Home Get Hired TCS CodeVita Coding Questions and Solutions for Freshers 2025

Table of content:

TCS CodeVita Coding Question Overview
Common Coding Question Categories in TCS CodeVita
TCS CodeVita Question Patterns
Top 5 TCS CodeVita Coding Questions & Solutions
Tips & Strategies for TCS CodeVita
Conclusion
Frequently Asked Questions (FAQs)

TCS CodeVita Coding Questions and Solutions for Freshers 2025

Prepare for the upcoming 2025 TCS CodeVita with top coding questions, expert solutions, strategies, and tips for excelling in the global programming competition. Read on for details.

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TCS CodeVita Coding Questions and Solutions for Freshers 2025

Tata Consultancy Services (TCS) CodeVita is one of the most prestigious and challenging coding competitions for freshers. For many aspiring candidates, participating in CodeVita represents a unique opportunity to showcase their coding skills, secure job offers from a leading IT company, and stand out in a highly competitive job market.

To succeed in TCS CodeVita, freshers must prepare for a range of coding questions that test problem-solving skills, algorithmic thinking, and technical knowledge. In this article, we’ll dive into the most common TCS CodeVita coding questions, the question patterns, sample questions, and essential preparation tips to help you succeed.

TCS CodeVita Coding Question Overview

TCS CodeVita typically tests participants on a variety of coding skills, including but not limited to data structures, algorithms, and problem-solving. The questions are designed to evaluate how well candidates can apply their programming knowledge to solve complex problems in a time-constrained environment.

Common Coding Question Categories in TCS CodeVita

Category	Focus	Example Questions
*Algorithmic Problems*	Testing efficiency and logic in solving algorithms	Find the longest increasing subsequence in an array.
*Mathematical Problems*	Problems related to number theory, GCD, LCM, and prime numbers	Find the sum of all prime numbers less than 1000.
*Greedy Algorithms*	Solving problems by making the locally optimal choice at each step	Find the minimum number of coins needed for a given sum.
*Dynamic Programming*	Using dynamic programming to solve optimisation problems	Find the nth Fibonacci number using dynamic programming.
*Data Structures*	Problems that test understanding of arrays, linked lists, stacks, queues, and trees	Implement a stack using two queues.
*String Manipulation*	Handling string-related operations like reversal, substring, or pattern matching	Write a program to check if two strings are anagrams of each other.

TCS CodeVita Question Patterns

TCS CodeVita generally follows a standard pattern, with a focus on problem-solving skills and coding proficiency. The competition typically consists of multiple rounds, with the first round focused on solving easier problems and the later rounds increasing in difficulty.

Round	Focus	Duration	Difficulty Level
Round 1 (Qualifying Round)	Basic problem-solving, algorithms, and data structures	2-3 hours	Moderate
Round 2 (Finals)	Advanced problem-solving coding challenges	2-3 hours	High

Are you looking for coding assessment questions related to job placement? Click here to access coding practice sessions from moderate to challenging levels.

Top 5 TCS CodeVita Coding Questions & Solutions

Here’s a look at some example coding questions that freshers can expect in TCS CodeVita. Solving such problems will help improve your chances of performing well in the competition.

Problem Statement 1

Steve is participating in competitive coding. Whenever he makes a submission at time 't', any submission made exactly 5000 seconds before 't' is removed from the queue.

Given N submissions, where each submission is associated with a distinct time 't', determine the maximum size of the submission queue at any point in time after all 'N submissions.

Note: It is guaranteed that every submission uses a strictly larger value of ‘t’ than the previous submission.

Input Format

The first line of input contains an integer N representing the size of the array, which is the number of submissions.

The second line contains N space-separated integers representing the array arr, which contains the timestamps of each submission.

Output Format

Print an integer in a single line containing the maximum size possible.

Constraints

1 <= N <= 10^5

1 <= arr[i] <= 10^8

Solution C++


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

int main()
{  
   int n ; 
      cin >> n ; 
     vector < int > v(n); // initialize the array
     for(int i = 0; i> v[i]; // input the element of an array
     }
    
     queue < int > q; 
    int mx = 0; 
         for(int i = 0; i<n; i++){ // run a for loop 
          
           while(!q.empty() && v[i] - q.front() >= 5000){ 
             q.pop(); 
           }
           q.push(v[i]); 
           int cnt = q.size(); 
           mx = max( mx , cnt); 
         }
         cout << mx ;// print the maxmimum size
    
    return 0; 
}

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Solution Java


            import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[] v = new int[n];
        
        for (int i = 0; i < n; i++) {
            v[i] = scanner.nextInt();
        }
        
        Queue q = new LinkedList<>();
        int mx = 0;
        
        for (int i = 0; i < n; i++) {
            while (!q.isEmpty() && v[i] - q.peek() >= 5000) {
                q.poll();
            }
            q.offer(v[i]);
            int cnt = q.size();
            mx = Math.max(mx, cnt);
        }
        
        System.out.println(mx);
    }
}

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Solution Python


            from collections import deque

def main():
    n = int(input())
    v = list(map(int, input().split()))
    
    q = deque()
    mx = 0
    
    for i in range(n):
        while q and v[i] - q[0] >= 5000:
            q.popleft()
        q.append(v[i])
        cnt = len(q)
        mx = max(mx, cnt)
    
    print(mx)

if __name__ == "__main__":
    main()

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Problem Statement 2

There are N suns in a galaxy, each sun has M planets, each of the M planets have some number of moons, denoted by galaxy(i)(j), where galaxy(i)(j) denotes the number of moons of the jth planet having the ith sun.

