Linked List Interview Questions and Answers For 2025

First of all if you want to check if the links are identical or not the data in the list must be same and also the position of the nodes must also be same for the linked list. To do that a simple idea you get is traverse both the lists in parallel and then while traversing keep comparing each element encountered in the way for both the lists.

A must-know for anyone heading into a linked list interview, this question is frequently asked in linked list interview questions for experienced. A linked list is a dynamic data structure, so at a point in time, you can change its size by adding or removing elements from it. Also, every element in a linked list is stored at random positions, and each element has a node and a reference to be stored in the memory. So, linked lists have a requirement for dynamic memory, and therefore for a linked list, memory is allocated from the heap section, which is like a free memory bank, and whenever a node is required to be created, the program you write asks from the memory manager to provide a free space after this there is a garbage collector which is used to collected deleted nodes, and hence this data structure is also called a dynamic data structure. There are memory spaces where the pointers are stored and are called AVAIL, which is a list of memory locations.

For sorting a linked list, you should use a merge sort, as it is the optimal solution for sorting a linked list. Return the head of the linked list's size is 1. Otherwise, The Tortoise and the Hare Method for Finding Mid Keep the following of mid in the right sub-linked list, or head2. Now Nullify the following midway. The new head of the left and right linked lists should be stored after calling mergeSort() recursively on both the left and right sub-linked lists. Call merge() with the updated heads of the left and right sub-linked lists as parameters, and then save the resultant head. Give the merged linked list's last head. Take a pointer, say merged, and place a fake node in it along with the merged list. Take a temp pointer and give it the merge property. Store head1 in the next available temp location and move it to the next head1 if the data in head1 is less than the data in head2.Alternatively, keep head2 at the next transient location and move it to the next head2.Transfer temp to the following temp.Till head1 and head2 are not equal to null, repeat the previous steps as necessary. The remaining nodes from the first or second linked lists should now be added to the combined linked list. Return the head and the next merged, which will disregard the dummy.

First, assume the node to be deleted as del, and if the node to be deleted is the head node itself, then modify the head pointer as the next current head and also set prev of next to del, again if next to del exists. Last set next of previous to del if previous to del exists.

Suppose there are N nodes in the list, and hence there will be N/k elements in every split part of the list and also the first N%k part will have extra remainder elements and to count the N we can simply use a loop. After this, for every part, you can calculate the number of elements each part will have by using the width of the part. 

(N/k + 1) And the left k-(N%k) groups will be of the size N/k .

A linked list is normally accessible only from its head node. From there, you can navigate from node to node until you reach the node you're looking for. So access is O(n). Similarly, searching for a specific value in a linked list requires iterating over all elements until that value is found. So the search is O(n). Inserting into a linked list requires the previous node (the node before the insertion point) to point to the inserted node, and the newly inserted node to point to the next node. So the insert is O(1). To remove from a linked list, the previous node (the node before the removed node) must be respecified as the next node (the node after the removed node). So cancellation is O(1).  

It is O(1). Note that there is no need to insert at the end of the list. Inserting at the beginning of a (unidirectionally linked) list is both O(1).Stack contains 1,2,3: [1]->[2]->[3] Push 5: [5]->[1]->[2]->[3] Pop: [1]->[2]->[3] // returning 5.

To convert a singly-linked list to a circularly-linked list, set the next pointer of the leaf node to the head pointer. Make a copy of the head pointer. Let it be temp. Loops through the linked list to the terminal node (last node) using a temporary pointer. Sets the next pointer of the leaf node to the head node. temp->next = head.

def convert(head): 
  # declare a node variable 
  # start and assign head 
  # node into start node. 
  start = head 
   
  # check that 
  while head.next 
  # not equal to null then head 
  # points to next node. 
  while(head.next is not None): 
   head = head.next 
   
  # 
  if head.next points to null 
  # then start assign to the 
  # head.next node. 
  head.next = start 
  return start 

Inserting a node in the middle of a linked list assumes you are already at the address where the node should be inserted. Indexing is considered a separate operation. So the insert itself is O(1), but getting to that intermediate node is O(n). So when appending to a linked list, no traversal is needed as long as the reference is known.

You can add two numbers represented by a LinkedList the same way you add two numbers manually. Iterate over the linked list, appending the appropriate elements, preserving the carry as you would when adding numbers by hand, and adding the carry from the previous addition to the running total. One of the most annoying parts of this problem is the number that is sent. If each pair of nodes sums up to a number less than 10, don't worry about "carrying" the number to the next node. However, adding a number such as 4 or 6 causes a carry. Even if one list is longer than the other, the nodes of the long list should be added to the solution, so the test should continue as long as the node is not null. This means that the while loop should execute as long as list 1 is not null or list 2 is not null.

Use a lockstep solution. The key to this algorithm is to first separate the two pointers p1 and p2 by n-1 nodes so that p2 points to the (n-1)th node from the beginning of the list, and then shift until p2 is at the end. That's it. You have reached a node in the list. As soon as p2 reaches the end of the list, p1 points to the Nth node from the end of the list. This lockstep approach generally improves cache utilization because the node hit by the front pointer may still be in the cache when the back pointer hits the node. For language implementations that use trace garbage collection, this approach also avoids the initial list (that is, the entire list) from being unnecessarily left active during operation. 

Another solution is to use a ring buffer. Keep a circular buffer of size x and add nodes as you iterate over the list. When the end of the list is reached, the x-th from the end is equal to the next entry in the ring buffer. The lockstep approach moves all array elements at each step, so the execution time is O(n x 2). On the other hand, the circular buffer approach uses circular arrays and runs in O(n). The circular buffer solution requires an extra x in memory. 

