, 35);``    ``first = push(first, 12);` `    ``Node second = ``null``;``    ` `    ``/* Second constructed linked list is:``    ``35 . 1 . 12 . 1 . 10 . 34 */``    ``second = push(second, 35);``    ``second = push(second, 1);``    ``second = push(second, 12);``    ``second = push(second, 1);``    ``second = push(second, 10);``    ``second = push(second, 34);` `    ``if` `(isPermutation(first, second))``    ``{``        ``Console.Write(``"Yes"``);``    ``}``    ``else``    ``{``        ``Console.Write(``"No"``);``    ``}``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output

```Yes

```

Complexity Analysis:

• Time Complexity: O(N) where N is the size of linked lists
• Auxiliary Space: O(1) because using constant space

### Check if two Linked Lists are permutations of each other using Hashing.

Here’s an approach to check if two linked lists are permutations of each other using a hash table:

• Traverse through both linked lists and store the frequency of each element in a hash table. The hash table can be implemented using an array of size equal to the range of the elements in the linked lists.
• Traverse through both linked lists again and compare the frequency of each element in the hash table. If the frequency is not equal, the linked lists are not permutations of each other.
• If the frequency of all elements is equal, the linked lists are permutations of each other.

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `// A linked list node``struct` `Node {``    ``int` `data;``    ``struct` `Node* next;``};` `// Function to check if two linked lists are permutations of each other``bool` `arePermutations(``struct` `Node* first, ``struct` `Node* second) {``    ``// Initialize hash table with all elements set to 0``    ``const` `int` `range = 1000; ``// Change range according to the range of elements in linked lists``    ``int` `hash[range] = {0};` `    ``// Traverse through first linked list and update hash table``    ``struct` `Node* temp = first;``    ``while` `(temp != NULL) {``        ``hash[temp->data]++;``        ``temp = temp->next;``    ``}` `    ``// Traverse through second linked list and compare hash table``    ``temp = second;``    ``while` `(temp != NULL) {``        ``if` `(hash[temp->data] == 0) {``            ``return` `false``;``        ``}``        ``hash[temp->data]--;``        ``temp = temp->next;``    ``}` `    ``// Check if all elements have the same frequency in both linked lists``    ``for` `(``int` `i = 0; i < range; i++) {``        ``if` `(hash[i] != 0) {``            ``return` `false``;``        ``}``    ``}` `    ``return` `true``;``}` `// Function to add a node at the beginning of Linked List``void` `push(``struct` `Node** head_ref, ``int` `new_data) {``    ``/* allocate node */``    ``struct` `Node* new_node = (``struct` `Node*)``malloc``(``sizeof``(``struct` `Node));` `    ``/* put in the data */``    ``new_node->data = new_data;` `    ``/* link the old list of the new node */``    ``new_node->next = (*head_ref);` `    ``/* move the head to point to the new node */``    ``(*head_ref) = new_node;``}` `// Driver program to test above function``int` `main() {``    ``struct` `Node* first = NULL;` `    ``/* First constructed linked list is:``    ``12 -> 35 -> 1 -> 10 -> 34 -> 1 */``    ``push(&first, 1);``    ``push(&first, 34);``    ``push(&first, 10);``    ``push(&first, 1);``    ``push(&first, 35);``    ``push(&first, 12);` `    ``struct` `Node* second = NULL;``    ``/* Second constructed linked list is:``    ``35 -> 1 -> 12 -> 1 -> 10 -> 34 */``    ``push(&second, 35);``    ``push(&second, 1);``    ``push(&second, 12);``    ``push(&second, 1);``    ``push(&second, 10);``    ``push(&second, 34);` `    ``if` `(arePermutations(first, second)) {``        ``cout << ``"Yes"` `<< endl;``    ``}``    ``else` `{``        ``cout << ``"No"` `<< endl;``    ``}` `    ``return` `0;``}`

## Javascript

 `// A linked list node``class Node {``    ``constructor(data) {``        ``this``.data = data;``        ``this``.next = ``null``;``    ``}``}``function` `GFG(first, second) {``    ``const range = 1000;``    ``const hash = Array.from({ length: range }, () => 0);``    ``// Traverse through first linked list and``    ``// update hash table``    ``let temp = first;``    ``while` `(temp !== ``null``) {``        ``hash[temp.data]++;``        ``temp = temp.next;``    ``}``    ``// Traverse through second linked list and``    ``// compare hash table``    ``temp = second;``    ``while` `(temp !== ``null``) {``        ``if` `(hash[temp.data] === 0) {``            ``return` `false``;``        ``}``        ``hash[temp.data]--;``        ``temp = temp.next;``    ``}``    ``// Check if all elements have the same frequency in``    ``// both linked lists``    ``for` `(let i = 0; i < range; i++) {``        ``if` `(hash[i] !== 0) {``            ``return` `false``;``        ``}``    ``}``    ``return` `true``;``}``// Function to add a node at the``// beginning of Linked List``function` `push(headRef, new_data) {``    ``const new_node = ``new` `Node(new_data);``    ``new_node.next = headRef;``    ``headRef = new_node;``    ``return` `headRef;``}` `    ``let first = ``null``;``// First constructed linked list``    ``first = push(first, 1);``    ``first = push(first, 34);``    ``first = push(first, 10);``    ``first = push(first, 1);``    ``first = push(first, 35);``    ``first = push(first, 12);``    ``let second = ``null``;``    ``second = push(second, 35);``    ``second = push(second, 1);``    ``second = push(second, 12);``    ``second = push(second, 1);``    ``second = push(second, 10);``    ``second = push(second, 34);``    ``if` `(GFG(first, second)) {``        ``console.log(``"Yes"``);``    ``} ``else` `{``        ``console.log(``"No"``);``    ``}`

Output

```Yes
```

Complexity Analysis:

• Time Complexity: O(N) where N is the size of linked lists
• Auxiliary Space: O(1) because using constant space