Linked lists are a fundamental data structure in computer science. They help you organize data efficiently, especially when you need to add or remove elements frequently. This guide covers 40 linked list interview questions, split between freshers and experienced candidates, to help you master this topic.
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Explore NowWhat is a linked list?
A linked list is a linear data structure where elements are stored in nodes. Each node contains data and a pointer (or reference) to the next node in the sequence. The last node’s pointer is usually NULL.
How do linked lists differ from arrays?
Arrays store elements in contiguous memory locations, allowing direct access by index (O(1) time). Linked lists store elements non-contiguously, connected by pointers. Accessing an element requires traversing from the beginning (O(n) time). Insertion and deletion are generally faster in linked lists (O(1) at ends, O(n) in middle) than in arrays (O(n)).
What is a node in a linked list?
A node is the basic building block of a linked list. It consists of two parts: the data stored at that node and a pointer (or reference) to the next node in the sequence.
What is the head and tail of a linked list?
The head is the first node in a linked list. It acts as the entry point to the list. The tail is the last node in the linked list. Its next pointer typically points to NULL.
Name different types of linked lists.
The main types are:
- Singly Linked List: Each node points only to the next node.
- Doubly Linked List: Each node has pointers to both the next and the previous nodes.
- Circular Linked List: The last node points back to the first node, forming a circle.
What is a singly linked list?
A singly linked list is a collection of nodes where each node contains data and a single pointer that points to the next node in the sequence. You can only traverse it in one direction, from head to tail.
What is a doubly linked list?
A doubly linked list is a collection of nodes where each node contains data, a pointer to the next node, and a pointer to the previous node. This allows traversal in both forward and backward directions.
What is a circular linked list?
A circular linked list is a linked list where the last node’s pointer points back to the first node. This creates a loop, meaning there is no NULL end.
What are the advantages of using a linked list?
Advantages include:
- Dynamic size: They can grow or shrink as needed.
- Efficient insertions and deletions: Adding or removing elements at certain positions is fast.
- Memory efficiency: Memory is allocated as needed.
What are the disadvantages of using a linked list?
Disadvantages include:
- No random access: You must traverse from the beginning to access an element.
- More memory overhead: Each node requires extra memory for pointers.
- Not cache-friendly: Nodes are not contiguous in memory, which can lead to slower access due to cache misses.
How do you create a simple linked list node?
struct Node {
int data;
Node* next;
Node(int val) : data(val), next(nullptr) {}
};How do you traverse a singly linked list?
You traverse a linked list by starting from the head and moving to the next node using its pointer until you reach NULL.
#include
void traverseList(Node* head) {
Node* current = head;
while (current != nullptr) {
std::cout << current->data << std::endl;
current = current->next;
}
} How do you insert a node at the beginning of a singly linked list?
Create a new node, set its next pointer to the current head, and then update the head to point to the new node.
Node* insertAtBeginning(Node* head, int data) {
Node* newNode = new Node(data);
newNode->next = head;
return newNode; // New head
}How do you insert a node at the end of a singly linked list?
Traverse to the last node. Set the last node’s next pointer to the new node. Handle the case of an empty list.
Node* insertAtEnd(Node* head, int data) {
Node* newNode = new Node(data);
if (head == nullptr) {
return newNode;
}
Node* current = head;
while (current->next != nullptr) {
current = current->next;
}
current->next = newNode;
return head;
}How do you delete the first node of a singly linked list?
Simply update the head to point to the second node. If the list is empty or has one node, handle those cases.
Node* deleteFirstNode(Node* head) {
if (head == nullptr) {
return nullptr;
}
Node* temp = head;
head = head->next;
delete temp;
return head; // New head
}How do you delete the last node of a singly linked list?
Traverse to the second-to-last node. Set its next pointer to NULL. Handle empty and single-node lists.
Node* deleteLastNode(Node* head) {
if (head == nullptr || head->next == nullptr) {
delete head;
return nullptr;
}
Node* current = head;
while (current->next->next != nullptr) {
current = current->next;
}
delete current->next;
current->next = nullptr;
return head;
}How do you search for an element in a singly linked list?
Traverse the list from the head. Compare each node’s data with the target value. Return True if found, False otherwise.
bool searchElement(Node* head, int target) {
Node* current = head;
while (current != nullptr) {
if (current->data == target) {
return true;
}
current = current->next;
}
return false;
}How do you find the length of a singly linked list?
