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

SDE - 1

PhenomPeople

2 rounds | 2 Coding problems

Sumanth Edara

Fresher | Dec 2021

Selected

Interview preparation journey

Preparation

Duration: 7 months

Topics: Data structures and Algorithms, Html, css, javascript, nodejs, java oops, DBMS

Tip

Tip 1 : DSA is the most important thing , practice as many quality questions as possible, practice should be consistent even after learning
Tip 2 : build good projects on both frontend and backend
Tip 3 : Revise concepts regularly and systematic way

Application process

Where: Coding Ninjas

Eligibility: B.tech only

Resume tip

Tip 1 : keep it simple and clear
Tip 2 : Only add about what matters to the job and can be added to it and add git links in resume

Interview rounds

Round

Easy

Online Coding Interview

Duration90 Minutes

Interview date22 Dec 2021

Coding problem1

1 Dsa questions with 16 Aptitude questions,
Aptitude is very basic ,
Coding questions are easy

1. Valid Parentheses

Easy

10m average time

80% success

0/40

Asked in companies

You're given a string 'S' consisting of "{", "}", "(", ")", "[" and "]" .

Return true if the given string 'S' is balanced, else return false.

For example:

'S' = "{}()".

There is always an opening brace before a closing brace i.e. '{' before '}', '(' before ').
So the 'S' is Balanced.

Problem approach

1)iterated through string ,i used a stack and pushed only open brackets into it,
2)but when i see closed brackets i pop element from stack
3) if i found a closing bracket and nothing in stack to pop i would say it's invalid,
Also if size is not zero after iteration its invalid, otherwise valid

Try solving now

Round

Easy

HR Round

Duration30 minutes

Interview date24 Dec 2021

Coding problem1

Asked to explain better way for the previous question,

1. Improve the solution of the previous question.

I checked a optimisation solution after exam, found a brilliant alternative which we length of string make it half And add it as open and closed variables count Ex:- 10 is length of string, openbracket:-10/2 Closed bracket=10/2, Now we iterate and start substract -1 each time we see open bracket from openbracket count, vice versa But if we find at any point close bracket count more than open we can consider invalid, or if we find either of open bracket or closed is not zero then also its invalid this solution has zero space complexity

Problem approach

Tip 1 : best approaches may be simple you may need to learn by solutions
Tip 2 : try to learn 150 quality questions perfectly instead of 500 to 1000 questions
Tip 3 : learning means you should identify similar question logic again and should solve it