presentation error tcs codevita

CodeVita Programming Contest

CodeVita 2013 Talks

Thursday, October 3, 2013

Adieu codevita2013.., fun facts of codevita 2013.

•   From Round-1 60K submissions we got some interesting facts about language inclination of Indian students. The pie chart below will show number of submissions in Round-1 w.r.t languages in which they were submitted

• CodeVita   season II was a huge success by having more than 1.3Lakh individual registrations in its account. If we divide number of user on basis of Recruitment branch of TCS. Mumbai Region was winner which defeated last year winner Chennai.

Sunday, August 11, 2013

Faqs-codevita 2k13.

• Firstly, ensure that you are using the same compiler that the CodeVita system is using. For e.g. do not use Turbo C and expect the code to be compiled on CodeVita systems which uses gcc.
• Secondly, even if you are using the same compiler, check if you are using a version higher than what CodeVita system is using. For e.g if CodeVita system is using Oracle JDK 1.6 and you are using  Oracle JDK 1.7, its likely that your code may fail to compile because new features of 1.7 are not available in 1.6.
• Finally, if you using same compiler of the same version that the CodeVita system is using, it is still possible to get a Compile Time Error.This is because CodeVita systems will use certain compilation flags which you may not have used in your system. In such cases, the Compile-time error message that CodeVita systems will depict should be carefully studied. If you pay enough attention to the error message, you should be able to figure out cause of CTE and then eliminate it.
• minimum time it takes to solve the given problem
• efficiency - ie minimum system resource consumption.
• If, A starts the contest at 3 PM, the contest end time for Team1 is 9 PM. It does not matter when B starts the contest.
• Similarly, If B starts the contest at 3 PM, the contest end time for Team1 is 9 PM. It does not matter when A starts the contest.
• Contest is started when the first member, either A or B, enters the Coding Arena.

Wednesday, August 7, 2013

Codevita process pseudo-code, tuesday, august 6, 2013, codevita 2k12 – a 1-2-4 experience.

• Read the problem carefully and consider all the cases (even the worst ones you can think of).
• File handling, Error handling, I/O operations and Input validations are the areas you should be comfortable with.
• Avoid wasting time searching for the solutions of the questions on internet. Rather try to work out on your own and write your own code.
• Plagiarism is prohibited, so don’t share you code even with your best friend as it may land up both of you in trouble and may even lead to disqualification.
• Use the same version of compilers as specified as those will be used on the online judge irrespective of the version you are using.
• Never be in hurry and don’t panic. Stay calm while solving the problems.

CodeVita 2012 - Team Chennai

Sunday, august 4, 2013, codevita guidelines from winners of 2012, thursday, august 1, 2013, a wonderful journey called codevita.

Useful Resources to Prepare for TCS CodeVita 2024

Team CodeQuotient / August 9, 2023 / Software Engineering Bootcamp

Are you a student who is passionate about coding and wants to showcase your skills on a global platform? If yes, then you should definitely participate in TCS CodeVita 2024 , one of the largest and most prestigious coding contests in the world.

TCS CodeVita is an initiative by Tata Consultancy Services (TCS) , a leading IT company, to identify and nurture talented coders and provide them with career opportunities in the tech industry.

However, with the increasing competitiveness and demand for coding skills, you need to prepare well to ace this contest and stand out from the crowd. According to a report by NASSCOM , of the 3 million joining the IT workforce, only 25% of graduates with engineering backgrounds are employable.

To triumph in TCS CodeVita, it’s essential to sharpen your programming skills and problem-solving prowess.

This article will guide you through the essentials of TCS CodeVita, the significance of thorough preparation, and the valuable resources available for your journey. Let’s get started!

Think, Code, Conquer: Understanding TCS CodeVita 2024

TCS CodeVita is a global coding contest for students from various disciplines and backgrounds. It aims to test your coding skills, logical thinking, and creativity through a series of challenging and fun problems.

The contest is open to students from all over the world who are enrolled in undergraduate or postgraduate programs. And the top three winners of the Grand Finale get a chance to win a cash prize of up to $20,000 .

