Brillio Coding Questions and Answers

May 16, 2023

Brillio Coding Questions with Solutions

Brillio Coding Questions and Answers page will help you find out the Coding Questions asked in Online Assessments and Technical Interviews of the Company’s Hiring process.

Go through this page to know more details about Brillio’s Hiring Process for 2 Engineer Job profiles in Digital Infrastructure and Product & Platform Engineering. And one more Job profile of a Data Analytics Practioner in the Data Analytics & Engineering Department is also discussed further on this page.

About Brillio

Brillio is a global digital transformation company that provides technology consulting, product engineering, and digital services to help businesses accelerate their digital initiatives. The company was founded in 2014 and is headquartered in Santa Clara, California, with operations in the United States, Europe, and Asia.

Brillio offers a range of services, including digital strategy consulting, customer experience design, enterprise data management, cloud transformation, and software engineering. Its clients are present in a variety of industries, including banking and financial services, healthcare, retail, and technology.

Details on Brillio Process 2023

Brillio has announced to conduct Hiring Process for 2023 freshers on 2 Engineer Job Profiles and a Data Analytics Practitioner post, we have mentioned the following steps which are included in this Hiring Process:

Online Aptitude + Coding Test
Group Discussion
Technical Interview
HR Interview

Different Skill sets are required for these Job profiles, go through this page to find a detailed explanation of all 3 of them.

We have mentioned further details of the Brillio Hiring Process in the following Tabular Form:

Brillio	Related Information
Position:	Engineer – Digital Infrastructure Engineer – Product & Platform Engineering Data & Analytics Practitioner
Course:	B.E./B.Tech – CSE, IT, EEE & ECE. Eligible Batch – 2023
Eligibility Criteria / Academic Qualification Required:	60% or equivalent CGPA in 10th, 12th, and Graduation. No Current Backlogs.
CTC Offered:	Basic Pay – ₹ 6 LPA ACE Bonus: ₹ 4 Lakh
Selection Process:	Online Aptitude & Coding Assessment Group Discussion Technical Interview HR Interview

FACTS ABOUT CTC OFFERED BY BRILLIO:

Baisic Pay: ₹ 6 LPA
ACE Bonus: ₹ 4 Lakh

Based on performance, ACE Bonus is further given 3 successive years.

1st Payout: ₹ 1 Lakh in 2nd Year of Employment.
2nd Payout: ₹ 1 Lakh in 3rd Year of Employment.
3rd Payout: ₹ 2 Lakh in 4th Year of Employment.

Important Note:

Steps involved in Recruitment Process can be changed on the basis of On-Campus, Off-Campus Drives, and also for different colleges.

Prime Course Trailer

Related Banners

Get PrepInsta Prime & get Access to all 200+ courses offered by PrepInsta in One Subscription

Insights on all 3 Brillio's Job Profiles

Engineer - Digital Infrastructure

ITIL/Cloud technologies like Azure basics, AWS,etc.
Experience/Knowledge on O365 within a network environment (permissions, calendar sharing, delegation).
Would be expected to work on core infra technologies and latest technologies such as cloud, DevOps and automation.
Ability to Understand Infrastructure set up on-premises and cloud environment.
1st line troubleshooting of IT related problems across different streams of support like Desktop/Laptop, Cloud ops and Network infra.

Practice Brillio Coding Questions with Solutions

Question 1 : Choco and chocolate

Problem Statement :

Choco, a chocolate lover, has N amount of money with him. He wants to buy as much chocolate as possible. So, he goes to a chocolate shop “Bandyman ”. Mike, the owner of “Bandyman ” has different types of chocolate in his store (represented by a character) placed in a row.

Mike, give an offer to Choco that he can buy a selected type of chocolate for free and need to pay for the other types of chocolates and Choco can only buy consecutive chocolates.

Now, you need to write a code to find the maximum amount of chocolates Choco can get by selecting the chocolates optimally.

Input format :

1st line contains 2 space separated integers A and B denoting the number of chocolates and the amount of money Choco has.

The 2nd line contains A chocolates represented by a string. All chocolates represented by lowercase alphabets.

