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Accenture
About Accenture Exam
Accenture is one of the top and the largest IT firms in the world, that offers unmatched services in Consulting, Strategy, Digital, Technology and Operations. It is headquartered at Dublin (Republic of Ireland). They have expertise across more than 40 industries and are partners with more than three-quarters of the Fortune Global 500 companies.
The company conducts recruitment exams for various roles and positions in the organization. The Accenture Exam is designed to assess the candidates' cognitive abilities, technical abilities, coding skills and communication skills.
Eligibility criteria for Accenture Exam
The eligibility criteria for Accenture's recruitment process include:
Academic Qualifications:
- Minimum 60% or above in Class 10th Standard
- Minimum 60% or above in Class 12th Standard
- Minimum 60% or above in College Graduation with a minimum CGPA of 6.5.
- Eligible passing year: 2023
College Qualification Required:
- B.E.
- B.Tech.
- MCA
- MSc (Computer Science)
Eligible Branches: All Engineering branches are available.
Other Important Criteria:
- Candidates must not have any backlogs at the time of the selection process.
- A maximum education gap of 1 year between 10th and graduation is allowed.
- Candidates must be from a full-time degree course recognized by the Central/State Government of India.
- Candidates must not have any pending attendance requirements with their college.
- Candidates must be Indian citizens or hold a PIO or OCI card if they hold a passport of any other country.
- Candidates must have a rest time of at least 6 months between their previous and present applications.
- Candidates who miss the opportunity to give an interview after applying, have the right to apply again and attend the selection process.
Please note that the above information is an example of the eligibility criteria and the actual criteria might vary.
Syllabus for Accenture EXAM
Verbal Ability
- Sentence Correction
- Prepositions
- Grammar
- Reading Comprehension
- Synonyms & Antonym
- Idioms and Phrases
- Speech and Tenses
- Article
- Sentence Selection
- Spotting Error
- Sentence Arrangement
Critical Reasoning
- Arrangements
- Analogies
- Blood Relations
- Statement & Conclusions
- Coding-Decoding
- Agree; Disagree Psychometric
- Inferred Meaning
- Logical Sequence
Abstract Reasoning
- Visual Reasoning
- Directional Sense
- Flowcharts-Visual Reasoning-DI
- Seating Arrangement
Technical Assessment
Fundamental of Networking
- Data and Computer Communication Networks
- Mobile & Wireless Networks
- Database Security
- Software Security
- Cryptography and Network Security
- Biometric Security
Common Applications and MS office
- Creating, editing, saving, and printing text documents
- Font and paragraph formatting
- Simple character formatting
- Inserting tables, smart art, page breaks
- Using lists and styles
- Mail Merge
- Spreadsheet basics
- Creating, editing, saving, and printing spreadsheets
- Working with functions & formulas
- Working with images
- Using Spelling and Grammar check
- Understanding document properties
- Modifying worksheets with color & autoformats
- Graphically representing data: Charts & Graphs
- Speeding data entry: Using Data Forms
- Analyzing data: Data Menu, Subtotal, Filtering Data
- Formatting worksheets
- Securing & Protecting spreadsheets
- Opening, viewing, creating, and printing slides
- Applying auto layouts
- Adding custom animation
- Using slide transitions
- Graphically representing data: Charts & Graphs
- Creating Professional Slide for Presentation
Pseudo Code
- Sequence
- While
- Repeat-until
- For
- If-then-else
- Case
Coding Syllabus
- C
- C++
- Dot Net
- JAVA
- Python
Accenture Exam Pattern
The Accenture online written test is divided into five sections, each with its own time limit and set of questions. These sections include:
- Logic and reasoning, which consists of 14 questions to be completed within 14 minutes.
- Aptitude test, which includes 16 questions to be completed within 16 minutes.
- English and comprehension, also known as verbal ability, which includes 22 questions to be completed within 18 minutes.
- Essay writing, in which candidates are asked to write one essay on a given topic, with a time limit of 20 minutes.
- Coding, which includes 2 problems that must be solved within 60 minutes.
The entire online test will take approximately 128 minutes to complete, and there is no negative marking for incorrect answers.
Accenture Selection Process:
The Accenture Selection Process for 2023 consists of four rounds of assessments, as outlined below:
Round 1: Cognitive and Technical Assessment
- Cognitive Assessment: This section includes Numerical Ability, Logical Reasoning and English Ability.
Technical Assessment: This section includes Pseudo Code, Common Applications & MS Office and Fundamentals of Networking.
- Time: 90 minutes
Round 2: Coding Assessment
Pure Coding Round (No sectional division): This section includes questions on C, C++, Dot Net, Java and Python.
- Time: 45 minutes
Round 3: Communication Test
No internal section. Candidates are to be purely judged on their communication skills. This section includes Sentence Mastery, Vocabulary, Fluency and Pronunciation.