Your task is to determine the maximum number of moons in any solar system. For each sun, calculate the total number of moons across all its planets and output the highest total number of moons found in a single solar system.

Input Format

The first line of input contains two space-separated integers, N and M, representing the number of suns and the number of planets per sun, respectively.

The next N lines of input contain M space-separated integers, representing the number of moons for each planet around the respective sun.

Output Format

Display a single integer denoting the maximum total number of moons in a solar system (i.e., across all planets orbiting the same sun).

Constraints

1<=N<=5*10^2

1<=M<=5*10^2

1<=galaxy(i)(j) <=10^4

Solution C++


            #include 
#include 
#include 
#include 
#include 
using namespace std;


int main() {
    int n,m;
    cin>>n>>m;
    int ans=0;
    for(int i=0;i<n;i++){
        int sum =0;
        for(int j=0;j<m;j++){
            int a;
            cin>>a;
            sum+=a;
        }
        ans=max(ans,sum);
    }   
    cout<<ans<<endl;
    return 0;
}

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Solution Java


            import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
       Scanner in=new Scanner(System.in);
       int a=in.nextInt();
       int b=in.nextInt();
       int i,j,sum=0,max=0;
       for(i=0;i<a;i++)
       {
        for(j=0;j<b;j++)
        {
            int k=in.nextInt();
            sum=sum+k;
        }
        if(sum>max)
        max=sum;
        sum=0;
       }
       System.out.println(max);
    }
}

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Solution Python


            # Enter your code here. Read input from STDIN. Print output to STDOUT
n,m=map(int,input().split())
gal=[]
for i in range(n):
    l=list(map(int,input().split()))
    gal.append(l)
max=0
s=0
for i in gal:
    s=sum(i)
    if s>max:
        max=s
print(max)

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Problem Statement 3

A toy manufacturing factory produced N toys in a day. The manager has a list of product IDs and another list containing the sizes of the corresponding toys, each of length N. In other words, the i-th integer in the product IDs list represents the product ID of the i-th toy, and the i-th integer in the sizes list represents the size of the i-th toy.

However, the manager accidentally combined these lists by placing the product IDs list in front of the sizes list, resulting in a single list of 2N elements.

You are an employee, and the manager wants you to recover the data and present it as another list of 2N elements such that the product ID of each toy is followed by its size.

Input Format

The first line of the input contains a single integer N, representing the number of toys produced.

The second line contains 2*N space-separated integers representing the merged list given to you by the manager.

Output Format

Print 2*N space-separated integers representing the list wanted by the manager

Constraints

1 <= N <= 10^5

1 <= ai, bi<= 10^2

Solution C++


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

int main() {
    // read inputs
    int n;
    cin >> n;
    int a[2 * n];
    for (int i = 0; i < 2 * n; i++)
        cin >> a[i];

    vector res; 

    // For the first n elements, find the corresponding element after n indices
    // and push both to the result vector
    for (int i = 0; i < n; i++) {
        res.push_back(a[i]);
        res.push_back(a[i + n]);
    }

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

    return 0; 
}

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Solution Java


            import java.util.Scanner;
import java.util.ArrayList;

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

        // Read inputs
        int n = scanner.nextInt();
        int[] a = new int[2 * n];
        for (int i = 0; i < 2 * n; i++) {
            a[i] = scanner.nextInt();
        }

        ArrayList res = new ArrayList<>();
        // For the first n elements, find the
        // size of the ith element and push them
        // at the end of result
        for (int i = 0; i < n; i++) {
            res.add(a[i]);
            res.add(a[i + n]);
        }

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

        scanner.close();
    }
}

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Solution Python


            # Enter your code here. Read input from STDIN. Print output to STDOUT
def main():
    import sys
    input = sys.stdin.read
    data = list(map(int, input().split()))

    n = data[0]
    a = data[1:]

    res = []
    # For the first n elements, find the
    # size of the ith element and push them
    # at the end of result
    for i in range(n):
        res.append(a[i])
        res.append(a[i + n])

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

if __name__ == "__main__":
    main()

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Problem Statement 4

Steve is participating in competitive coding. Whenever he makes a submission at time 't', any submission made exactly 5000 seconds before 't' is removed from the queue.

Given N submissions, where each submission is associated with a distinct time 't', determine the maximum size of the submission queue at any point in time after all 'N submissions.

Note: It is guaranteed that every submission uses a strictly larger value of ‘t’ than the previous submission.

Input Format

The first line of input contains an integer N representing the size of array which is number of submissions.

The second line contains N space separated integers representing the array arr which contains the timestamps of each submission.

Output Format

Print an integer in a single line containing the maximum size possible.