The 3rd line represents 26 space separated integers representing the cost to buy the chocolates.
[First integer represents the cost of the chocolate of type ‘a’, 2nd integer represents the cost of the chocolates of type ‘b’ and so on]

Output format :
Print the maximum number of chocolates Choco can buy.

Constraints :
1<=A<=10^5
1<=B<=10^9
1<=cost of chocolate<=10^9

Sample input 1 :
6 10
aabcda
5 4 4 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Sample output 1 :
4

Explanation :

Choco can select the chocolate of type ‘a’ for free and start buying from index 0 and if he buys “aabc” then he has to pay less (0+0+4+4=8) than the total money he has.

This is the maximum number of chocolates he can get in this case.

C++

JAVA

Python

C++

Run

#include<bits/stdc++.h>

using namespace std;

int main() {
    int n, m;
    cin >> n >> m;
    string s, s1;
    cin >> s;
    cin.ignore();
    getline(cin, s1);
    istringstream ss(s1);
    vector < int > a;
    vector < char > a1;
    while (ss) {
        string w;
        ss >> w;
        if (w == "") break;
        a.push_back(stoi(w));
    }
    map < char, int > mm;
    for (auto i: s)
        if (mm[i] == 0) {
            mm[i]++;
            a1.push_back(i);
        }
    int t = 0;
    for (auto i: a1) {
        vector < int > a2;
        for (auto i1: s) {
            if (i1 == i) a2.push_back(0);
            else a2.push_back(a[i1 - 97]);
        }
        int l = 0, su = 0;
        for (int i2 = 0; i2 < n; i2++) { if (l == i2) su += a2[i2]; else su += a2[i2]; if (su > m) {
                while (su > m && l <= i2) { su -= a2[l]; l++; } } if (i2 - l + 1 > t) t = i2 - l + 1;
        }
    }
    cout << t;
}

JAVA

Run

import java.util.*;

class Main {
    public static int solve(int n, int amnt, String s, int[] price) {
        int[] freq = new int[26];
        char[] ch = s.toCharArray();
        int temp = 0, ans = 0, maxFreq = 0, st = 0;
        for (int i = 0; i < n; i++) { int indx = ch[i] - 'a'; freq[indx] += price[indx]; temp += price[indx]; maxFreq = Math.max(maxFreq, freq[indx]); if (temp - maxFreq > amnt) {
                while (temp > amnt) {
                    boolean b = false;
                    int tempIndx = ch[st] - 'a';
                    if (maxFreq == freq[tempIndx]) {
                        b = true;
                    }
                    temp -= price[tempIndx];
                    freq[tempIndx] -= price[tempIndx];
                    st++;
                    if (b) {
                        maxFreq = 0;
                        for (int j = 0; j < 26; j++) {
                            maxFreq = Math.max(freq[j], maxFreq);
                        }
                    }
                    if (temp - maxFreq < amnt) {
                        break;
                    }
                }
            }
            ans = Math.max(ans, i - st + 1);
        }
        return ans;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int amnt = sc.nextInt();
        String s = sc.next();
        int[] price = new int[26];
        for (int i = 0; i < n; i++) {
            price[i] = sc.nextInt();
        }
        System.out.println(solve(n, amnt, s, price));
    }
}

Python

Run

n, m = map(int, input().split())
s = input().strip()
a = list(map(int, input().split()))

s1 = set()
for i in s:
    if i not in s1:
        s1.add(i)

s1 = list(s1)
t = 0

for i in s1:
    a1 = []
    for i1 in s:
        if i1 == i:
            a1.append(0)

        else:
            a1.append(a[ord(i1) - 97])

    l = 0
    su = 0

    for i2 in range(n):
        if l == i2:
            su = a1[i2]
        else:
            su += a1[i2]

        if su > m:
            while su > m and l <= i2:
                su -= a1[l]
                l += 1 
                if (i2 - l + 1) > t:
                    t = i2 - l + 1

print(t)

Question 2 : Vowel Encryption

Problem Statement :

There is an encryption game going on. You will be given a number. If a digit is prime, it will take a vowel. Otherwise it will take a consonant value.
By this process, you have to make the string the lexicographically smallest possible. For a given number, print the output as a string.;

Input Format:
An integer n denoting the number.
Output Format:
The encrypted word.