- Time: 20 minutes
Round 4: Interview
Learning Agility and Communication – Tested by panel members. The time for this round varies from candidate to candidate.
- Time: 30 minutes
It is important to note that the information given above is an example of the selection process and the actual process might vary.
Popular Questions
#Popular Question 1
Bubble Sort
Given an Integer N and a list arr. Sort the array using bubble sort algorithm.
#Example 1:
```sh
Input:
N = 5
arr[] = {4, 1, 3, 9, 7}
Output:
1 3 4 7 9
```
#Coding Question 1
The function def differenceofSum(n. m) accepts two integers n, m as arguments Find the sum of all numbers in range from 1 to m(both inclusive) that are not divisible by n. Return difference between sum of integers not divisible by n with sum of numbers divisible by n.
Assumption:
- n>0 and m>0
- Sum lies between integral range
#Solution 1
```sh
n = int(input())
m = int(input())
sum1 = 0
sum2 = 0
for i in range(1,m+1):
if i % n == 0:
sum1+=i
else:
sum2+=i
print(abs(sum2-sum1))
```
#Popular Question 2
Insertion Sort
The task is to complete the insert() function which is used to implement Insertion Sort.
#Example 1:
```sh
Input:
N = 5
arr[] = { 4, 1, 3, 9, 7}
Output:
1 3 4 7 9
```
#Coding Question 2
You are required to implement the following Function def LargeSmallSum(arr).
The function accepts an integers arr of size ’length’ as its arguments you are required to return the sum of second largest largest element from the even positions and second smallest from the odd position of given ‘arr’.
Assumption:
- All array elements are unique
- Treat the 0th position a seven
NOTE
- Return 0 if array is empty
- Return 0, if array length is 3 or less than 3
#Answer 2
```sh
length = int(input())
arr = list(map(int, input().split()))
even_arr = []
odd_arr = []
for i in range(length):
if i % 2 == 0:
even_arr.append(arr[i])
else:
odd_arr.append(arr[i])
even_arr = sorted(even_arr)
odd_arr = sorted(odd_arr)
print(even_arr[len(even_arr)-2] + odd_arr[len(odd_arr)-2])
```
#Popular Question 3
Write code Check if the given string is Palindrome or not
#Coding Question 3
The function def ProductSmallestPair(sum, arr) accepts an integers sum and an integer array arr of size n. Implement the function to find the pair, (arr[j], arr[k]) where j!=k, Such that arr[j] and arr[k] are the least two elements of array (arr[j] + arr[k] <= sum) and return the product of element of this pair
NOTE
- Return -1 if array is empty or if n<2
- Return 0, if no such pairs found
- All computed values lie within integer
#Solution 3
```sh
n = int(input())
sum1 = int(input())
arr = list(map(int, input().split()))
if n < 2:
print('-1')
arr = sorted(arr)
for i in range(n-1):
if arr[i] + arr[i+1] < sum1:
print(arr[i] * arr[i+1])
break
else:
print('0')
```
#Coding Question 4
You are given a function, Void *ReplaceCharacter(Char str[], int n, char ch1, char ch2);
The function accepts a string ‘ str’ of length n and two characters ‘ch1’ and ‘ch2’ as its arguments . Implement the function to modify and return the string ‘ str’ in such a way that all occurrences of ‘ch1’ in original string are replaced by ‘ch2’ and all occurrences of ‘ch2’ in original string are replaced by ‘ch1’.
Assumption: String Contains only lower-case alphabetical letters.
Note:
- Return null if string is null.
- If both characters are not present in string or both of them are same , then return the string unchanged.
#Solution 4
```sh
def swap (user_input, str1, str2):
result = ''
if user_input != None:
result = user_input.replace(str1, '*').replace(str2, str1).replace('*', str2)
return result
return 'Null'
user_input = input()
str1, str2 = map(str,input().split())
print(swap(user_input, str1, str2))
```
#Coding Question 5
You are required to implement the following function, Int Calculate(int m, int n);
The function accepts 2 positive integer ‘m’ and ‘n’ as its arguments.You are required to calculate the sum of numbers divisible both by 3 and 5, between ‘m’ and ‘n’ both inclusive and return the same.
Note:
0 < m <= n
#Answer 5
```sh
m = int(input("M:"))
n = int(input("N:"))
def calculate(m, n):
sum = 0
for i in range(m,n+1,1):
if i%3 == 0 and i%5 == 0:
sum = sum + i
print(sum)
calculate(m,n)
```
Testimonials
Ria
Sakila
I started practising on Edyst platform since my 3rd year of college focused on placements & always liked the way they helped us when we were stuck at a particular problem.
Thank you, Edyst for all the assistance and amazing support!
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Dileep
Being a mechanical student and getting into an IT company is very tough. One of the main reason I could able to crack TCS CodeVita is because of Edyst.