Traverse the list from the head, incrementing a counter for each node.
int getLength(Node* head) {
int count = 0;
Node* current = head;
while (current != nullptr) {
count++;
current = current->next;
}
return count;
}How do you reverse a singly linked list?
Iterate through the list, changing the next pointer of each node to point to its previous node. You need three pointers: prev, current, and next_node.
Node* reverseList(Node* head) {
Node* prev = nullptr;
Node* current = head;
while (current != nullptr) {
Node* nextNode = current->next;
current->next = prev;
prev = current;
current = nextNode;
}
return prev; // New head
}How do you find the middle element of a singly linked list?
Use two pointers, slow and fast. slow moves one step at a time, fast moves two steps. When fast reaches the end, slow will be at the middle.
int findMiddle(Node* head) {
Node* slow = head;
Node* fast = head;
while (fast != nullptr && fast->next != nullptr) {
slow = slow->next;
fast = fast->next->next;
}
return slow->data;
}How do you detect a cycle in a singly linked list?
Use Floyd’s Cycle-Finding Algorithm (Tortoise and Hare). If slow and fast pointers meet at any point, a cycle exists.
bool hasCycle(Node* head) {
Node* slow = head;
Node* fast = head;
while (fast != nullptr && fast->next != nullptr) {
slow = slow->next;
fast = fast->next->next;
if (slow == fast) {
return true;
}
}
return false;
}How do you remove duplicates from a sorted singly linked list?
Traverse the list. If the current node’s data is the same as the next node’s data, skip the next node by linking the current node to the one after the duplicate.
Node* removeDuplicatesSorted(Node* head) {
Node* current = head;
while (current != nullptr && current->next != nullptr) {
if (current->data == current->next->data) {
Node* temp = current->next;
current->next = current->next->next;
delete temp;
} else {
current = current->next;
}
}
return head;
}How do you merge two sorted singly linked lists?
Create a dummy node. Compare the data of the current nodes from both lists and append the smaller one to the merged list. Advance the pointer of the list from which the node was taken.
Node* mergeTwoSortedLists(Node* l1, Node* l2) {
Node dummy(0);
Node* tail = &dummy;
while (l1 != nullptr && l2 != nullptr) {
if (l1->data < l2->data) {
tail->next = l1;
l1 = l1->next;
} else {
tail->next = l2;
l2 = l2->next;
}
tail = tail->next;
}
if (l1 != nullptr) {
tail->next = l1;
} else if (l2 != nullptr) {
tail->next = l2;
}
return dummy.next;
}How do you remove the Nth node from the end of a singly linked list?
Use two pointers, first and second. Move first N steps ahead. Then move both pointers one step at a time until first reaches the end. second will then be at the node before the Nth node from the end.
Node* removeNthFromEnd(Node* head, int n) {
Node dummy(0);
dummy.next = head;
Node* first = &dummy;
Node* second = &dummy;
for (int i = 0; i < n + 1; i++) {
first = first->next;
}
while (first != nullptr) {
first = first->next;
second = second->next;
}
Node* temp = second->next;
second->next = second->next->next;
delete temp;
return dummy.next;
}How do you check if a singly linked list is a palindrome?
You can reverse the second half of the list and then compare it with the first half. Alternatively, use a stack to store the first half and compare it with the second half.
#include
bool isPalindrome(Node* head) {
std::stack s;
Node* slow = head;
Node* fast = head;
while (fast != nullptr && fast->next != nullptr) {
s.push(slow->data);
slow = slow->next;
fast = fast->next->next;
}
if (fast != nullptr) { // Odd number of nodes
slow = slow->next;
}
while (slow != nullptr) {
if (s.top() != slow->data) {
return false;
}
s.pop();
slow = slow->next;
}
return true;
} How do you find the intersection point of two singly linked lists?
Find the lengths of both lists. Move the pointer of the longer list ahead by the difference in lengths. Then, traverse both lists simultaneously. The first common node is the intersection point.