Constraints

1 <= N <= 10^5

1 <= arr[i] <= 10^8

Solution C++


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

int main()
{  
   int n ; 
      cin >> n ; 
     vector < int > v(n); // initialize the array
     for(int i = 0; i> v[i]; // input the element of an array
     }
    
     queue < int > q; 
    int mx = 0; 
         for(int i = 0; i<n; i++){ // run a for loop 
          
           while(!q.empty() && v[i] - q.front() >= 5000){ 
             q.pop(); 
           }
           q.push(v[i]); 
           int cnt = q.size(); 
           mx = max( mx , cnt); 
         }
         cout << mx ;// print the maxmimum size
    
    return 0; 
}

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Solution Java


            import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[] v = new int[n];
        
        for (int i = 0; i < n; i++) {
            v[i] = scanner.nextInt();
        }
        
        Queue q = new LinkedList<>();
        int mx = 0;
        
        for (int i = 0; i < n; i++) {
            while (!q.isEmpty() && v[i] - q.peek() >= 5000) {
                q.poll();
            }
            q.offer(v[i]);
            int cnt = q.size();
            mx = Math.max(mx, cnt);
        }
        
        System.out.println(mx);
    }
}

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Solution Python


            from collections import deque

def main():
    n = int(input())
    v = list(map(int, input().split()))
    
    q = deque()
    mx = 0
    
    for i in range(n):
        while q and v[i] - q[0] >= 5000:
            q.popleft()
        q.append(v[i])
        cnt = len(q)
        mx = max(mx, cnt)
    
    print(mx)

if __name__ == "__main__":
    main()

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Problem Statement 5

One day, Jack finds a string of characters. He is very keen to arrange the characters in reverse order, i.e., first characters become the last characters, second characters become the second-last characters, and so on. Now, he wants your help to find the kth character from the new string formed after reversing the original string.

Note: String contains only lowercase Latin letters.

Input Format

The first line contains two integers n, k — the length of array and the value of k respectively.

The second line contains a string containing n characters.

Output Format

Print a single line containing the kth character of the string.

Constraints

1 ≤ k ≤ n≤ 10^6

Solution C++


            #include<bits/stdc++.h> 
using namespace std; 
  
  
  int main(){
       int n , k; 
        cin >> n >> k; 
          string s; cin >> s; 
            
            int i = 1 , j = n-1; 
          
            cout << s[n-k]; 
         
         return 0; 
  }

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Solution Java


            import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        
        // Input reading
        int n = sc.nextInt();
        int k = sc.nextInt();
        String s = sc.next();
        
        // Output the kth character from the reversed string
        System.out.println(s.charAt(n - k));
        
        sc.close();
    }
}

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Solution Python


            # Input reading
n, k = map(int, input().split())
s = input()

# Output the kth character from the reversed string
print(s[n - k])

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Tips & Strategies for TCS CodeVita

To improve your chances of success in TCS CodeVita, freshers should focus on the following preparation strategies:

Preparation Area	Tips
Coding Platforms	Practice regularly on available coding platforms, or you can even look at the company's official website.
Data Structures & Algorithms	Master core concepts like arrays, linked lists, dynamic programming, greedy algorithms, and trees.
Problem-Solving	Solve a variety of problems in different difficulty levels to improve speed and accuracy.
Mock Tests	Take mock tests to simulate the actual exam environment and practice time management.

Conclusion

TCS CodeVita is one of the most prestigious and competitive coding competitions, providing a gateway to opportunities at TCS. The key to success in CodeVita is consistent practice and a deep understanding of algorithms, data structures, and problem-solving techniques.

By solving real-world coding problems and improving your coding efficiency, you can significantly boost your chances of securing a spot at TCS. Stay focused, practice regularly, and keep solving diverse coding problems — and you’ll be ready to conquer TCS CodeVita!

Frequently Asked Questions (FAQs)

1. What are the common types of coding questions asked in TCS CodeVita?

Expect algorithmic problems, dynamic programming, greedy algorithms, and questions based on data structures like arrays, stacks, and queues.

2. How difficult are the TCS CodeVita coding questions?

The difficulty increases as the rounds progress. The first round is moderate, but the final round can be highly challenging.

3. What is the best way to prepare for TCS CodeVita coding questions?

Practice solving problems on coding platforms, focus on algorithm design and participate in mock tests to improve speed and accuracy.

4. Can I use multiple programming languages in TCS CodeVita?

Yes, TCS CodeVita supports various programming languages, including C, C++, Java, and Python.

5. How long does it take to prepare for TCS CodeVita?

The preparation time depends on your existing coding skills. On average, 2-3 months of regular practice should be sufficient for freshers to perform well.

Keep the greatest view for the people arriving a bit late so that we don't lose the essence of touch with the crowd with the passage of the time in question.

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.

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Edited by

Kaihrii Thomas

Senior Associate Content Writer

Instinctively, I fall for nature, music, humor, reading, writing, listening, traveling, observing, learning, unlearning, friendship, exercise, etc., all these from the cradle to the grave- that's ME! It's my irrefutable belief in the uniqueness of all. I'll vehemently defend your right to be your best while I expect the same from you!

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