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Table of content:

What is the Cognizant GenC Elevate Coding Assessment?
Top 5 Cognizant GenC Elevate Coding Questions with Answers
Preparation Strategies for GenC Elevate Coding Assessment
Conclusion
Frequently Asked Questions (FAQs)

Cognizant GenC Elevate Coding Questions & Answers for Freshers 2025

Practice for the Cognizant GenC Elevate coding test with selected questions and answers. Get the best tips to ace the coding round and secure your spot at Cognizant.

9 mins read

Cognizant GenC Elevate Coding Questions & Answers for Freshers 2025

As part of the GenC Elevate program, Cognizant offers freshers a fantastic opportunity to build a career in technology. However, the selection process is highly competitive, especially the coding assessment. To help you excel and secure your place in this prestigious program, this article will provide you with some important aspects you need to know about Cognizant GenC Elevate coding questions.

We'll also break down key strategies, share important tips, and highlight crucial topics, ensuring you're prepared to succeed.

What is the Cognizant GenC Elevate Coding Assessment?

The Cognizant GenC Elevate coding assessment is designed to test your problem-solving and coding abilities, evaluating your skills in algorithms, data structures, and programming languages. Here’s an overview of what you can expect:

Aspect	Details
Total Questions	4 coding questions
Time Duration	60 minutes
Difficulty Level	Moderate to High Difficulty
Languages Allowed	Java, Python, C, C++ (most commonly Java and Python)
Focus Areas	Data Structures, Algorithms, Problem-solving, Object-Oriented Concepts

Key Topics

The coding questions in the GenC Elevate assessment focus on core programming concepts. Here's a list of topics you need to be well-versed in:

Data Structures and Algorithms (DSA)
Understanding DSA is critical for solving problems efficiently.

Arrays: Searching, sorting, and manipulation techniques.
Linked Lists: Operations such as insertion, deletion, and reversal.
Stacks & Queues: Used to solve problems related to backtracking and recursion.
Trees & Graphs: Traversals (preorder, inorder, postorder), Binary Search Trees (BST), and Graph traversal algorithms.
Searching & Sorting Algorithms: Binary search, Merge Sort, Quick Sort, and Heap Sort.
Dynamic Programming (DP): Solve optimization problems like Knapsack, Longest Common Subsequence, etc.

Object-Oriented Programming (OOP)
OOP is a cornerstone of software engineering, and proficiency in its concepts is crucial for solving complex problems.

Classes and Objects: The building blocks of any OOP language.
Inheritance, Polymorphism, Encapsulation, and Abstraction: Principles that allow for modular and reusable code.

Database Management Systems (DBMS)
Though it’s secondary to coding, having a basic understanding of SQL and databases can be helpful.

SQL Queries: Creating, updating, and querying databases.
Normalization: Ensuring data integrity and avoiding redundancy.
Joins and Subqueries: Mastery of these concepts is crucial for database-related coding problems.

Mathematics and Logic

Bit Manipulation: Performing operations at the bit level.
Number Theory: GCD, LCM, prime number generation, etc.

Top 5 Cognizant GenC Elevate Coding Questions with Answers

Problem Statement 1

A sequence of positive integers is called "great" for a number X if we can divide it into pairs such that, for each pair, multiplying the first number by X equals the second number. More formally, a sequence A of size N is "great" for X if N is even and there exists a permutation P of size N such that for each i (1 ≤ i ≤ (N/2)), we can rearrange it so that for every pair (Ai, Aj), we have ( Ai * x = Aj ).

Ram has a sequence A and a number X. Help Ram make the sequence "great" by determining the minimum number of positive integers that should be added to the sequence A so that it becomes "great" for the number X.

Input Format

The first line of input contains two integers N, X representing the size of the sequence and the integer X respectively.

The next line of input contains N space-seperated integers a1,a2,……an, representing the elements of the sequence A.

Output Format

Print a single integer representing the minimum number of integers that need to be appended to the end of list A in order to make it a great sequence for the number X.

