Dell Maintenance Electrician Interview Questions and Answers

Q1. What is sensors and it's type?? And briefing about maintenance
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Q1. Finally project and previous company detailed.l
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Q1. Self introduction
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Q2. Technical knowledge
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Q3. Critical thinking
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Q4. Pressure work environment
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Q5. Environment health safety
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Q1. How to Any breakdown handaly?
Ans. 
To handle any breakdown, it is important to follow a systematic approach and have a good understanding of the equipment and its components.
• Identify the cause of the breakdown by conducting thorough inspections and diagnostic tests.

• Refer to equipment manuals and documentation to understand the functioning and troubleshooting procedures.

• Use appropriate tools and equipment to repair or replace faulty components.

• Ensure saf...read more
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Q2. Electrical and cooling system
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Q3. Aircondinar lowest cooling ?
Ans. 
The lowest cooling capacity of an air conditioner depends on its BTU rating and the ambient temperature.
• The cooling capacity of an air conditioner is measured in BTUs (British Thermal Units).

• The lowest cooling capacity is determined by the BTU rating of the air conditioner.

• The ambient temperature also affects the lowest cooling capacity.

• Air conditioners with lower BTU ratings may have a lower lowest cooling capacity.

• Fo...read more
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Round duration - 180 Minutes
Round difficulty - Medium

Round was held in the morning at 10 am.

Q1. 
M - Coloring Problem Statement
Given an undirected graph with 'N' nodes in the form of an adjacency matrix and an integer 'M', determine if it is possible to color the vertices of the graph using at most ...read more
Ans. 
The problem involves determining if a given graph can be colored with at most 'M' colors without adjacent vertices sharing the same color.
• Create a function that takes the adjacency matrix, number of nodes 'N', and maximum number of colors 'M' as input.

• Implement a graph coloring algorithm such as backtracking or greedy coloring to check if the graph can be colored with at most 'M' colors.

• Check if adjacent vertices have ...read more
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Round duration - 40 Minutes
Round difficulty - Medium

The round was held in the evening

Q1. 
Trie Data Structure Implementation
Design and implement a Trie (prefix tree) to perform the following operations:
• insert(word): Add a string "word" to the Trie.
• search(word): Verify if the string "w...read more
Ans. 
Implement a Trie data structure to insert, search, and check for prefixes in strings.
• Create a TrieNode class with children and isEndOfWord attributes.

• Implement insert, search, and startsWith methods in the Trie class.

• Use a Trie to efficiently store and search for strings based on prefixes.

• Example: insert 'apple', search 'apple' returns true, startsWith 'app' returns true, search 'app' returns false.
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Round duration - 10 Minutes
Round difficulty - Easy

Was held in the morning around 9 am

Round duration - 60 minutes
Round difficulty - Easy

This was coding round and was conducted on Cocubes platform. It is not IDE but kind of white board coding platform. C/C++, Java and Python were only allowed languages.

Round duration - 30 minutes
Round difficulty - Easy

This was pen and paper round. Total 40 shortlisted candidates were grouped into 8 groups (each of 5). Each group was given a coding question which they have to solve on paper and explain it to the recruiter. 2 to 3 from each group were selected for the next round.

Round duration - 60 minutes
Round difficulty - Easy

This was general face to face Data Structures and Algorithms based round.

Q1. How do you perform a spiral order traversal of a binary tree? Please provide an explanation or code to print the binary tree in spiral order.
Ans. 
Spiral order traversal of a binary tree involves printing nodes level by level alternating between left to right and right to left.
• Start by pushing the root node into a queue.

• While the queue is not empty, pop a node, print its value, and push its children into the queue.

• For each level, alternate between printing nodes from left to right and right to left.

• Repeat until all nodes are printed in spiral order.
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Round duration - 45 minutes
Round difficulty - Easy

It was the last round.

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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Round duration - 180 minutes
Round difficulty - Easy

This was Online Coding round on Samsung Software. There were 50 test cases and all test cases should be passed to get selected.

Round duration - 30 minutes
Round difficulty - Easy

This was also Face to Face coding Interview totally based on Data Structures and Algorithms.

Q1. The Longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order.
Ans. 
The Longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence with increasing order.
• Use dynamic programming to solve the LIS problem efficiently.

• Maintain an array to store the length of the LIS ending at each element.

• Iterate through the array and update the LIS length based on previous elements.

• Example: For input [10, 22, 9, 33, 21, 50, 41, 60, 80], the LIS is [10
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Round duration - 30 minutes
Round difficulty - Easy

This Face to Face Interview was Based on coding and projects.

Q1. Implement the Depth First Search (DFS) algorithm for a graph.
Ans. 
DFS is a graph traversal algorithm that explores as far as possible along each branch before backtracking.
• Start at a node and explore as far as possible along each branch before backtracking

• Use a stack to keep track of nodes to visit

• Mark visited nodes to avoid revisiting them

• Recursive implementation is common
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Round duration - 15 minutes
Round difficulty - Easy

HR round was mainly based on questions about myself and my knowledge about the company.