Samsung Software Developer Interview Questions, Process, and Tips

Round duration - 120 Minutes
Round difficulty - Medium

This was an online coding round where we had 3 questions to solve under 120 minutes. The questions were of medium to hard difficulty level.

Q1. 
Minimum Time in Wormhole Network
Determine the minimum time required to travel from a starting point to a destination point in a two-dimensional coordinate system, considering both direct movement and the...read more
Ans. 
Find the minimum time to travel from a starting point to a destination point using direct movement and wormholes.
• Calculate the time taken for direct movement from source to destination.

• Consider using each wormhole to see if it reduces the total travel time.

• Choose the path with the minimum total time to reach the destination.
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Q2. 
Substrings Differ by One Problem Statement
Ninja needs help in a battle against the string man. Given two strings, 'S' and 'T', the task is to find the number of substrings in 'S' that differ from some su...read more
Ans. 
The task is to find the number of substrings in 'S' that differ from some substrings of 'T' by exactly one character.
• Iterate through all substrings of 'S' and 'T' and compare them character by character to find the ones that differ by exactly one character.

• Use nested loops to generate all possible substrings of 'S' and 'T'.

• Count the number of substrings that differ by exactly one character and return the total count.
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Q3. 
Power of Two Problem Statement
Determine whether a given integer N is a power of two. Return true if it is, otherwise return false.
Explanation
An integer 'N' is considered a power of two if it can be e...read more
Ans. 
Check if a given integer is a power of two or not.
• Check if the given integer is greater than 0.

• Check if the given integer has only one bit set in its binary representation.

• Return true if the above conditions are met, else return false.
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Round duration - 60 Minutes
Round difficulty - Medium

This round had 2 questions related to DSA where I was first expected to explain my approaches and then discuss the time and space complexities of my solution. After that , I was asked some core concepts related to OS.

Q1. 
Reverse Linked List Problem Statement
Given a singly linked list of integers, return the head of the reversed linked list.
Example:
Initial linked list: 1 -> 2 -> 3 -> 4 -> NULL
Reversed link...read more
Ans. 
Reverse a singly linked list of integers and return the head of the reversed linked list.
• Iterate through the linked list and reverse the pointers to point to the previous node instead of the next node.

• Use three pointers - prev, current, and next to reverse the linked list in O(N) time and O(1) space complexity.

• Update the head of the reversed linked list as the last node encountered during reversal.
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Q2. 
Cycle Detection in Undirected Graph Problem Statement
You are provided with an undirected graph containing 'N' vertices and 'M' edges. The vertices are numbered from 1 to 'N'. Your objective is to determi...read more
Ans. 
Detect cycles in an undirected graph.
• Use Depth First Search (DFS) to detect cycles in the graph.

• Maintain a visited array to keep track of visited vertices.

• If a visited vertex is encountered again during DFS, a cycle exists.

• Check for cycles in each connected component of the graph.

• Example: For input N=3, Edges=[[1, 2], [2, 3], [1, 3]], output is Yes.
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Q3. What are the differences between a mutex and a semaphore?
Ans. 
Mutex is used for exclusive access to a resource by only one thread at a time, while semaphore can allow multiple threads to access a resource simultaneously.
• Mutex is binary semaphore with ownership, used for mutual exclusion.

• Mutex is typically used to protect critical sections of code.

• Semaphore is a signaling mechanism, used for synchronization between multiple threads.

• Semaphore can have a count greater than 1, allowi...read more
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Q4. What is meant by multitasking and multithreading in operating systems?
Ans. 
Multitasking refers to the ability of an operating system to run multiple tasks concurrently, while multithreading involves executing multiple threads within a single process.
• Multitasking allows multiple processes to run simultaneously on a single processor, switching between them quickly to give the illusion of parallel execution.

• Multithreading involves breaking a process into multiple threads that can be executed ind...read more
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Round duration - 60 Minutes
Round difficulty - Medium

This round had 2 Algorithmic questions wherein I was supposed to code both the problems after discussing their
approaches and respective time and space complexities . After that , I was grilled on some OOPS concepts related to C++.

Q1. 
Trapping Rain Water Problem Statement
You are given a long type array/list ARR of size N, representing an elevation map. The value ARR[i] denotes the elevation of the ith bar. Your task is to determine th...read more
Ans. 
Calculate the total amount of rainwater that can be trapped between given elevations in an array.
• Iterate through the array and calculate the maximum height on the left and right of each bar.

• Calculate the amount of water that can be trapped at each bar by taking the minimum of the maximum heights on the left and right.

• Sum up the trapped water at each bar to get the total trapped water for the entire array.
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Q2. 
Find The Repeating And Missing Number Problem Statement
You are provided with an array nums which contains the first N positive integers. In this array, one integer appears twice, and one integer is missi...read more
Ans. 
Given an array of first N positive integers with one number repeating and one missing, find the repeating and missing numbers.
• Iterate through the array and keep track of the sum of elements and sum of squares to find the missing and repeating numbers.

• Use a set to identify the repeating number and calculate the missing number based on the sum of elements.

• Example: For nums = [1, 2, 3, 4, 4, 5], the repeating number is 4
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Q3. What is the Diamond Problem in C++ and how can it be resolved?
Ans. 
Diamond Problem in C++ is a common issue in multiple inheritance where a class inherits from two classes that have a common base class.
• Diamond Problem occurs when a class inherits from two classes that have a common base class, leading to ambiguity in accessing members.

• It can be resolved in C++ using virtual inheritance, where the common base class is inherited virtually to avoid duplicate copies of base class members.

• ...read more
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Q4. What are friend functions in C++?
Ans. 
Friend functions in C++ are functions that are not members of a class but have access to its private and protected members.
• Friend functions are declared inside a class with the 'friend' keyword.

• They can access private and protected members of the class.

• They are not member functions of the class, but have the same access rights as member functions.

• Friend functions are useful for implementing operators that are not membe...read more
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