Solution C++

#include <bits/stdc++.h>

using namespace std;

signed main() {
    ios_base::sync_with_stdio(false);
    //cin.tie(NULL);
    //#ifndef ONLINE_JUDGE
    //freopen("IO files/Sample_input_01.txt", "r", stdin);
    //freopen("IO files/Sample_output_01.txt", "w", stdout);
    //#endif

    int n;
    int64_t x;
    cin >> n >> x;

    vector<int> ar(n);
    for (auto& it : ar)
        cin >> it;

    sort(ar.begin(), ar.end());

    vector<bool> vis(n);
    int j = 0, q = 0;
    int ans = 0;

    for (int i = 0; i < n; ++i) {
        if (vis[i])
            continue;

        if (ar[i] * x > ar[j]) {
            while (ar[i] * x >= ar[j] && j < n)
                q = ++j;
            q = --j;
        }

        if (i < q && ar[i] * x == ar[q])
            vis[q--] = 1;
        else
            ans++;
    }

    cout << ans << "\n";

    return 0;
}

Solution Java

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        // Read inputs
        int n = scanner.nextInt();
        long x = scanner.nextLong();

        int[] ar = new int[n];
        for (int i = 0; i < n; i++) {
            ar[i] = scanner.nextInt();
        }

        // Sort the array
        Arrays.sort(ar);

        boolean[] vis = new boolean[n];
        int j = 0, q = 0;
        int ans = 0;

        for (int i = 0; i < n; i++) {
            if (vis[i]) {
                continue;
            }

            if (ar[i] * x > ar[j]) {
                while (j < n && ar[i] * x >= ar[j]) {
                    q = ++j;
                }
                q = --j;
            }

            if (i < q && ar[i] * x == ar[q]) {
                vis[q--] = true;
            } else {
                ans++;
            }
        }

        System.out.println(ans);
    }
}

Solution Python

def main():
    # Read inputs
    n, x = map(int, input().split())
    ar = list(map(int, input().split()))  # Input array elements

    # Sort the array
    ar.sort()

    # Initialize a list to track visited elements
    vis = [False] * n

    j = 0
    ans = 0

    # Iterate through the array
    for i in range(n):
        if vis[i]:
            continue

        # Ensure j is within bounds before accessing ar[j]
        while j < n and ar[i] * x > ar[j]:
            j += 1

        # If the current element multiplied by x matches ar[j]
        if j < n and ar[i] * x == ar[j]:
            vis[j] = True
            j += 1
        else:
            ans += 1

    print(ans)

if __name__ == "__main__":
    main()

Problem Statement 2

Polycarp was learning a new data structure called an array. He found that arrays are a continuous allocation of the same type of data. He later began playing with different integer arrays. He was solving problems involving integer arrays from a famous problem portal on the internet when he got stuck on a problem. He approached you for help.

You are given an array arr; you need to find the number of empty subarrays in the given array. An empty subarray is defined as a non-empty array whose sum is zero. Help Polycarp in solving the problem.

Input Format

The first line of input contains an integer N representing the number of elements in the array.

The following line contains N integers arr[i] denoting the elements of the given array.

Output Format

Print the number of empty subarrays present in the given array.

Solution C++

#include<bits/stdc++.h>
using namespace std;
 
int main()
{
   int n;
   cin>>n;
   vector<int> arr(n);
   for(int i=0; i<n; i++)
      cin>>arr[i];

   int cnt = 0;
   unordered_map<int,int> mp; //map to store the prefix sum 
   mp[0]++; //storing the sum of empty array
   int pref = 0; //for calculation of the prefix
   for(int i=0; i<n; i++)
   {
      pref += arr[i]; 
      if(mp.count(pref)) //if previous pref existed means there exists a subarray whose sum is 0
         cnt += mp[pref]; //incrementing the number of such subarrays
      mp[pref]++; //hashing the current value of prefix
   }
   cout<<cnt<<endl; //printing the ans
   return 0;
}

Solution Java

import java.util.HashMap;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        
        // Input the size of the array
        int n = sc.nextInt();
        
        // Input the array elements
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = sc.nextInt();
        }
        
        int cnt = 0;  // Initialize the count of subarrays with sum 0
        
        // HashMap to store the prefix sum and its frequency
        HashMap<Integer, Integer> mp = new HashMap<>();
        
        mp.put(0, 1);  // Storing the sum of an empty array
        
        int pref = 0;  // For calculating the prefix sum
        
        for (int i = 0; i < n; i++) {
            pref += arr[i];  // Calculating prefix sum
            
