Lexmark International Software Developer Intern Interview Questions and Answers

The aptitude round had the topics of train and distance and time , OOPS , Core Java , JavaScript questions , Operating System questions and DBMS questions.

Q1. Had some data structures and algorithm questions and graph , dynamic programming questions
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Q2. Questions in the datastructures
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90 min MCQ +coding test on hackerrank.

Q1. Sort colors DSA question
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Q2. Detect cycle in a linked list
Ans. 
Use Floyd's Tortoise and Hare algorithm to detect cycle in a linked list.
• Initialize two pointers, slow and fast, at the head of the linked list.

• Move slow pointer by one step and fast pointer by two steps.

• If they meet at some point, there is a cycle in the linked list.
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Q1. Was asked about why you want to join the company,my experience etc.
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Round duration - 90 minutes
Round difficulty - Medium

Q1. 
Cycle Detection in a Singly Linked List
Determine if a given singly linked list of integers forms a cycle or not.
A cycle in a linked list occurs when a node's next points back to a previous node in the ...read more
Ans. 
The task is to determine if a given singly linked list forms a cycle or not.
• A cycle occurs when a node's next points back to a previous node in the list.

• To solve this problem, we can use the Floyd's Cycle-Finding Algorithm.

• The algorithm uses two pointers, one moving at a normal pace and the other moving twice as fast.

• If there is a cycle, the fast pointer will eventually catch up to the slow pointer.

• If the fast pointer ...read more
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Round duration - 60 minutes
Round difficulty - Medium

Q1. 
Minimum Depth of a Binary Tree
Determine the minimum depth of an integer-based binary tree. The minimum depth is defined as the number of nodes present along the shortest path from the root node down to t...read more
Ans. 
The minimum depth of a binary tree is the number of nodes along the shortest path from the root node down to the nearest leaf node.
• The minimum depth can be found by performing a breadth-first search (BFS) traversal of the binary tree

• During the BFS traversal, keep track of the current level and increment the depth by 1 for each level

• Stop the BFS traversal when a leaf node is encountered and return the depth as the minim
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Q2. 
Intersection of Two Arrays Problem Statement
Given two arrays A and B with sizes N and M respectively, both sorted in non-decreasing order, determine their intersection.
The intersection of two arrays in...read more
Ans. 
The problem is to find the intersection of two sorted arrays.
• Use two pointers to iterate through the arrays.

• Compare the elements at the current pointers and move the pointers accordingly.

• If the elements are equal, add it to the intersection array and move both pointers.

• If the element in the first array is smaller, move the first pointer.

• If the element in the second array is smaller, move the second pointer.

• Repeat until...read more
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Q3. What is the difference between a mutex and a semaphore?
Ans. 
A mutex is a binary semaphore used for mutual exclusion, while a semaphore is a generalized synchronization primitive.
• Mutex is used to protect a critical section of code, allowing only one thread to access it at a time.

• Semaphore is used to control access to a shared resource, allowing multiple threads to access it simultaneously.

• Mutex has ownership, meaning the thread that locks it must unlock it.

• Semaphore does not hav...read more
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Round duration - 60 minutes
Round difficulty - Easy

Round duration - 25 minutes
Round difficulty - Easy

Q1. What are oops concepts in java, explain real time scenario
Ans. 
OOPs concepts in Java include inheritance, polymorphism, encapsulation, and abstraction.
• Inheritance allows a subclass to inherit properties and methods from a superclass.

• Polymorphism allows objects to take on multiple forms and behave differently based on their context.

• Encapsulation hides the implementation details of an object and only exposes necessary information.

• Abstraction allows for the creation of abstract class...read more
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Q2. Uses of interface, inheritance
Ans. 
Interfaces define contracts for behavior, while inheritance allows for code reuse and polymorphism.
• Interfaces allow for loose coupling and abstraction, enabling multiple implementations of the same behavior.

• Inheritance allows for code reuse and extension of existing classes, reducing code duplication.

• Polymorphism allows objects of different classes to be treated as if they were of the same class, simplifying code and i
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Q3. SQL query for join of tables
Ans. 
SQL query for joining tables
• Use JOIN keyword to combine two or more tables based on a related column

• Specify the columns to be selected using SELECT keyword

• Use ON keyword to specify the condition for joining the tables

• Different types of joins include INNER JOIN, LEFT JOIN, RIGHT JOIN, and FULL OUTER JOIN
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Q4. Java concepts used in your project
Ans. 
Used Java concepts such as inheritance, polymorphism, and exception handling in my project.
• Implemented inheritance to create a base class and derived classes with specific functionalities.

• Utilized polymorphism to allow objects of different classes to be treated as if they were of the same class.

• Implemented exception handling to handle errors and prevent program crashes.

• Used interfaces to define a set of methods that a ...read more
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Q5. Overloading vs overriding, practical uses
Ans. 
Overloading is having multiple methods with the same name but different parameters. Overriding is having a method in a subclass with the same name and parameters as a method in the superclass.
• Overloading is used to provide different ways to call a method with different parameters

• Overriding is used to provide a specific implementation of a method in a subclass

• Overloading is resolved at compile-time while overriding is r...read more
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Normal apptitude queations were asked in this round

Q1. Questions on javascript Basic of HTML, CSS
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Q1. Lamda experience
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Q1. Question are about interview
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Q2. Answers about interview
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