Sample Input: 123421
Sample Output: baecab

C++

JAVA

Python

C++

Run

#include<bits/stdc++.h>

using namespace std;
map < char, char > M;
map < int, bool > prime;

int main() {
    prime[2] = true;
    prime[3] = true;
    prime[5] = true;
    prime[7] = true;
    char vow[] = {
        'a',
        'e',
        'i',
        'o',
        'u'
    };
    char con[] = {
        'b',
        'c',
        'd',
        'f',
        'g',
        'h'
    };
    int jv = 0, jc = 0;
    string s;
    cin >> s;
    for (auto i: s) {
        if (prime[i - '0']) {
            if (M[i]) {} else {
                M[i] = vow[jv];
                jv++;
            }
        } else if (M[i]) {} else {
            M[i] = con[jc];
            jc++;
        }
    }
    for (auto i: s) cout << M[i];
}

JAVA

Run

import java.util.*;
class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String vowel = "aeiou";
        char arr[] = new char[10];
        arr[2] = 'a';
        arr[3] = 'e';
        arr[5] = 'i';
        arr[7] = 'o';
        char ch = 'b';
        for (int i = 1; i < arr.length; i++) {
            if (vowel.indexOf(ch) != -1) {
                ch++;
                continue;
            } else if (arr[i] == 0)
                arr[i] = (char) ch++;
        }
        int temp = n;
        int res = 0;
        while (temp != 0) {
            res = res * 10 + temp % 10;
            temp = temp / 10;
        }
        int i = 1;
        while (res != 0) {
            System.out.print(arr[res % 10]);
            res = res / 10;
        }
    }

}

Python

Run

s = input()
res = []
x, y = 0, 0
v = "aeiou"
c = "bcdfgh"
p = "2357"
ans = sorted(set(s))
for i in ans:
    if i in p:
        res.append(v[x])
        x += 1
    else:
        res.append(c[y])
        y += 1
for i in s:
    print(res[ans.index(i)], end="")

Question 3 : Spiral Matrix

Problem Statement :

You will be given a 2d matrix. Write the code to traverse the matrix in a spiral format. Check the input and output for better understanding.

Sample Input :
Input :
5 4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
17 18 19 20

Output :

[1, 2, 3, 4, 8, 12, 16, 20, 19, 18, 17, 13, 9, 5, 6, 7, 11, 15, 14, 10]

C++

JAVA

Python

C++

Run

#include <bits/stdc++.h>

using namespace std;

void Spiral(vector < vector < int >> a) {
  if (a.size() == 0) return;
  int m = a.size(), n = a[0].size();
  int i, k = 0, l = 0;
  while (k < m && l < n) {
    for (i = l; i < n; ++i) cout << a[k][i] << " ";
    k++;
    for (i = k; i < m; ++i) cout << a[i][n - 1] << " ";
    n--;
    if (k < m) { for (i = n - 1; i >= l; --i) cout << a[m - 1][i] << " ";
      m--;
    }
    if (l < n) { for (i = m - 1; i >= k; --i) cout << a[i][l] << " "; l++; } } } int main() { int r, c; cin >> r >> c;
  vector < vector < int >> mat(r, vector < int > (c));
  for (int i = 0; i < r; i++)
    for (int j = 0; j < c; j++) cin >> mat[i][j];
  Spiral(mat);
}