            // If the prefix sum already exists, increment the count
            if (mp.containsKey(pref)) {
                cnt += mp.get(pref);
            }
            
            // Update the frequency of the current prefix sum
            mp.put(pref, mp.getOrDefault(pref, 0) + 1);
        }
        
        // Output the count of subarrays with sum 0
        System.out.println(cnt);
    }
}

Solution Python

# Input the size of the array
n = int(input())

# Input the array elements
arr = list(map(int, input().split()))

cnt = 0  # Initialize the count of subarrays with sum 0

# Dictionary to store the prefix sum and its frequency
mp = {0: 1}  # Storing the sum of an empty array

pref = 0  # For calculating the prefix sum

for i in range(n):
    pref += arr[i]  # Calculating prefix sum
    
    # If the prefix sum already exists, increment the count
    if pref in mp:
        cnt += mp[pref]
    
    # Update the frequency of the current prefix sum
    mp[pref] = mp.get(pref, 0) + 1

# Output the count of subarrays with sum 0
print(cnt)

Problem Statement 3

You have N boxes numbered from 1 to N. Each box contains a certain number of balls. Additionally, you have M spare balls, and you can choose to put all of them into any one of the boxes.

For each box, determine if adding all the spare balls to that box will make it the box with the largest number of balls. Note that multiple boxes can have the maximum number of balls.

Input Format

The first line of the input contains two integers N (number of boxes) and M (number of spare balls).

The second line of the input contains N space-separated integers a1, a2, …, an, where ai represents the number of balls in the ith box.

Output Format

For each box, print “1” if it will be the box with the largest number of balls after adding the spare balls and “0” if not.

Solution C++

#include <bits/stdc++.h>
using namespace std;

int main(){
    // read inputs
    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    for(int i = 0; i < n; i++)
        cin >> a[i];
    
    // vector to store results
    vector<int> res;

    // maximum element in the given array
    int mx = *max_element(a.begin(), a.end());

    // find the answer for each element
    for(int i = 0; i < n; i++){
        bool ok = true;

        // if the maximum element in the array
        // is more than the sum of current element
        // and spare balls, answer will be false
        if(a[i] + m < mx)
            ok = false;

        // based on the result, push 1 or 0 in
        // the resultant vector
        if(ok)
            res.push_back(1);
        else
            res.push_back(0);
    }

    // print the result
    for(int i: res)
        cout << i << ' ';
    return 0;
}

Solution Java

import java.util.*;

public class Main {
    public static void main(String[] args) {
        // Read inputs
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int[] a = new int[n];

        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }

        // Array to store results
        int[] res = new int[n];

        // Find the maximum element in the array
        int mx = Arrays.stream(a).max().getAsInt();

        // Find the answer for each element
        for (int i = 0; i < n; i++) {
            boolean ok = true;

            // If the maximum element in the array
            // is more than the sum of the current element
            // and spare balls, answer will be false
            if (a[i] + m < mx)
                ok = false;

            // Based on the result, store 1 or 0 in the result array
            res[i] = ok ? 1 : 0;
        }

        // Print the result
        for (int r : res) {
            System.out.print(r + " ");
        }
    }
}

Solution Python

def main():
    import sys
    input = sys.stdin.read
    data = input().split()

    # Read inputs
    n = int(data[0])
    m = int(data[1])
    a = list(map(int, data[2:2+n]))

    # List to store results
    res = []

    # Find the maximum element in the array
    mx = max(a)

    # Find the answer for each element
    for balls in a:
        # If the maximum element in the array
        # is more than the sum of the current element
        # and spare balls, answer will be false
        if balls + m >= mx:
            res.append(1)
        else:
            res.append(0)

    # Print the result
    print(' '.join(map(str, res)))

if __name__ == "__main__":
    main()

Problem Statement 4

There are n forts in an enemy kingdom with different power levels in the array arr. The army must:

Not attack forts with a power of 0 or less.

Attack forts in increasing order of power where the power difference between the last attacked fort and the next to be attacked is exactly 1.

The army can attack any fort at any time but can only attack each fort once.

Your task is to maximize the number of attacked forts and determine the absolute difference between (last attacked fort power + 1) and the fort with the maximum power. If no fort is attacked, print -1.