The format of the contest is as follows:

1. Pre-Qualifier Zonal Round

This round is conducted online and consists of six hours of coding . You can choose any six-hour slot from the available time slots. You have to solve 6 problems of varying difficulty levels using any of the 15 programming languages supported by the contest platform . The top performers from each zone qualify for the next round.

2. Qualifier Round

This round is also conducted online and consists of six hours of coding . You must solve 9 problems of varying difficulty levels using any of the 15 programming languages supported by the contest platform. The top performers from this round qualify for the Grand Finale.

3. Grand Finale

This round is conducted offline at a designated venue and consists of six hours of coding . You have to solve 10 problems of varying difficulty levels using any of the 15 programming languages supported by the contest platform. The top three performers from this round are declared the contest winners .

The rules of the contest are as follows:

• You must register for the contest on the official website using your college email ID.
• You have to participate individually rather than as a team.
• You have to adhere to the code of conduct and avoid any form of plagiarism or cheating.
• You have to submit your solutions within the stipulated time limit and follow the input/output specifications.
• You have to pass all the test cases for each problem to get full marks.

The evaluation criteria for each problem are based on the following:

• Accuracy: The number of test cases passed by your solution.
• Time complexity: The efficiency of your solution in terms of time taken to execute.
• Memory usage: The amount of memory consumed by your solution during execution.
• Code quality: The readability, modularity, and maintainability of your code.

Mastering CodeVita: How Preparation Can Make the Difference in Your Performance

Preparing for TCS CodeVita 2024 is not only necessary but also essential if you want to excel in this contest and impress potential employers. By preparing well, you can:

• Enhance your coding skills and learn new concepts and techniques.
• Improve your problem-solving abilities and logical thinking.
• Gain exposure to real-world coding problems and scenarios.
• Boost your confidence and self-esteem as a coder.
• Showcase your talent and potential on a global platform.
• Win attractive prizes and rewards.
• Get an opportunity to work with TCS or other reputed IT companies.

How to Prepare for TCS CodeVita 2024: Online Resources to Boost Your Coding Skills

There are many resources available online that can help you prepare for TCS CodeVita effectively. Some of these resources are:

1. Online Coding Platforms

One of the best ways to practice coding is by using online coding platforms that offer a wide range of coding challenges for different skill levels and topics.

These platforms allow you to test your code against various test cases, get instant feedback, compare your solutions with others, and learn from experts.

Some of the popular online coding platforms that you can use are:

I. LeetCode

LeetCode is one of the most popular and trusted online coding platforms that provides over 2000 coding problems covering various topics such as data structures, algorithms, databases, system design, and more. You can choose from 14 programming languages and use the built-in code editor and debugger to write and run your code.

II. HackerRank

HackerRank is another leading online coding platform that provides over 1500 coding problems covering various domains such as data structures, algorithms, mathematics, artificial intelligence, databases, and more. You can choose from 35 programming languages and use the online code editor and compiler to write and test your code.

2. TCS CodeVita 2024 Official Resources

Another useful resource for preparing for TCS CodeVita 2024 is the official website of the contest. Here you can find all the information related to the contest, such as registration, schedule, format, rules, syllabus, prizes, and more.

You can also access the participants’ sample problems and practice sets provided by TCS. These problems are similar to the ones that appear in the actual contest and can help you get familiar with the difficulty level and style of the questions. You can also check your solutions against the test cases and get feedback on your performance.

3. Previous Year Questions

One of the best ways to prepare for any exam or contest is by solving the previous year’s questions. This helps you understand the pattern and trends of the questions, identify your strengths and weaknesses, and improve your speed and accuracy. You can find previous year’s CodeVita questions on various online platforms such as:

I. GeeksforGeeks

This is a comprehensive online platform that provides tutorials, articles, videos, quizzes, and interview questions on various topics related to computer science and programming. You can find previous years’ CodeVita questions, along with their solutions and explanations, on this platform.

II. CodeVita Zone

CodeVita Zone is a dedicated online platform that provides previous years’ CodeVita questions along with their solutions and test cases. You can also find tips and tricks, FAQs, blogs, and videos related to CodeVita on this platform.

