Samsung Interview Questions, Process, and Tips

Round duration - 180 Minutes
Round difficulty - Medium

It is a coding round that contains only 1 coding problem, and the timing for this test is 10 A.M. to 1 A.M. The test is on full-screen mode. You can choose Java or Cpp as a language to code the particular problem. The problem contains 50 test cases and you have only 10 chances to submit your code. The interviewer was very friendly.

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
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Round duration - 90 Minutes
Round difficulty - Easy

In this round lot of questions from the topics core java, java backend, react, and node js. The timing of this round is 10 A.M. to 11:30 A.M. The interviewer was very friendly.

1 coding questions and aptitude reasoning questions MCQ - 90 minutes duration

2 coding questions - 1 hrs

Q1. Basics of programing & Data structures with coding questions
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Q1. Why they hired you
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Q1. Why they hired you
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Q1. Overall career & experience review
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Q1. About project and skills
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Q2. Languages know and communication skills
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Q3. Salary expectation and about there company
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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. 
Triplets with Given Sum Problem
Given an array or list ARR consisting of N integers, your task is to identify all distinct triplets within the array that sum up to a specified number K.
Explanation:
A t...read more
Ans. 
Approach : 

1) Sort the array in ascending order.

2) Now since we want triplets such that x + y + z = ‘K’, we have x+ y = ‘K’ - z and now we can fix z as arr[i]. So we want to find the sum of two numbers x and y as ‘K’ - arr[i] in the array.

3) We will use two pointers, one will start from i+1, and the other will start from the end of the array.

4) Let the two pointers be ‘FRONT’ and ‘BACK’, where ‘FRONT’ = i + 1 and ...read more
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Q2. 
Minimum Operation Needed to Convert to the Given String
You are given two strings str1 and str2. Determine the minimum number of operations required to transform str1 into str2.
Explanation:
An operatio...read more
Ans. 
Approach : I solved this problem using 2-pointer technique. 

1) Initialize the first pointer to the start of str1 and the second pointer to the start of the str2.

2) Now increase the first pointer till the character pointed by the first pointer in str1 matches the character at the second pointer in str2. And the number of character which did not match should be placed at the end of str1. Therefore increase our answe...read more
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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. 
Approach : If a number N is power of 2 then bitwise & of N and N-1 will be zero. We can say N is a power of 2 or not based on the value of N&(N-1). The expression N&(N-1) will not work when N is 0.
So,handle that case separately.

TC : O(1)
SC : O(1)
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Round duration - 60 Minutes
Round difficulty - Medium

Q1. 
Merge K Sorted Arrays Problem Statement
Given 'K' different arrays that are individually sorted in ascending order, merge all these arrays into a single array that is also sorted in ascending order.
Inpu...read more
Ans. 
Approach 1 (Brute Force) : 

Create an output array and and one by one copy all arrays to it. Finally, sort the output array using. This approach takes O(N Logn N) time where N is count of all elements.

Approach 2 (Using Min-Heap) :

1. Create an output array. 
2. Create a min heap of size k and insert 1st element in all the arrays into the heap 
3. Repeat following steps while priority queue is not empty. ...read more
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Q2. 
Spiral Order Traversal of a Binary Tree
Given a binary tree with N nodes, your task is to output the Spiral Order traversal of the binary tree.
Input:
The input consists of a single line containing elem...read more
Ans. 
Approach 1 (Using Level Order Traversal) : 

Steps : 
1) Initialize an array ‘spiral’ to store the result.
2) Initialize a variable ‘direction’ to keep track of traversing direction with 1(denoting left to right)
3) Find the height of the given tree and store it in a variable ‘height’.
4) Run a loop - level : 1 to height, to traverse through every level.
4.1) Find level order traversal in the current direction.and s...read more
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Q3. What is the difference between a structure and a union, and what are the pros and cons of each?
Ans. 
Answer : 

Structure : Structure is a user-defined data type in C programming language that combines logically related data items of different data types together.
All the structure elements are stored at contiguous memory locations. Structure type variable can store more than one data item of varying data types under one name.

Union : Union is a user-defined data type, just like a structure. Union combines objects of...read more
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Round duration - 60 Minutes
Round difficulty - Medium

This round had 2 questions related to Trees and Graphs and some questions revolving around types of Operating Systems their pros and cons.

Q1. 
Binary Tree Diameter Problem Statement
You are given a Binary Tree, and you need to determine the length of the diameter of the tree.
The diameter of a binary tree is the length of the longest path betwe...read more
Ans. 
Approach 1 :

Steps : 

1) If the ‘root’ node is NULL, return 0. The diameter of an empty tree is 0.
2) Recur for the left subtree and store the diameter of the left subtree in a variable ‘leftDiameter’, i.e. 
‘leftDiameter’ = getDiameter(left child of the root node)
3) Similarly, recur for the right subtree and store the diameter of the right subtree in a variable 
‘rightDiameter’ i.e. ‘rightDiameter’ = getDiam...read more
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Q2. 
Shortest Path in a Binary Matrix Problem Statement
Given a binary matrix of size N * M where each element is either 0 or 1, find the shortest path from a source cell to a destination cell, consisting only...read more
Ans. 
Approach : I solved this problem using BFS.

Breadth-First-Search considers all the paths starting from the source and moves ahead one unit in all those paths at the same time. This makes sure that the first time when the destination is visited, it is the path with the shortest length.

Steps : 

1) Create an empty queue and enqueue source cell and mark it as visited
2) Declare a ‘STEPS’ variable, to keep track of the le...read more
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