Note: If there are multiple orders with the maximum number of attacked forts, then the last attacked fort will be the minimum from all such orders.

Input Format

The first line contains a single integer n denoting the number of elements in array arr.

The second line contains n space-separated integers denoting elements of array arr.

Output Format

Print the integer denoting the absolute difference between (last attacked fort power + 1) and the fort with the maximum power. If no fort is attacked, print -1.

Solution C++

#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <cstdlib> // For std::abs

using namespace std;

void calculate_difference(vector<int>& arr) {
    // Filter out forts with power less than or equal to 0
    vector<int> filtered_forts;
    for (int power : arr) {
        if (power > 0) {
            filtered_forts.push_back(power);
        }
    }
    
    if (filtered_forts.empty()) {
        cout << -1 << endl;
        exit(0);
    }
    
    // Sort the filtered forts in increasing order
    sort(filtered_forts.begin(), filtered_forts.end());
    
    int max_length_sequence = 0;
    int current_length = 0;
    int last_attacked = -1;
    int final_last_attacked = -1;
    
    for (int power : filtered_forts) {
        if (last_attacked == -1 || power - last_attacked == 1) {
            current_length++;
            last_attacked = power;
        } else {
            current_length = 1;
            last_attacked = power;
        }
        
        if (current_length > max_length_sequence) {
            max_length_sequence = current_length;
            final_last_attacked = last_attacked;
        }
    }
    
    int max_power = filtered_forts.back();
    int difference = abs(final_last_attacked + 1 - max_power);
    
    cout << difference << endl;
}

int main() {
    int n;
    cin >> n;
    
    set<int> arr_set;
    for (int i = 0; i < n; ++i) {
        int power;
        cin >> power;
        arr_set.insert(power);
    }
    
    vector<int> arr(arr_set.begin(), arr_set.end());
    
    // Calculating the difference
    calculate_difference(arr);
    
    return 0;
}

Solution Java

import java.util.*;

public class Main {

    public static void calculateDifference(List<Integer> arr) {
        // Filter out forts with power less than or equal to 0
        List<Integer> filteredForts = new ArrayList<>();
        for (int power : arr) {
            if (power > 0) {
                filteredForts.add(power);
            }
        }

        if (filteredForts.isEmpty()) {
            System.out.println(-1);
            return;
        }

        // Sort the filtered forts in increasing order
        Collections.sort(filteredForts);

        int maxLengthSequence = 0;
        int currentLength = 0;
        Integer lastAttacked = null;
        int finalLastAttacked = 0;

        for (int power : filteredForts) {
            if (lastAttacked == null || power - lastAttacked == 1) {
                currentLength += 1;
                lastAttacked = power;
            } else {
                currentLength = 1;
                lastAttacked = power;
            }

            if (currentLength > maxLengthSequence) {
                maxLengthSequence = currentLength;
                finalLastAttacked = lastAttacked;
            }
        }

        int maxPower = Collections.max(filteredForts);
        int difference = Math.abs((finalLastAttacked + 1) - maxPower);

        System.out.println(difference);
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        
        // Taking input from the user
        int n = scanner.nextInt();
        Set<Integer> arrSet = new HashSet<>();
        for (int i = 0; i < n; i++) {
            arrSet.add(scanner.nextInt());
        }
        
        List<Integer> arr = new ArrayList<>(arrSet);

        // Calculating the difference
        calculateDifference(arr);

        scanner.close();
    }
}

Solution Python

def calculate_difference(arr):
    # Filter out forts with power less than or equal to 0
    filtered_forts = [power for power in arr if power > 0]
    
    if not filtered_forts:
        print(-1)
        exit(0)
    
    # Sort the filtered forts in increasing order
    filtered_forts.sort()
    
    max_length_sequence = 0
    current_length = 0
    last_attacked = None
    
    for power in filtered_forts:
        if last_attacked is None or power - last_attacked == 1:
            current_length += 1
            last_attacked = power
        else:
            current_length = 1
            last_attacked = power
        
        if current_length > max_length_sequence:
            max_length_sequence = current_length
            final_last_attacked = last_attacked
    
    max_power = max(filtered_forts)
    difference = abs((final_last_attacked + 1) - max_power)
    
    print(difference)


# Taking input from the user
n = int(input())
arr = list(map(int, input().split()))
arr = set(arr)
arr = list(arr)



# Calculating the difference
calculate_difference(arr)

Problem Statement 5

The teacher needs to choose a class representative for her class. She asks you to find the eligible students based on their marks. You are given the array arr that represents the marks of the N students and an integer K.