JAVA

Run

import java.util.*;
public class Main
{
    public static List < Integer > solve(int [][]matrix,int row,int col)
   {
      List < Integer > res=new ArrayList < Integer > ();
      boolean[][] temp=new boolean[row][col];
      int []arr1={0,1,0,-1};
      int []arr2={1,0,-1,0};
      int di=0,r=0,c=0;
      for(int i=0;i < row*col; i++) { res.add(matrix[r][c]); temp[r][c]=true; int count1=r+arr1[di]; int count2=c+arr2[di]; if(count1 >= 0 && row > count1 && count2 >= 0 && col > count2 && !temp[count1][count2]){
	r=count1;
                  c=count2; 
            } 
            else 
            {
                 di=(di+1)%4;
                 r+=arr1[di];
                 c+=arr2[di];
             }
      }  
     return res;
    }
    public static void main(String[] args)
   {
         Scanner sc=new Scanner(System.in);
         int m=sc.nextInt();
         int n=sc.nextInt();
         int matrix[][]=new int[m][n];
         for(int i=0;i < m;i++)
         {
                 for(int j=0;j < n;j++)
                      matrix[i][j]=sc.nextInt();
         }
       System.out.println(solve(matrix,m,n));
  }
}

Python

Run

def spiralOrder(arr):
    ans = []
    while arr:
        ans += arr.pop(0)
        arr = (list(zip(*arr)))[::-1]
    return ans
arr = []
n, m = map(int, input().split())
for i in range(n):
    arr.append(list(map(int, input().split())))
print(*spiralOrder(arr))

Question 4 : Array Subarray

Problem Statement :
You are given an array, You have to choose a contiguous subarray of length ‘k’, and find the minimum of that segment, return the maximum of those minimums.
Sample input :
1 → Length of segment x =1
5 → size of space n = 5
1 → space = [ 1,2,3,1,2]
2
3
1
2

Sample output :
3
Explanation :
The subarrays of size x = 1 are [1],[2],[3],[1], and [2],Because each subarray only contains 1 element, each value is minimal with respect to the subarray it is in. The maximum of these values is 3. Therefore, the answer is 3

C++

JAVA

Python

C++

Run

#include <bits/stdc++.h>
using namespace std;
vector < int > arr;
int prevmin=-1;
int flag=0;
int x,n,q;

int sorting(int start,int end)
{
    if(start+1==n) {start=0;end=end-n;}
    if(start==end) return arr[start];
    return min(arr[start],sorting(start+1,end));
}

int func(int start,int end)
{
    if(flag==0) {flag++;return prevmin=sorting(start,end);}
    if(arr[start-1]==prevmin) return prevmin;
    return prevmin=(arr[end] <= prevmin)?prevmin:sorting(start,end); } int main() { cin >> x >> n;
    int ans=0;
    for(int i=0;i < n;i++) {cin >> q;arr.push_back(q);}
    for(int i=0;i < n;i++)
    {
       ans=max(ans,func(i,i+x-1));
    }
    cout << ans;
}

JAVA

Run

import java.util.*;
public class DiskSpace
{
  public static void main(String[] args)
{
  Scanner sc=new Scanner(System.in);
  int x=sc.nextInt();
  int n=sc.nextInt();
  int arr[]=new int[n];
  for(int i=0;i < n;i++)
   arr[i]=sc.nextInt();

  int min=Integer.MAX_VALUE;
  int max=Integer.MIN_VALUE;

  for(int i=0;i <= n-x;i++)
 {
    min=Integer.MAX_VALUE;
    for(int j=i;j < (i+x);j++)
     min=Math.min(min,arr[j]);
   max=Math.max(min,max);     
 }
 System.out.println(max);
}
}

Python

Run

n=int(input())
arr=[]
for i in range(n):
    arr.append(int(input()))
s,a=0,0
for i in arr:
    s=s+i   
    if(s < 1):
        a=a+(-1*s) 
        s=0 
print(a)

Question 5 : Order check

Problem Statement: In an IT company there are n number of Employees , they are asked to stand in ascending order according to their heights. But some employees are not currently standing in their correct position.

Your task is to find how many employees are there who are not standing in their correct positions.

Example

height=[1,2,1,3,3,4,3]

The 4 employees at indices 1,2,5 and 6 are not in the right positions. The correct positions are (1,1,2,3,3,3,4).Return 4.

Function Description

Complete the function countEmployees in the editor below.

count Employee has the following parameter(s):

int height[n]:an array of heights in the order the employees are standing

Returns:

Int : the number of employees not standing in the correct positions

Constraints

1<=n<=10^5
1<=height[i]<=10^9

Sample case 0

Sample input 0

7 ->height[] size n=7

Sample output 0:

Explanation

The four employees who are not standing in the correct positions are at indices [1,2,5,6] return 4. The correct positions are [1,1,2,3,3,3,4].

