Nagarro SE is a global digital engineering and technology consulting company headquartered in Munich, Germany. Established in 2020 as a spin-off from Allgeier SE, Nagarro has rapidly grown to become a significant player in the IT services sector. As of 2023, the company employs over 18,000 professionals across 38 countries, including a substantial presence in India, with offices in cities like Hyderabad .
Nagarro specializes in digital product engineering, offering a comprehensive suite of services such as:(Nagarro)
Application development and management
Artificial intelligence and machine learning solutions
Cloud and infrastructure services
Enterprise resource planning (ERP) and customer relationship management (CRM) consulting
User experience (UX) and design services
Intelligent automation and Internet of Things (IoT) solutions
The company serves a diverse range of industries, including automotive, banking and financial services, energy and utilities, gaming and entertainment, insurance, life sciences and healthcare, media and publishing, public sector, retail and consumer packaged goods (CPG), software and high-tech, telecommunications, and travel and logistics.
In 2023, Nagarro reported revenues exceeding €1 billion, marking significant growth since its inception. The company's inclusion in the SDAX and TecDAX indices of the Frankfurt Stock Exchange underscores its financial stability and market presence.
Nagarro prides itself on a culture defined by its "CARING" values—Customer-centric, Agile, Responsible, Intelligent, Non-hierarchical, and Global. This ethos fosters an environment that emphasizes innovation, collaboration, and a human-centric approach to technology solutions.
With operations in 38 countries, Nagarro maintains a robust global footprint. Its distributed delivery model enables the company to provide agile and scalable solutions tailored to the unique needs of clients worldwide.
Nagarro's recruitment process for software engineering roles is structured to assess candidates comprehensively across various competencies. Here's an overview of the typical stages involved:
This initial assessment comprises three sections:
General Aptitude Test: Approximately 30 multiple-choice questions focusing on logical reasoning, verbal ability, and mathematical reasoning.
Technical Ability Test: Around 20 questions covering topics such as operating systems, data structures, algorithms, SQL, Java, C, and networking.
Coding Test: 2–3 coding problems to be solved within approximately 75 minutes.
The total duration for this round is about 120–140 minutes.
Candidates who clear the first round proceed to a second written test, which is entirely coding-based. This round typically includes 3–4 coding problems of medium difficulty, to be solved within 40–50 minutes. The test may be conducted on-site or online via platforms like HackerRank.
In this round, interviewers delve into:
Optimization of solutions provided in previous coding rounds.
In-depth questions on data structures, algorithms, SQL, and relevant programming languages.
Problem-solving through coding puzzles and real-world scenarios.
The interview typically lasts between 30–45 minutes.
The final round assesses:
Communication skills and cultural fit.
Motivation and alignment with Nagarro's values.
Discussion of resume details and career aspirations.
This round usually spans 15–20 minutes and is generally non-eliminative.
To be eligible for Nagarro's recruitment process:
A Bachelor's degree (B.Tech/B.E.) in Computer Science, IT, ECE, or related fields.
A minimum aggregate of 60% in 10th, 12th, and graduation.
No active backlogs at the time of application.
Aspiring candidates can apply through:
Direct Application: Via Nagarro's.
Campus Recruitment Drives: Nagarro conducts recruitment drives at various institutes and universities.
Employee Referrals: Leveraging connections within the company for referrals.
Practice Coding: Regularly solve problems on platforms like HackerRank, LeetCode, and GeeksforGeeks.
Revise Core Subjects: Ensure a strong grasp of data structures, algorithms, operating systems, and networking concepts.
Mock Interviews: Engage in mock interviews to build confidence and improve communication skills.
Stay Updated: Keep abreast of the latest technologies and industry trends relevant to the role.
Cache memory is a small, high-speed memory located close to the CPU (Central Processing Unit) that stores frequently accessed data and instructions. Its primary purpose is to reduce the time it takes for the CPU to access data from the main memory (RAM), which is comparatively slower.
Faster than RAM: It operates at much higher speeds than regular RAM.
Smaller in size: Usually ranges from a few kilobytes to several megabytes.
Volatile: Loses data when the power is off, just like RAM.
Costlier: More expensive per byte than RAM due to its speed and design.
L1 Cache (Level 1): Smallest and fastest, located directly on the CPU chip.
L2 Cache (Level 2): Slightly larger and slower than L1, can be on the CPU or nearby.
L3 Cache (Level 3): Larger and slower than L2, shared among cores in multi-core processors.
When the CPU needs to read data:
It first checks L1 cache.
If not found (a cache miss), it checks L2, then L3, and finally the main memory.
If found (a cache hit), it retrieves the data much faster.
public void next permutation(int[] nums) {
int peakIndex = 0;
//Finding the peak index
for(int i = 1; i < nums.length; i++){
if(nums[i] > nums[i-1])
peakIndex = i;
}
//When the index is 0 the next permutation
//will be the first. So return by sorting.
if(peakIndex == 0){
Arrays.sort(nums);
return;
}
//Adjusting Pointers
int prevIndex = peakIndex - 1;
int index = prevIndex;
int currIndex = peakIndex;
//Finding the next greater element index
while(currIndex < nums.length && nums[currIndex] > nums[index]){
prevIndex = currIndex;
currIndex++;
}
//swapping the value of that location
int temp = nums[index];
nums[index] = nums[prevIndex];
nums[prevIndex] = temp;
//Changing the values by sorting
Arrays.sort(nums, index+1, nums.length);
}(n*n + n) / 2. This calculates the actual sum that needs to be in the array. And if we sum the array. Then the difference will be the answer.//Method that returns the missing value
public int missingNumber(int[] nums) {
int sumArray = 0;
int n = nums.length;
// Finding the sum of the array
for(int i = 0; i<n; i++)
sumArray += nums[i];
// Calculating actual sum.
int ActualSum = (n*n + n)/2;
return ActualSum - sumArray;
}public boolean areOccurrencesEqual(String s) {
int max = 0;
char[] hash = new char[26];
//Counting the frequency and getting the max frequency
for(int i = 0; i < s.length(); i++){
hash[s.charAt(i)-'a']++;
max = Math.max(max, hash[s.charAt(i)-'a']);
}
//Checking if the frequency matches with the max number
for(int i = 0; i < 26; i++){
if(hash[i] != 0 && hash[i] != max)
return false;
}
return true;
}class Solution {
int trib(int n, int [] dp){
//Base case for terminating the recursive call
if(n == 0 || n == 1)
return n;
if(n == 2)
return 1;
//Finding the result not already computed
if(dp[n] == 0){
dp[n] = trib(n-1, dp) + trib(n-2, dp) + trib(n-3, dp);
}
//Returning already computed result or the result that
//latest computed
return dp[n];
}
public int tribonacci(int n) {
int []dp = new int[n+1];
//Calling helper method that computes and returns the result.
return trib(n, dp);
}
}INSERT), updated (UPDATE), or deleted (DELETE).