Aneeq sir, your doubt clearing sessions, daily assignments & incredible mentors support really brushed up my skills.
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When I joined the Edyst courses I received personalized mentoring on how to crack coding rounds of different companies. Through a combination of coding skills and great projects, I received multiple offers above 6+ lakhs per annum. Finally I joined for 8+ Lakhs package. Thanks for all the support, from Edyst Team.
Vaishnavi
Edyst's training style completely resonated with me. I approached programming as more than a subject. Thanks to Edyst team for the guidance!
Coding Interview Round Sample Questions
#Popular Question 1
Bubble Sort
Given an Integer N and a list arr. Sort the array using bubble sort algorithm.
#Example 1:
```sh
Input:
N = 5
arr[] = {4, 1, 3, 9, 7}
Output:
1 3 4 7 9
```
#Coding Question 1
The function def differenceofSum(n. m) accepts two integers n, m as arguments Find the sum of all numbers in range from 1 to m(both inclusive) that are not divisible by n. Return difference between sum of integers not divisible by n with sum of numbers divisible by n.
Assumption:
- n>0 and m>0
- Sum lies between integral range
#Solution 1
```sh
n = int(input())
m = int(input())
sum1 = 0
sum2 = 0
for i in range(1,m+1):
if i % n == 0:
sum1+=i
else:
sum2+=i
print(abs(sum2-sum1))
```
#Popular Question 2
Insertion Sort
The task is to complete the insert() function which is used to implement Insertion Sort.
#Example 1:
```sh
Input:
N = 5
arr[] = { 4, 1, 3, 9, 7}
Output:
1 3 4 7 9
```
#Coding Question 2
You are required to implement the following Function def LargeSmallSum(arr).
The function accepts an integers arr of size ’length’ as its arguments you are required to return the sum of second largest largest element from the even positions and second smallest from the odd position of given ‘arr’.
Assumption:
- All array elements are unique
- Treat the 0th position a seven
NOTE
- Return 0 if array is empty
- Return 0, if array length is 3 or less than 3
#Answer 2
```sh
length = int(input())
arr = list(map(int, input().split()))
even_arr = []
odd_arr = []
for i in range(length):
if i % 2 == 0:
even_arr.append(arr[i])
else:
odd_arr.append(arr[i])
even_arr = sorted(even_arr)
odd_arr = sorted(odd_arr)
print(even_arr[len(even_arr)-2] + odd_arr[len(odd_arr)-2])
```
#Popular Question 3
Write code Check if the given string is Palindrome or not
#Coding Question 3
The function def ProductSmallestPair(sum, arr) accepts an integers sum and an integer array arr of size n. Implement the function to find the pair, (arr[j], arr[k]) where j!=k, Such that arr[j] and arr[k] are the least two elements of array (arr[j] + arr[k] <= sum) and return the product of element of this pair
NOTE
- Return -1 if array is empty or if n<2
- Return 0, if no such pairs found
- All computed values lie within integer
#Solution 3
```sh
n = int(input())
sum1 = int(input())
arr = list(map(int, input().split()))
if n < 2:
print('-1')
arr = sorted(arr)
for i in range(n-1):
if arr[i] + arr[i+1] < sum1:
print(arr[i] * arr[i+1])
break
else:
print('0')
```
#Coding Question 4
You are given a function, Void *ReplaceCharacter(Char str[], int n, char ch1, char ch2);
The function accepts a string ‘ str’ of length n and two characters ‘ch1’ and ‘ch2’ as its arguments . Implement the function to modify and return the string ‘ str’ in such a way that all occurrences of ‘ch1’ in original string are replaced by ‘ch2’ and all occurrences of ‘ch2’ in original string are replaced by ‘ch1’.
Assumption: String Contains only lower-case alphabetical letters.
Note:
- Return null if string is null.
- If both characters are not present in string or both of them are same , then return the string unchanged.
#Solution 4
```sh
def swap (user_input, str1, str2):
result = ''
if user_input != None:
result = user_input.replace(str1, '*').replace(str2, str1).replace('*', str2)
return result
return 'Null'
user_input = input()
str1, str2 = map(str,input().split())
print(swap(user_input, str1, str2))
```
#Coding Question 5
You are required to implement the following function, Int Calculate(int m, int n);
The function accepts 2 positive integer ‘m’ and ‘n’ as its arguments.You are required to calculate the sum of numbers divisible both by 3 and 5, between ‘m’ and ‘n’ both inclusive and return the same.
Note:
0 < m <= n
#Answer 5
```sh
m = int(input("M:"))
n = int(input("N:"))
def calculate(m, n):
sum = 0
for i in range(m,n+1,1):
if i%3 == 0 and i%5 == 0:
sum = sum + i
print(sum)
calculate(m,n)
```
Technical Interview Round Sample Questions
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