Java

Python

Java

Run

import java.util.*;
public class Main
{
  public static int countEmployees (int arr[])
  {
    int n = arr.length;
    int temp[] = new int[n];
    for (int i = 0; i < n; i++)
        temp[i] = arr[i];
      Arrays.sort (arr);
    int count = 0;
    for (int i = 0; i < n; i++)
      if (arr[i] != temp[i])
	  count++;
      return count;
  }
  public static void main (String[]args)
  {
    Scanner sc = new Scanner (System.in);
    int n = sc.nextInt ();
    int arr[] = new int[n];
    for (int i = 0; i < n; i++)
      arr[i] = sc.nextInt ();
    System.out.println (countEmployees (arr));
  }
}

Python

Run

def countEmployees(n,arr):
    dup_arr=arr[:]
    dup_arr.sort()
    count=0
    for i in range(n):
        if arr[i]!=dup_arr[i]:
            count+=1 
        else:
            pass
    return count
n=int(input())
arr=[ ]
for i in range(n):
    arr.append(int(input()))
print(countEmployees(n,arr))

FAQs on Brillio Coding Questions with Solutions

Question 1: What kind of services does Brillio provide?

Brillio provides a range of digital transformation services, including technology consulting, product engineering, and digital services such as digital strategy consulting, customer experience design, enterprise data management, cloud transformation, and software engineering.

Question 2: What are some major fields in which Brillio offers Jobs?

There are some major field in which Brillio offers Jobs:

Platform & Product Engineering
Cloud Technology Deployment
Data Analytics, AI & ML
Customer Experience
Support System

Get over 200+ course One Subscription

Courses like AI/ML, Cloud Computing, Ethical Hacking, C, C++, Java, Python, DSA (All Languages), Competitive Coding (All Languages), TCS, Infosys, Wipro, Amazon, DBMS, SQL and others

Unlock this article for Free,
by logging in

Brillio Coding Questions and Answers

Brillio Coding Questions with Solutions

About Brillio

Details on Brillio Process 2023

We have mentioned further details of the Brillio Hiring Process in the following Tabular Form:

Prime Course Trailer

Related Banners

Insights on all 3 Brillio's Job Profiles

Engineer - Digital Infrastructure

Data & Analytics Practitioner

Engineer - Product & Platform Engineering

Practice Brillio Coding Questions with Solutions

Question 1 : Choco and chocolate

Question 2 : Vowel Encryption

Question 3 : Spiral Matrix

Question 4 : Array Subarray

Question 5 : Order check

FAQs on Brillio Coding Questions with Solutions

Question 1: What kind of services does Brillio provide?

Question 2: What are some major fields in which Brillio offers Jobs?

Get over 200+ course One Subscription

Checkout list of all the video courses in PrepInsta Prime Subscription

Checkout list of all the video courses in PrepInsta Prime Subscription

30+ Companies are Hiring

Unlock this article for Free, by logging in

Brillio Coding Questions and Answers

Brillio Coding Questions with Solutions

About Brillio

Details on Brillio Process 2023

We have mentioned further details of the Brillio Hiring Process in the following Tabular Form:

Prime Course Trailer

Related Banners

Insights on all 3 Brillio's Job Profiles

Engineer - Digital Infrastructure

Data & Analytics Practitioner

Engineer - Product & Platform Engineering

Practice Brillio Coding Questions with Solutions

Question 1 : Choco and chocolate

Question 2 : Vowel Encryption

Question 3 : Spiral Matrix

Question 4 : Array Subarray

Question 5 : Order check

FAQs on Brillio Coding Questions with Solutions

Question 1: What kind of services does Brillio provide?

Question 2: What are some major fields in which Brillio offers Jobs?

Get over 200+ course One Subscription

Checkout list of all the video courses in PrepInsta Prime Subscription

Checkout list of all the video courses in PrepInsta Prime Subscription

30+ Companies are Hiring

Unlock this article for Free,
by logging in