Node* getIntersectionNode(Node* headA, Node* headB) {
int lenA = 0;
Node* current = headA;
while (current != nullptr) {
lenA++;
current = current->next;
}
int lenB = 0;
current = headB;
while (current != nullptr) {
lenB++;
current = current->next;
}
Node* ptrA = headA;
Node* ptrB = headB;
if (lenA > lenB) {
for (int i = 0; i < lenA - lenB; i++) {
ptrA = ptrA->next;
}
} else {
for (int i = 0; i < lenB - lenA; i++) {
ptrB = ptrB->next;
}
}
while (ptrA != nullptr && ptrB != nullptr) {
if (ptrA == ptrB) {
return ptrA;
}
ptrA = ptrA->next;
ptrB = ptrB->next;
}
return nullptr;
}How do you swap nodes in pairs in a singly linked list?
Iterate through the list, swapping adjacent nodes. You might need a dummy node to handle the head.
Node* swapPairs(Node* head) {
Node dummy(0);
dummy.next = head;
Node* prev = &dummy;
while (prev->next != nullptr && prev->next->next != nullptr) {
Node* first = prev->next;
Node* second = prev->next->next;
prev->next = second;
first->next = second->next;
second->next = first;
prev = first;
}
return dummy.next;
}How do you rotate a singly linked list by K places?
Find the length of the list. Adjust k to be k % length. Find the (length – k)-th node from the beginning, which will be the new tail. The node after it becomes the new head. Connect the original tail to the original head.
Node* rotateRight(Node* head, int k) {
if (head == nullptr || head->next == nullptr || k == 0) {
return head;
}
int length = 1;
Node* tail = head;
while (tail->next != nullptr) {
tail = tail->next;
length++;
}
k %= length;
if (k == 0) {
return head;
}
int newTailPos = length - k;
Node* newTail = head;
for (int i = 1; i < newTailPos; i++) {
newTail = newTail->next;
}
Node* newHead = newTail->next;
newTail->next = nullptr;
tail->next = head;
return newHead;
}How do you flatten a multi-level doubly linked list?
This typically involves a list where nodes have both next and child pointers. You flatten it by treating child lists as continuations of the main list, adjusting prev and next pointers accordingly. This is a common recursive problem.
struct Node {
int data;
Node* prev;
Node* next;
Node* child;
Node(int val) : data(val), prev(nullptr), next(nullptr), child(nullptr) {}
};
Node* flatten(Node* head) {
Node* curr = head;
while (curr != nullptr) {
if (curr->child != nullptr) {
Node* nextNode = curr->next;
Node* childTail = curr->child;
while (childTail->next != nullptr) {
childTail = childTail->next;
}
curr->next = curr->child;
curr->child->prev = curr;
curr->child = nullptr;
childTail->next = nextNode;
if (nextNode != nullptr) {
nextNode->prev = childTail;
}
}
curr = curr->next;
}
return head;
}How do you reverse a doubly linked list?
Iterate through the list. For each node, swap its prev and next pointers. Update the head to be the last node from the original list.
struct DoublyNode {
int data;
DoublyNode* prev;
DoublyNode* next;
DoublyNode(int val) : data(val), prev(nullptr), next(nullptr) {}
};
DoublyNode* reverseDoublyList(DoublyNode* head) {
DoublyNode* current = head;
DoublyNode* prev = nullptr;
while (current != nullptr) {
DoublyNode* tempNext = current->next;
current->next = prev;
current->prev = tempNext;
prev = current;
current = tempNext;
}
return prev; // New head
}How do you find the starting node of a cycle in a singly linked list?
First, detect the cycle using Floyd’s Cycle-Finding Algorithm (Tortoise and Hare). This algorithm uses two pointers, slow and fast. slow moves one step at a time, while fast moves two steps. If there’s a cycle, they will eventually meet.
Once slow and fast meet, a cycle is confirmed. To find the starting node of the cycle, move slow back to the head of the list. Keep fast at the meeting point. Now, move both slow and fast one step at a time. The point where they meet again will be the beginning of the cycle.
struct Node {
int data;
Node* next;
Node(int val) : data(val), next(nullptr) {}
};
Node* detectCycleStart(Node* head) {
if (head == nullptr || head->next == nullptr) {
return nullptr; // No cycle
}
Node* slow = head;
Node* fast = head;
// Step 1: Detect if a cycle exists
while (fast != nullptr && fast->next != nullptr) {
slow = slow->next;
fast = fast->next->next;
if (slow == fast) {
break; // Cycle detected
}
}
if (fast == nullptr || fast->next == nullptr) {
return nullptr; // No cycle
}
// Step 2: Find the starting node
slow = head;
while (slow != fast) {
slow = slow->next;
fast = fast->next;
}
return slow; // Starting node of the cycle
}How do you sort a singly linked list (e.g., using Merge Sort)?
