ITRONIX SOLUTIONS interview experience Real time questions & tips from candidates to crack your interview

SDE - Intern

ITRONIX SOLUTIONS
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2 rounds | 4 Coding problems

Interview preparation journey

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Preparation
Duration: 3 months
Topics: Data Structures, Pointers, OOPS, System Design, Algorithms, Dynamic Programming
Tip
Tip

Tip 1 : Practice coding from Leetcode, Interview bit, at least 100 questions 
Tip 2 : Practice any one automation framework includes design patterns

Application process
Where: Campus
Eligibility: Above 7 CGPA
Resume Tip
Resume tip

Tip 1 : Restrict your resume to 1 page and write your practical work only
Tip 2 : Mention top topics like Selenium, Rest Assured Automation, Language used Java etc

Interview rounds

01
Round
Easy
Video Call
Duration45 minutes
Interview date17 Oct 2021
Coding problem2

1. Insertion Sort in Linked List

Easy
10m average time
90% success
0/40
Asked in companies
Disney + HotstarIntuitAmazon

You are given an arbitrary linked list consisting of 'N' nodes having integer values. You need to perform insertion sort on the linked list and print the final list in sorted order.

In other words, you are given a singly linked list, you need to perform insertion sort on it.

Insertion Sort is a sorting algorithm that removes one element from the input data, finds the location it belongs within the sorted list and inserts it there. It repeats until no input elements remain.
Problem approach

You are given an arbitrary linked list consisting of 'N' nodes having integer values. You need to perform insertion sort on the linked list and print the final list in sorted order.
In other words, you are given a singly linked list, you need to perform insertion sort on it.

Try solving now

2. Puzzle Question

Give two boxes B1 and B2 one have 50 red marbles and other have 50 blue marbles. A ball is selected randomly from any of the box and the task is to maximize the probability of selecting a red ball, by reshuffling marbles in both the boxes.

Problem approach

Let P(R) be the probability of picking a red marble.

P(R) = P(B1) * P(B1 | J1) + P(B2) * P(B2 | J2)

Here, P(B1) and P(B2) refers to selecting B1 and B2 and the probability of selecting each box is \frac{1}{2} J1 and J2 refers to number of total balls in B1 and B2 respectively.

P (R) = ((1 / 2) * (1 / 1)) + ((1 / 2) * (49 / 99)) = 0.747474

02
Round
Easy
Video Call
Duration45 minutes
Interview date17 Oct 2021
Coding problem2

1. Next Greater Element

Easy
10m average time
90% success
0/40
Asked in companies
IBMInfo Edge India (Naukri.com)Amazon

You are given an array 'a' of size 'n'.



The Next Greater Element for an element 'x' is the first element on the right side of 'x' in the array, which is greater than 'x'.


If no greater elements exist to the right of 'x', consider the next greater element as -1.


For example:
Input: 'a' = [7, 12, 1, 20]

Output: NGE = [12, 20, 20, -1]

Explanation: For the given array,

- The next greater element for 7 is 12.

- The next greater element for 12 is 20. 

- The next greater element for 1 is 20. 

- There is no greater element for 20 on the right side. So we consider NGE as -1.
Problem approach

For a given array/list of integers of size N, print the Next Greater Element(NGE) for every element. The Next Greater Element for an element X is the first element on the right side of X in the array, which is greater than X. If no greater elements exist to the right of X, consider the next greater element as -1.

Try solving now

2. Puzzle Question

There are 100 doors in a row, all doors are initially closed. A person walks through all doors multiple times and toggle (if open then close, if close then open) them in the following way: 

In the first walk, the person toggles every door 

In the second walk, the person toggles every second door, i.e., 2nd, 4th, 6th, 8th, … 

In the third walk, the person toggles every third door, i.e. 3rd, 6th, 9th, … 

Likewise,

In the 100th walk, the person toggles the 100th door. 

Which doors are open in the end?

Problem approach

A door is toggled in an ith walk if i divide door number. For example, door number 45 is toggled in the 1st, 3rd, 5th, 9th,15th, and 45th walks.
The door is switched back to an initial stage for every pair of divisors. For example, 45 is toggled 6 times for 3 pairs (5, 9), (15, 3), and (1, 45). 

It looks like all doors would become closed at the end. But there are door numbers that would open, for example, in 16, the divisors are (1,2,4,8,16) and as the pair(4,4) contributes only one divisor making the number of divisors odd, it would become open at the end. Similarly, all other perfect squares like 4, 9,…, and 100 would become open. Now, for prime numbers like 2,3,5,7… the divisors are (1, that number) and it is a pair, so they will remain closed at the end. And for all other numbers divisors are always in pairs, e.g. 15 = (1,15),(3,5), they will also remain closed.

So the answer is 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100.

Here's your problem of the day

Solving this problem will increase your chance to get selected in this company

Skill covered: Programming

What is recursion?

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