A simple solution is to clone the linked list, reverse it and check if both linked lists are the same. This method requires repeating the linked list three times and requires additional space to store the duplicates. A better solution is to use recursion. The idea is to reach the end of the linked list by recursion and compare if the last node has the same value as the first node and the two nodes before have the same value as his second node. That's it. The pointer at the head of the linked list is used.

We know that the height of a binary tree is the total number of nodes along the longest path from the root to a leaf node. The idea is to traverse the tree after ordering and calculate the heights of the left and right subtrees. The height of the subtree rooted at any node is one greater than the maximum height of the subtrees to its left and right. Applies this property to all tree nodes recursively from bottom to top (post order method) and returns the maximum height of the subtree rooted at this node. In a normal binary tree, the left and right children of leaf nodes are null pointers. But here the left and right children of the leaf node behave like previous and next pointers in a circularly doubly linked list. For a node to be a leaf node, ensure that its left, right and left sides point to itself.

First traverse a linked list into an array. Thereafter construct a monotonically decreasing stack from right to left. If the topmost value of each element is greater than the current value, update the topmost value to the next greater value than the current value and also push it on the stack. Otherwise, open the stack until the topmost value is greater than the current value or the stack is empty. For an empty stack, there is no next max value for the current value, simply set 0 as the next max value as needed for your problem.

Insertion sort is simply done like you sort a deck of playing cards. You take a card and look for a place to insert it so that it’s ordered. To implement this in a linked list you should first create an empty linked list then if there is no next element then make this element as head and return, else Goto the next node and compare the value with the previous one and see if it this is smaller than the head node then make this node as the head node and insert this at the start of the empty list. Now select the next node and traverse through the new empty list looking for a value greater than this value and as you find such a node you insert the node prior to this node and keep traversing in such a manner and at the end return this new list.

Traversing is the most basic operation in any data structure, as you use it in performing various other operations. So in the case of a linked list, you need a pointer for traversing. First of all, you start with the head and point the pointer at the head. Then you start a loop with a condition to continue looping until the pointer you created is pointing Null, and keep changing the value of the pointer as next, and once the pointer points at Null, the loop will terminate, and the traversal is completed.

Yes, Java.util is a package in Java that is used to import data structure classes and many other components like date & time, enumeration, exception handling components etc. It has pre-written code blocks for complete data structures, which you can simply import into your code to avoid writing those whole sets of code lines repeatedly. It also contains a certain set of operations for these packages for data structures; you have a stream that is used to traverse and many more. This was simply created to make the code look cleaner, manageable and easy to understand.

For this, you can use iterative reversal of the linked list. You simply traverse through the linked list and reverse every prefix of the linked list starting from the left, and in this way, you will find the lengthiest common linked list, which must start from the reversed prefix and the list existing just after the reversed prefix.

You can simply get the element at any given index from a linked list by simply traversing through the linked list and checking for the given index. For this, you will have to first initiate a counter with value 0 and loop through the linked list if the value of the counter equals the value of the index of the required node, then terminate the loop and return the current node; else, continue the loop and increase the counter by 1 and also change the pointer to the next node and keep looping until the values of the counters become equal to the required index.

To add two polynomials using a linked list, you need to have a linked list whose nodes store three values viz; the coefficient, the power of the variable and the address to the next node.

Though a circular linked list is different from a singly linked list inserting a node in it is similar to singly linked lists only. First of all, you create a node to store the value of the node to be inserted, then traverse through the list until you find the node, after which this new node has to be inserted and then point the pointer of the new node to the next to the current node and the current nodes next to the new node in this way you can simply insert a new node at any point of the circular linked list.

Given a linked list where each node has a child pointer in addition to the next pointer, which may or may not point to a different list. As indicated in the image below, these child lists may have one or more children of their own, and so on, to create a multilayer data structure. The head of the list's initial level is supplied to you. Flatten the list to make a single-level linked list with all of the nodes. You must flatten the list such that all of the first-level nodes appear first, followed by those at the second level, and so forth.

To group all the odd and the even nodes together, first, you need to find out where will the odd node start and where it ends, and the same goes for the even nodes; after that, connect the end of the odd list to the beginning of the event list. And after that, mark the end of the even linked list as Null. This code has a time complexity of order O(N) as you use a loop to traverse, and the space complexity will be O(1) as no extra space is needed, and only a few pointers are stored.

You can use hashing technique to remove duplicate elements from a linked list. For this, you traverse through the list and check if this element is already existing in the hash table or not; if yes, remove the node else, and push the node to the hash table. To implement this, you can create a set in order to create a track of the visited elements, then traverse through the list and check for each node if it's present in the hashset or not and do accordingly.

• If you compare an array and a linked list of the same size, the memory used by the linked list will be more as each node of the linked list also stores additional data of pointers, so when you want to save memory, you can avoid linked lists. 
• If you want to traverse in a linked list, you don't have indexes, so you have to traverse from the start to the end; hence it makes traversing slower in a linked list. 
• If, in any scenario, you have to traverse in the reverse direction, it's not possible in a singly linked list. 

• Gallery; you have linked images and can go to images by previous and next options.
• Media players; similar to galleries, you can play any media from the list and access them.
• You can use linked lists to schedule tasks like in Robin Scheduling.
• Redo as well as undo operations can be performed
• Dynamic management for the memory of an operating system can be performed.