Sorting a linked list efficiently often involves Merge Sort. Merge Sort is a divide-and-conquer algorithm. You recursively divide the linked list into two halves until you have single-node lists (which are inherently sorted). Then, you merge these sorted halves back together.
The process involves three main steps:
- Find the Middle: Use the fast and slow pointer approach to find the middle of the linked list.
- Split the List: Break the list into two halves at the middle.
- Merge Sorted Halves: Recursively sort each half and then merge the two sorted sub-lists into a single sorted list.
struct Node {
int data;
Node* next;
Node(int val) : data(val), next(nullptr) {}
};
Node* mergeTwoSortedLists(Node* l1, Node* l2) {
Node dummy(0);
Node* tail = &dummy;
while (l1 != nullptr && l2 != nullptr) {
if (l1->data < l2->data) {
tail->next = l1;
l1 = l1->next;
} else {
tail->next = l2;
l2 = l2->next;
}
tail = tail->next;
}
if (l1 != nullptr) {
tail->next = l1;
} else if (l2 != nullptr) {
tail->next = l2;
}
return dummy.next;
}
Node* sortList(Node* head) {
if (head == nullptr || head->next == nullptr) {
return head; // Already sorted
}
Node* slow = head;
Node* fast = head->next;
while (fast != nullptr && fast->next != nullptr) {
slow = slow->next;
fast = fast->next->next;
}
Node* mid = slow->next;
slow->next = nullptr;
Node* leftSorted = sortList(head);
Node* rightSorted = sortList(mid);
return mergeTwoSortedLists(leftSorted, rightSorted);
}How do you reverse a linked list in groups of K?
This problem requires you to reverse segments of the linked list, each containing K nodes. You connect the reversed segments to form the final list. This can be done iteratively or recursively.
The core idea is:
- Check if there are at least K nodes remaining in the current segment. If not, return the current head as there’s nothing to reverse.
- If K nodes are present, recursively call the function for the next K group (starting from the K+1-th node). This will give you the head of the already reversed subsequent part.
- Reverse the current K nodes and link the last node of this reversed segment to the head returned from the recursive call.
struct Node {
int data;
Node* next;
Node(int val) : data(val), next(nullptr) {}
};
Node* reverseKGroup(Node* head, int k) {
Node* current = head;
int count = 0;
while (current != nullptr && count < k) {
current = current->next;
count++;
}
if (count == k) {
Node* reversedRest = reverseKGroup(current, k);
Node* prev = nullptr;
Node* curr = head;
for (int i = 0; i < k; i++) {
Node* nextNode = curr->next;
curr->next = prev;
prev = curr;
curr = nextNode;
}
head->next = reversedRest;
return prev;
}
return head;
}How do you design and implement an LRU (Least Recently Used) Cache using a doubly linked list and a hash map?
An LRU cache evicts the least recently used item when the cache reaches its capacity. To implement this efficiently:
- Doubly Linked List: Stores the actual key-value pairs. The most recently used items are at the head of the list, and the least recently used items are at the tail. This allows O(1) time for adding/removing items from both ends and for moving items to the front.
- Hash Map (Dictionary/Hash Table): Maps the keys to their corresponding nodes in the doubly linked list. This allows O(1) lookup of a key.
Operations:
- get(key): If the key exists, retrieve the value. Move the corresponding node to the head of the doubly linked list to mark it as most recently used.
- put(key, value): If the key already exists, update its value and move its node to the head. If the key is new, create a new node. If the cache is full, remove the tail node (least recently used) from the doubly linked list and also remove its entry from the hash map. Add the new node to the head of the doubly linked list. Add the new key-node mapping to the hash map.