III. GitHub repositories

GitHub is a web-based platform that allows developers to host and share their code projects using version control systems. Several GitHub repositories contain previous year’s CodeVita questions along with their solutions in different programming languages.

4. Online Tutorials and Coding Courses

Another useful resource for preparing for TCS CodeVita is online tutorials and coding courses that can help you learn new concepts and techniques related to coding.

These tutorials and courses are designed by experts and instructors who have experience in teaching coding and competitive programming. Some of the popular online platforms that offer tutorials and courses are:

I. Coursera

Coursera is one of the largest online learning platforms, offering courses on data structures, algorithms, competitive programming, and more.

edX presents courses on data structures, algorithms, competitive programming, etc., on this platform.

Udemy is an online marketplace that lets you pursue courses on data structures, algorithms, competitive programming, etc., on this platform.

Seize the CodeVita Moment: Prepare, Participate, Prosper!

TCS CodeVita 2024 is an excellent opportunity for students passionate about coding who want to showcase their skills on a global platform. It is also a valuable experience that can help you enhance your coding proficiency and problem-solving abilities.

However, to succeed in this contest, you need to prepare well and use the resources and strategies discussed in this article. Another option you can consider is joining the CodeQuotient School of Technology .

Our Bachelor of Computer Application (BCA) + UG Program in Software Engineering can help you learn data structures, algorithms, web development, etc., from experienced mentors and instructors.

Get in touch with us now and take your coding skills to the next level!

Get UGC Approved, BCA Degree And Earn While You Do That

Join CodeQuotient's Under-Graduate Program In Software Engineering

Get Paid Internship with Our Hiring Partners to Sponsor Your Fees

100% Placement Assistance

You will also like:

• Fueling the Passion for Programming: TCS CodeVita and Its Role
• All you need to know about Codevita!
• TCS CodeVita – Decoding The Coding Challenge Syllabus
• 10 Last Minute Preparation Tips for TCS CodeVita
• TCS CodeVita 2023: All Frequently Asked Questions (Answered)

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Hours to prove your mettle

USD in prize money

Global Ranking List to top coders

Get ready to unlock the world of programming excitement with CodeVita! It's more than just a contest; it's a vibrant culture that brings people from diverse backgrounds and cultures together, shattering physical and cultural boundaries. At TCS, we believe that programming is a thrilling adventure, and that's why we created CodeVita!

Started in 2012 in India, this coding extravaganza aimed to showcase the countless applications of coding while identifying brilliant programming talent. The overwhelming response from the first two domestic seasons fueled our ambition, and in 2014, we proudly went global. The inaugural global edition featured 6 teams in the finals, proving that CodeVita had become a truly global phenomenon.

Fast forward to Season 10, and we witnessed a jaw-dropping 3,05,000+ registrations from 87 countries, representing over 3500 institutes. Imagine participants from 10 regions engaging in a thrilling battle during the grand finale! Are you intrigued? Brace yourself for more exhilarating programming challenges and unforgettable experiences with CodeVita!

What's in it for students?

• Top 3 coders to win total prize money of USD 20,000/-
• Chance to explore exciting careers * with one of the world's most powerful brands
• Chance to compete with some of the best coders in the world
• Platform to showcase your programming skills
• Finalists stand a chance to travel ** to India for the season 11 live grand finale experience.

* as applicable in the respective geographies

** subject to TCS guidelines

Eligibility

• With year of completion 2024, 2025, 2026, and 2027.
• From institutes across the globe.
• Pursuing undergraduate/diploma/postgraduate disciplines from engineering/science background with any specialization.

The Journey

Journey of a thousand miles begins with one step.

Registration

The first step in your CodeVita journey. Click on the register button to get started.

MockVitas are just like actual rounds to give you a demo of the actual contest.

Clear the Actual Rounds to move further in your CodeVita journey.

Win prize money and prove that you are the best coder.

Sample Questions

Sorting boxes, sport stadium, water cistern, square free numbers, codu and sum love, obstacle game.

