Miracle Software Systems Intern Interview Questions and Answers

Hackerrrank and for 1 hour 3 LC medium

Q1. Aggresive cow medium dsa q
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Q2. Binary search on ans type
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Networking core java , oops . data interpretation , data analysis , English and basic Maths question

Java questions , Dsa questions . timing was approx 2 hours

Q1. Oops concepts and data structures
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Q2. Dbms questions where based on normalization
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Two questions were asked and answered both

Two programs on Java was asked on loops and collections

Round duration - 60 Minutes
Round difficulty - Medium

Interview was scheduled in afternoon. It was more on OOPs and DSA.

Q1. 
Check If Linked List Is Palindrome
Given a singly linked list of integers, determine if the linked list is a palindrome.
Explanation:
A linked list is considered a palindrome if it reads the same forwar...read more
Ans. 
I gave O(n) both time and space solution using stacks but interviewer asked to optimize it and give clue as well.

I understood the clue and was able to optimise the code. I was not asked to code in this round just had an discussion but few of my peers were asked to write code as well.
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Round duration - 20 Minutes
Round difficulty - Easy

She asked me more about myself and few projects in resume. It was more of formal discussion than an interview round. I was very comfortable and she aksed me about my location preference at end of. Interview.

The first round was a mcq test based on aptitude and statstics

2nd round was a coding round. Three questions were given . Two were easy and one was medium

Q1. Two leetcode easy level questions
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Q2. One leetcode medium level question
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It was not good, it contain mostly question of java programming.

Q1. Print factorial of n through recursion Sum of all n numbers
Ans. 
Factorial of n and sum of n numbers using recursion
• Create a recursive function to calculate factorial of n

• Use a recursive function to calculate the sum of all n numbers

• Handle base cases for both factorial and sum calculations

• Example: Factorial of 5 = 5 * 4 * 3 * 2 * 1 = 120, Sum of first 5 numbers = 1 + 2 + 3 + 4 + 5 = 15
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