Eligibility criteria:

Only K number of students having greater marks will be chosen.

Students whose marks are multiples of 5 are not eligible.

You are required to print the marks of the eligible students.

Note: You need to return the marks in ascending order.

Input Format

The first line contains a single integer N.

The second line contains N space-separated integers denoting the array arr.

The third line contains an integer K.

Output Format

Print the marks in ascending order.

Solution C++

#include <bits/stdc++.h>
using namespace std;
// Function to swap integers present at ith and jth index
void swap(vector<int> &arr, int i, int j) 
{
    int tmp = arr[i];
    arr[i] = arr[j];
    arr[j] = tmp;
}

// This function will put Kth largest element at its correct position similar to quicksort partition algorithm
int choose(vector<int> &arr, int low, int high, int k) 
{
    int pivot = low;
    for (int j = low; j < high; j++) 
    {
        if (arr[j] <= arr[high]) 
        {
            swap(arr, pivot++, j);
        }
    }
    
    swap(arr, pivot, high);
    // Count the numbers that are greater then pivot from high
    int count = high - pivot + 1;
    // If pivot is the Kth largest then return it
    if (count == k) 
    {
        return arr[pivot];
    }
    // If pivot is too small, so Kth largest must be on the right
    if (count > k) 
    {
        return choose(arr, pivot + 1, high, k);
    } 
    else
    {
        // If pivot is too big, so Kth larges must be on the left
        return choose(arr, low, pivot - 1, k - count);
    }
}

vector<int> eligible(vector<int> &arr, int k, int n)
{
    // It will store K largest element
    vector<int> answer(k);

    // Put the Kth largest at its correct position and partition the array around it
    choose(arr, 0, n - 1, k);

    // Putting K largest element in answer
    for (int i = n - k, j = 0; j < k; ++i, ++j)
    {
        answer[j] = arr[i];
    }
    
    // But these integers are present in random order so we need to sort them
    sort(answer.begin(), answer.end());

    return answer;
}

vector<int> remove_divisible(vector<int>&arr)
{
  vector<int>temp;
  for(auto i : arr)
  {
    if(i%5!=0)
    temp.push_back(i);
  }
  return temp;
}

int main() 
{
   int n;
   cin>>n;
   vector<int>ans(n);
   for(int i=0;i<n;i++)
   cin>>ans[i];
   int k;
   cin>>k;
   ans = remove_divisible(ans);
   vector<int>answer;
   answer=eligible(ans,k,n);
   for(int i=0;i<answer.size();i++)
   {
     cout<<answer[i]<<" ";
   }
   return 0;
}

Solution Java

import java.util.*;

public class Main {

    // Function to swap elements in the array
    public static void swap(List<Integer> arr, int i, int j) {
        int temp = arr.get(i);
        arr.set(i, arr.get(j));
        arr.set(j, temp);
    }

    // Quickselect to find the kth largest element
    public static int choose(List<Integer> arr, int low, int high, int k) {
        int pivot = low;
        for (int j = low; j < high; j++) {
            if (arr.get(j) <= arr.get(high)) {
                swap(arr, pivot++, j);
            }
        }
        swap(arr, pivot, high);
        int count = high - pivot + 1;

        if (count == k) {
            return arr.get(pivot);
        }
        if (count > k) {
            return choose(arr, pivot + 1, high, k);
        } else {
            return choose(arr, low, pivot - 1, k - count);
        }
    }

    // Function to get k largest eligible marks
    public static List<Integer> eligible(List<Integer> arr, int k) {
        int n = arr.size();
        List<Integer> answer = new ArrayList<>(k);
        
        // Put the k largest elements at their correct positions
        choose(arr, 0, n - 1, k);

        // Copy the last k elements (the largest ones) to the answer
        for (int i = n - k; i < n; i++) {
            answer.add(arr.get(i));
        }

