Accenture interview experience Real time questions & tips from candidates to crack your interview
SDE - 1
Accenture
2 rounds | 3 Coding
problems
Selected Interview preparation journey
Preparation
Duration: 6 Months
Topics: DSA , OOPS , OS, CN, Algorithms, DBMS , Dynamic Programming , Graphs
Tip 1 : Improve Communication Skills
Tip 2 : Do Extra Subjects
Tip 3 : Prepare some Projects
Application process
Where: Campus
Eligibility: 7 CGPA
Tip 1 : Do Mention coding profiles in resume
Tip 2 : Do add summary of Projects
Interview rounds
01
Round
Easy
Online Coding Interview
Duration90 Minutes
Interview date16 Dec 2021
Coding problem2
There were 10-15 Aptitude questions and 2 Coding questions
1. Add Two Numbers As Linked Lists
Moderate
20m average time
80% success
0/80
Asked in companies
You are given two non-negative numbers 'num1' and 'num2' represented in the form of linked lists.
The digits in the linked lists are stored in reverse order, i.e. starting from least significant digit (LSD) to the most significant digit (MSD), and each of their nodes contains a single digit.
Calculate the sum of the two numbers and return the head of the sum list.
Example :
Input:
'num1' : 1 -> 2 -> 3 -> NULL
'num2' : 4 -> 5 -> 6 -> NULL
Output: 5 -> 7 -> 9 -> NULL
Explanation: 'num1' represents the number 321 and 'num2' represents 654. Their sum is 975.
Problem approach
1. Traverse the two linked lists in order to add preceding zeros in case a list is having lesser digits than the other one
.
2. Start from the head node of both lists and call a recursive function for the next nodes.
3. Continue it till the end of the lists.
4. Creates a node for current digits sum and returns the carry.
2. 3 Divisors
Moderate
25m average time
75% success
0/80
Asked in companies
You are given a positive integer ‘N’. Your task is to find all the positive integers between 1 and ‘N’ (both inclusive) that have exactly 3 divisors. Print these numbers in increasing order.
Note :
It is guaranteed that there is at least one such integer under given constraints.
Problem approach
We can simply iterate over the numbers from 1 - > Sqrt(Number) and check whether the it is a factor or not and also its divisor is a factor or not
T.C -> O(Sqrt(Number))
02
Round
Easy
HR Round
Duration15 Minutes
Interview date5 Jan 2022
Coding problem1
This Round mainly focused on Projects
1. Basic HR Questions
I was asked to explain my Project and what problems i faced while making that
What are your hobbies?
Problem approach
Tip 1 : Prepare a brief Summary of your Project
Tip 2 : Don't Copy paste code in your Project
Tip 3 : Prepare a Decent Project and not the ones which are readily available on Google
Here's your problem of the day 
Solving this problem will increase your chance to get selected in this company
Skill covered: Programming
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