[Part - 2 One Shot] Accenture Actual Coding Assessment Question 2025|Accenture Previous Year Coding

[Part - 2 One Shot] Accenture Actual Coding Assessment Question 2025 | Accenture Previous Year Coding Ques

#accenture #accenturecoding #accenturehiring #coding #accentureassessment #accenturejobs

Prepare for your Accenture coding assessment with this video featuring actual coding questions from previous years. Get ready to ace your assessment with this one shot tutorial!

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Accenture Coding Questions :

TimeStamp:
0:00 Intro
2:00 1st Question
5:57 2nd Question
15:49 3rd question
30:43 4th question
43:38 5th question
1:12:22 6th question
1:25:47 7th question
1:35:48 8th question
1:44:54 9th question
1:51:48 10th question

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accenture coding round questions , accenture coding round questions in java

[namrathagk]

does accenture even look into time complexity of our code?

[adityavardhanjain]

For the card question:

n = int(input())
b = 0
gap = 1
ans = 0

for i in range(1, n + 1):
a = i * i
b += gap
ans = a + b
gap += 1

print(ans)

[PremShinde-pq]

01:01:55
There's one edge case, if 2nd barrier is smaller than 1st then it will not work properly

eg:-
Xi Yi Di
2 3 8
4 4 2

then our starting and ending points will be 2, 6 according to your logic but they should be 2, 10.

[anandsinha]

#include <iostream>
using namespace std;

int RemainderMod(string &s) {
int size = s.size();
int ans = 0, digit;
//asci value of 1 is 49 and 0 is 48 so 49-48=10
//asci value of 2 is 50 and and 0 is 48 so 50-48=2

for (int i = 0; i < size; i++) {
digit = s[i] - '0';
ans = ans * 10 + digit;
}

int mod=(ans%11);
return mod;
}

int main() {
string s = "1345";
int ans = RemainderMod(s);
cout << "Answer is " << ans;

return 0;
}
This is the code for question three much simpler

[NavnitKeshav]

on campus questions or off campus questions?

[BTADMANAVA]

For Third Question you can use this also and simple method:
#include<stdio.h>
int main()
{
int n;
scanf("%d", &n);
if(n%11==0)
printf("0");
else
printf("%d", n%11);
}

[THIRUSELVAMPTSIR]

def count(str1):
d = {}
for i in range(len(str1)):
if str1[i] in d:
d[str1[i]]+=1
else:
d[str1[i]] =1
string_list = []
for key, value in d.items():

ans = " "
for i in string_list:
ans = ans+i
return ans
str1 = "aabbccddeeff"
print(count(str1))
python code for question 4

[beinspired-themotivational]

question 6
= 2**n+1 -1 kr sakte heina ?

[lucky-]

I think in Accenture assessment we have to write whole code

[nehashetty]

#include<iostream>
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n, value=0;
cin>>n;
vector<int>arr;
vector<int>res;

for(int i=0;i<n;i++)
{
cin>>value;
arr.push_back(value);
}
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
res.push_back(arr[i]);
}
}

for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
res.push_back(arr[i]);
}

}
for(int i =0;i<n;i++)
{
cout<<res[i]<<" ";
}
return 0;
} Sir is this correct?? for question 8??

[harinicherala]

does these solutions really satisfy all the test cases

[kunjmaheshwari]

In my college Accenture is offering 11LPA for the same job role for which accenture gives 6 LPA in other colleges and in offcampus drive.So what would be the difficulty level of the coding questions ? will it remain the same or we can expect something more difficult?

[rks-jktr]

Are These questions actual PYQs ?
or you are posting it as most important and expected for Accenture ?