        // Sort the answer before returning
        Collections.sort(answer);
        return answer;
    }

    // Function to remove marks divisible by 5
    public static List<Integer> removeDivisible(List<Integer> arr) {
        List<Integer> temp = new ArrayList<>();
        for (int mark : arr) {
            if (mark % 5 != 0) {
                temp.add(mark);
            }
        }
        return temp;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        List<Integer> arr = new ArrayList<>(n);
        for (int i = 0; i < n; i++) {
            arr.add(sc.nextInt());
        }
        
        int k = sc.nextInt();
        
        // Remove marks divisible by 5
        arr = removeDivisible(arr);
        
        // Get the eligible marks
        List<Integer> answer = eligible(arr, k);
        
        // Print the result
        for (int mark : answer) {
            System.out.print(mark + " ");
        }
    }
}

Solution Python

def swap(arr, i, j):
    arr[i], arr[j] = arr[j], arr[i]

# Quickselect algorithm to find the kth largest element
def choose(arr, low, high, k):
    pivot = low
    for j in range(low, high):
        if arr[j] <= arr[high]:
            swap(arr, pivot, j)
            pivot += 1
    swap(arr, pivot, high)
    count = high - pivot + 1

    if count == k:
        return arr[pivot]
    if count > k:
        return choose(arr, pivot + 1, high, k)
    else:
        return choose(arr, low, pivot - 1, k - count)

# Function to return k largest eligible marks
def eligible(arr, k):
    n = len(arr)
    # Get the kth largest element using quickselect
    choose(arr, 0, n - 1, k)

    # Collect the k largest elements
    answer = arr[-k:]

    # Sort the answer before returning
    answer.sort()
    return answer

# Function to remove marks divisible by 5
def remove_divisible(arr):
    return [x for x in arr if x % 5 != 0]

def main():
    # Input number of students
    n = int(input())
    
    # Input the marks of the students
    arr = list(map(int, input().split()))
    
    # Input the value of k
    k = int(input())
    
    # Remove marks divisible by 5
    arr = remove_divisible(arr)
    
    # Get the eligible marks
    answer = eligible(arr, k)
    
    # Print the eligible marks in ascending order
    print(" ".join(map(str, answer)))

if __name__ == "__main__":
    main()

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Here are some proven strategies to help you prepare effectively:

Learn by Solving Problems:
Regular practice on free, available online platforms to improve your problem-solving speed and increase your familiarity with common coding patterns.
Master the Basics:
Ensure you have a strong foundation in data structures (arrays, linked lists, trees) and algorithms (searching, sorting, dynamic programming). These concepts are essential for tackling difficult coding problems.
Mock Assessments:
Taking mock tests simulates real exam conditions and improves your time management.
Review Past Coding Questions:
Analyzing previous years’ questions will help you recognize recurring patterns and types of problems. This also gives you an idea of the difficulty level.
Understand Problem-Solving Techniques:
It’s not just about writing code. Break down each problem, develop a plan, and then code it. Try to write clean, efficient code, as readability is important.

Conclusion

The journey from a fresher to a full-fledged corporate professional in a prestigious company like Cognizant is filled with learning and growth. The GenC Elevate coding assessment serves as an important milestone in this journey. By focusing on the right topics, practicing consistently, and honing your coding skills, you can ace this assessment and unlock a world of opportunities in one of the leading tech companies globally.

Stay focused, practice smartly, and success will follow. Best of luck with your GenC Elevate journey!

Frequently Asked Questions (FAQs)

1. What is the difficulty level of the GenC Elevate coding assessment?

The questions range from moderate to high difficulty. They test your coding ability, as well as your problem-solving skills and efficiency.

2. How many coding questions are there in the GenC Elevate assessment?

There are four coding questions to be solved within 60 minutes. Time management is key to success.

3. Which programming languages are allowed in the assessment?

Java, Python, C, and C++ are allowed, though it's recommended to be proficient in Java or Python.

4. How do I improve my coding skills for this assessment?

Regular practice on coding platforms, solving mock assessments, and revisiting core data structures and algorithms are the most effective preparation methods.

Disclaimer: While we strive for accuracy, we do not guarantee its completeness or reliability. Readers are encouraged to verify all facts and statistics from the official company website or check independently before making decisions.