#include
struct Node {
int key;
int value;
Node* prev;
Node* next;
Node(int k, int v) : key(k), value(v), prev(nullptr), next(nullptr) {}
};
class LRUCache {
private:
int capacity;
std::unordered_map cache;
Node* head;
Node* tail;
void addNode(Node* node) {
node->prev = head;
node->next = head->next;
head->next->prev = node;
head->next = node;
}
void removeNode(Node* node) {
node->prev->next = node->next;
node->next->prev = node->prev;
}
void moveToFront(Node* node) {
removeNode(node);
addNode(node);
}
public:
LRUCache(int cap) : capacity(cap) {
head = new Node(0, 0);
tail = new Node(0, 0);
head->next = tail;
tail->prev = head;
}
int get(int key) {
if (cache.find(key) != cache.end()) {
Node* node = cache[key];
moveToFront(node);
return node->value;
}
return -1;
}
void put(int key, int value) {
if (cache.find(key) != cache.end()) {
Node* node = cache[key];
node->value = value;
moveToFront(node);
} else {
Node* newNode = new Node(key, value);
cache[key] = newNode;
addNode(newNode);
if (cache.size() > capacity) {
Node* lru = tail->prev;
removeNode(lru);
cache.erase(lru->key);
delete lru;
}
}
}
}; How do you deep copy a linked list with next and random pointers?
A deep copy creates a completely new, independent copy of the linked list, including the random pointers. Simply copying node by node won’t work because random pointers might point to nodes that haven’t been copied yet.
A common approach involves three steps:
- Create and Interleave Nodes: Iterate through the original list. For each original node A, create a new copied node A’ and insert A’ right after A.
- Assign Random Pointers: Iterate through the interleaved list. For each original node A, its random pointer (if not None) points to some original node X. The copied node A”s random pointer should then point to X’ (which is X.next in the interleaved list).
- Separate Lists: Separate the original and copied lists by adjusting their next pointers.
struct Node {
int val;
Node* next;
Node* random;
Node(int x) : val(x), next(nullptr), random(nullptr) {}
};
Node* copyRandomList(Node* head) {
if (head == nullptr) {
return nullptr;
}
// Step 1: Create interleaved list
Node* current = head;
while (current != nullptr) {
Node* newNode = new Node(current->val);
newNode->next = current->next;
current->next = newNode;
current = newNode->next;
}
// Step 2: Assign random pointers
current = head;
while (current != nullptr) {
if (current->random != nullptr) {
current->next->random = current->random->next;
}
current = current->next->next;
}
// Step 3: Separate lists
Node* copiedHead = head->next;
Node* currCopied = copiedHead;
Node* currOriginal = head;
while (currOriginal != nullptr) {
currOriginal->next = currCopied->next;
currOriginal = currOriginal->next;
if (currOriginal != nullptr) {
currCopied->next = currOriginal->next;
currCopied = currCopied->next;
} else {
currCopied->next = nullptr;
}
}
return copiedHead;
}How do you flatten a sorted linked list where each node can have a down pointer, and all sub-lists are also sorted?
This problem is similar to merging multiple sorted linked lists. Each node has a next pointer (to the next node in the main list) and a down pointer (to the head of a sorted sub-list). The goal is to flatten it into a single sorted linked list.
The most common and efficient approach is to recursively merge sub-lists.
- Base Case: If the list is empty or has no next node, it’s already flattened and sorted.
- Recursive Step: Flatten the rest of the main list (head.next). This will give you a single sorted linked list for the right part.
- Merge: Merge the current head’s down list (if it exists) with the already flattened and sorted right part.
struct Node {
int data;
Node* next;
Node* down;
Node(int val) : data(val), next(nullptr), down(nullptr) {}
};
Node* mergeTwoSortedDownLists(Node* list1, Node* list2) {
Node dummy(0);
Node* current = &dummy;
while (list1 != nullptr && list2 != nullptr) {
if (list1->data < list2->data) {
current->down = list1;
list1 = list1->down;
} else {
current->down = list2;
list2 = list2->down;
}
current = current->down;
}
if (list1 != nullptr) {
current->down = list1;
}
if (list2 != nullptr) {
current->down = list2;
}
return dummy.down;
}
Node* flattenMultiLevelSortedList(Node* head) {
if (head == nullptr || head->next == nullptr) {
return head;
}
Node* flattenedNext = flattenMultiLevelSortedList(head->next);
return mergeTwoSortedDownLists(head, flattenedNext);
}How do you reorder a linked list so that it alternates between the first half and the reversed second half?
This problem involves three main steps:
- Find the Middle: Use the fast and slow pointer approach to locate the middle of the linked list.
- Reverse the Second Half: Separate the list into two halves and reverse the second half.
