Given a linked list. We need to find unique elements in the linked list i.e, those elements which are not repeated in the linked list or those elements whose frequency is 1. If No such elements are present in list so Print ” No Unique Elements”.

Examples:

Input : 1 -> 4 -> 4 -> 2 -> 3 -> 5 -> 3 -> 4 -> 5 Output :1 2 Input :4 -> 5 -> 2 -> 5 -> 1 -> 4 -> 1 -> 2 Output :No Unique Elements

** Method 1 (Using Two Loops) ** This is the simple way where two loops are used. Outer loop is used to pick the elements one by one and inner loop compares the picked element with rest of the elements. If Element is not equal to other elements than Print that Element. Time Complexity : O(N * n)

** Method 2 (Sorting) : ** Sort the elements using Merge Sort. O(n Log n). Now Traverse List in linear time and check if current element is not equal to previous element then Print O(N)

Please note that this method doesn’t preserve the original order of elements.

Time Complexity: O(NLogN)

**Method 3 (Hashing)**

We use the concept of Hash table Here, We traverse the link list from head to end. For every newly encountered element, we put it in the hash table after that we again traverse list and Print those elements whose frequency is 1.Time Complexity : O(N)

Below is the Implementation of this

// C++ Program to Find the Unique elements in // linked lists #include <bits/stdc++.h> using namespace std; /* Linked list node */ struct Node { int 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 Find the unique elements in linked lists void uniqueElements(struct Node* head) { // Initialize hash array that store the // frequency of each element of list unordered_map<int, int> hash; for (Node *temp=head; temp!=NULL; temp=temp->next) hash[temp->data]++; int count = 0; for (Node *temp=head; temp!=NULL; temp=temp->next) { // Check whether the frequency of current // element is 1 or not if (hash[temp->data] == 1) { cout << temp->data << " "; count++; } } // If No unique element in list if (count == 0) cout << " No Unique Elements "; } // Driver program to test above int main() { struct Node* head = NULL; // creating linked list push(&head, 5); push(&head, 4); push(&head, 3); push(&head, 5); push(&head, 3); push(&head, 2); push(&head, 4); push(&head, 4); push(&head, 1); uniqueElements(head); return 0; }

**Output:**

1 2

Time Complexity : O(N)

Auxiliary Space : O(N)

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