Nvidia DFT Engineer Interview Questions and Answers

2 coding questions and a set of mcq questions based on oops, dbms, networking and os

Q1. Reverse a linked list
Ans. 
Reverse a linked list
• Iteratively swap the next and previous pointers of each node

• Use three pointers to keep track of the current, previous, and next nodes

• Update the head pointer to the last node after reversing
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Q2. What is the point of BST?
Ans. 
BST is a data structure used for efficient searching, insertion and deletion of elements in a sorted manner.
• BST stands for Binary Search Tree.

• It has a root node and every node has at most two children.

• The left subtree of a node contains only nodes with keys lesser than the node's key.

• The right subtree of a node contains only nodes with keys greater than the node's key.

• BST allows for efficient searching, insertion and d...read more
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Round duration - 60 Minutes
Round difficulty - Easy

This was the initial round consisting of some aptitude questions,technical multiple choice questions and one coding question.
The Multiple Choice Questions were pretty straight forward.

Q1. 
Snake and Ladder Problem Statement
Given a 'Snake and Ladder' board with N rows and N columns, where positions are numbered from 1 to (N*N) starting from the bottom left, alternating direction each row, f...read more
Ans. 
Find the minimum number of dice throws required to reach the last cell on a 'Snake and Ladder' board.
• Start from the bottom left cell and move according to dice outcomes (1-6).

• Utilize snakes and ladders to reach the last cell faster.

• Keep track of the minimum number of throws required to reach the last cell.

• If unreachable, return -1 as output.
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Round duration - 45 Minutes
Round difficulty - Medium

I had two interviewers in this round.Some Problem solving questions were asked in this round.

Q1. 
Anagram Pairs Verification Problem
Your task is to determine if two given strings are anagrams of each other. Two strings are considered anagrams if you can rearrange the letters of one string to form the...read more
Ans. 
Check if two strings are anagrams of each other by comparing their sorted characters.
• Sort the characters of both strings and compare them.

• Use a dictionary to count the frequency of characters in each string and compare the dictionaries.

• Ensure both strings have the same length before proceeding with comparison.

• Example: For input 'spar' and 'rasp', after sorting both strings, they become 'aprs' which are equal, so return
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Q2. Can you explain all the main Object Oriented Programming concepts?
Ans. 
Object Oriented Programming concepts include encapsulation, inheritance, polymorphism, and abstraction.
• Encapsulation: Bundling data and methods that operate on the data into a single unit (class). Example: Class Car with properties like color and methods like drive().

• Inheritance: Creating new classes based on existing classes, inheriting their attributes and methods. Example: Class SUV inheriting from class Car.

• Polymor...read more
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Q3. Can you explain one of your projects?
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Round duration - 45 Minutes
Round difficulty - Easy

Number Of Interviewers : 2
I was asked two problem solving questions and I was asked to explain my projects.

Round duration - 60 minutes
Round difficulty - Medium

The test was in the evening which was not too late. It was conducted on Hackerrank which is a very common platform and was hence familiar. There were 26 questions:
Q1: Coding Question
3 languages were allowed: c, java8, python3
Q2 - Q26: Multiple Choice Questions

Q1. 
Intersection of Linked List Problem
You are provided with two singly linked lists containing integers, where both lists converge at some node belonging to a third linked list.
Your task is to determine t...read more
Ans. 
Find the node where two linked lists merge, return -1 if no merging occurs.
• Traverse both lists to find the lengths and the last nodes

• Align the starting points of both lists by moving the longer list's pointer ahead by the difference in lengths

• Traverse both lists simultaneously until a common node is found, which is the merging point
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