- Merge Alternately: Merge the first half and the reversed second half by taking one node from each alternately.
struct Node {
int data;
Node* next;
Node(int val) : data(val), next(nullptr) {}
};
Node* reverseList(Node* head) {
Node* prev = nullptr;
Node* current = head;
while (current != nullptr) {
Node* nextNode = current->next;
current->next = prev;
prev = current;
current = nextNode;
}
return prev;
}
void reorderList(Node* head) {
if (head == nullptr || head->next == nullptr) {
return;
}
Node* slow = head;
Node* fast = head;
while (fast != nullptr && fast->next != nullptr) {
slow = slow->next;
fast = fast->next->next;
}
Node* head2 = slow->next;
slow->next = nullptr;
head2 = reverseList(head2);
Node* p1 = head;
Node* p2 = head2;
while (p1 != nullptr && p2 != nullptr) {
Node* temp1 = p1->next;
Node* temp2 = p2->next;
p1->next = p2;
p2->next = temp1;
p1 = temp1;
p2 = temp2;
}
}Explain the use cases where linked lists are preferred over arrays.
Linked lists offer specific advantages that make them preferred over arrays in certain scenarios:
- Dynamic Size and Flexibility: Linked lists can grow or shrink in size as needed during runtime. You don’t need to pre-allocate a fixed amount of memory.
- Efficient Insertions and Deletions at Arbitrary Positions: Inserting or deleting at the beginning of a linked list is an O(1) operation. For a doubly linked list or a singly linked list with a tail pointer, insertion/deletion at the end is also O(1).
- Implementing Abstract Data Types: Linked lists serve as efficient underlying data structures for implementing stacks (push/pop O(1)) and queues (enqueue/dequeue O(1) with tail pointer).
- Memory Management: Linked lists can make more efficient use of memory when elements are dynamically allocated and deallocated.
- Representing Sparse Data or Polynomials: Linked lists can store only the non-zero terms, saving space.
Describe the time complexity of various linked list operations (insertion, deletion, search, access).
Understanding the time complexity of operations is crucial for choosing the right data structure.
Singly Linked List:
- Insertion at Beginning: O(1) – You only need to create a new node and update the head pointer.
- Insertion at End: O(n) – You must traverse the entire list to find the last node.
- Insertion in Middle: O(n) – You must traverse to the node before the insertion point.
- Deletion at Beginning: O(1) – You only need to update the head pointer.
- Deletion at End: O(n) – You must traverse to the second-to-last node to update its next pointer.
- Deletion in Middle: O(n) – You must traverse to the node before the deletion point.
- Search for an Element: O(n) – In the worst case, you might need to traverse the entire list.
- Access by Index: O(n) – You must traverse from the head to reach the Nth element.
Doubly Linked List:
- Insertion at Beginning: O(1) – Update head and new node’s prev/next pointers.
- Insertion at End: O(1) – If you maintain a tail pointer, you can directly add to the end.
- Insertion in Middle: O(n) – Still requires traversal to the insertion point.
- Deletion at Beginning: O(1) – Update head and new head’s prev pointer.
- Deletion at End: O(1) – If you maintain a tail pointer, you can directly delete from the end.
- Deletion in Middle: O(n) – Still requires traversal to the node to be deleted.
- Search for an Element: O(n) – Same as singly linked list.
- Access by Index: O(n) – Same as singly linked list.
Discuss how linked lists are used in real-world applications.
Linked lists are fundamental to many practical applications:
Operating Systems:
- Memory Management: Operating systems use linked lists to manage free memory blocks.
- Process Scheduling: Circular linked lists are often used to implement round-robin scheduling algorithms.
Web Browsers:
- History (Back/Forward Buttons): Doubly linked lists are perfect for managing browser history, allowing efficient navigation backward and forward.
Music/Video Players:
- Playlists: Playlists can be implemented using linked lists, allowing easy addition, removal, or reordering of songs.
Image Viewers:
- Image Navigation: Linked lists are used to navigate between images in a folder.
Implementing Abstract Data Structures:
- Stacks and Queues: Linked lists are commonly used to build stacks (LIFO) and queues (FIFO) efficiently.
- Representing Polynomials: Linked lists can represent polynomials by storing only non-zero terms.
- Undo/Redo Functionality: In text editors or graphic design software, a doubly linked list can store the history of actions.