A solid cube of 10 cm x 10cm x 10 cm rests on the ground.   It has a beetle on it, and some sweet honey spots at various locations on the surface of the cube.  The beetle starts at a point on the surface of the cube, and goes to the honey spots in order along the surface of the cube.

Problem Description

1.       If it goes from a point to another point on the same face (say X to Y), it goes in an arc of a circle that subtends an angle of 60 degrees at the centre of the circle

2.       If it goes from one point to another on a different face, it goes by the shortest path on the surface of the cube, except that it never travels along the bottom of the cube

The beetle is a student of Cartesian geometry, and knows the coordinates (x, y, z) of all the points it needs to go to.   The origin of coordinates it uses is one corner of the cube on the ground, and the z axis points up.   Hence, the bottom surface (on which it does not crawl) is z=0, and the top surface is z=10.   The beetle keeps track of all the distances travelled, and rounds the distance travelled to two decimal places once it reaches the next spot, so that the final distance is a sum of the rounded distances from spot to spot.

The first line gives an integer N, the total number of points (including the starting point) the beetle visits

The second line is a set of 3N comma separated non-negative numbers, with up to two decimal places each.   These are to be interpreted in groups of three as the x, y, z coordinates of the points the beetle needs to visit in the given order.

One line with a number giving the total distance travelled by the beetle accurate to two decimal places.   Even if the distance travelled is an integer, the output should have two decimal places.

Constraints

None of the points the beetle visits is on the bottom face (z=0) or on any of the edges of the cube (the lines where two faces meet)

2<=N<=10

Difficulty Level

Time limit (secs).

1,1,10,2,1,10,0,1,9

Explanation

There are three points visited by the beetle (N=3). The beetle starts on the top face of the cube (z=10) at point (1,1,10) and goes to another point on the same face (2,1,10).   Though the straight line distance is 1, it travels on the arc of a circle subtending an angle of 60 degrees at the centre of the circle, and hence travels (2*pi)/6 or 1.05 (note that it rounds the distance at each leg of the journey).   It then travels from (2,1,10) on the face z=10 to (0,1,9) on the face x=0 along the surface of the cube. This is a distance of 3.   The total distance travelled is 1.05+3=4.05.   The output is 4.05

1,1,10,2,1,10,0,5,9

There are three points visited by the beetle (N=3). The beetle starts on the top face of the cube (z=10) at point (1,1,10) and goes to another point on the same face (2,1,10).   As before. This distance is 1.05.    It then travels from (2,1,10) on the face z=10 to (0,5,9) on the face x=0 along the surface of the cube. The shortest distance on the surface of the cube between these points is 5.   The total distance travelled is 1.05+5=6.05.   The output is 6.05.

The parcel section of the Head Post Office is in a mess.   The parcels that need to be loaded to the vans have been lined up in a row in an arbitrary order of weights.   The Head Post Master wants them to be sorted in the increasing order of the weights of the parcels, with one exception.   He wants the heaviest (and presumably the most valuable) parcel kept nearest his office.

  You and your friend try to sort these boxes and you decide to sort them by interchanging two boxes at a time.   Such an interchange needs effort equals to the product of the weights of the two boxes.  

The objective is to reposition the boxes as required with minimum effort.

The first line consists of two space separated positive integers giving the number of boxes (N) and the position of the Head Post Master’s office (k) where the heaviest box must be.

The second line consists of N space separated positive integers giving the weights of the boxes.   You may assume that no two weights are equal.

The output is one line giving the total effort taken to get the boxes in sorted order, and the heaviest in position k.

Weights <= 1000

20 50 30 80 70

There are 5 boxes (N=5) and the heaviest box must be in position 2 (k=2).   If we look at the final order (sorted, with the heaviest at position 2), it should be 20 80 30 50 70.   If we look at this, we notice that only the 50 and the 80 parcels need to be exchanged.   As this takes effort of the product of the weights, the effort is 4000.    

Further reduction can be obtained if we use the smallest package (20) as an intermediary.   If we exchange 20 with 50 (effort 1000), then with 80 (effort 1600) and back with 50 again (effort 1000), the effect is the same, with a total effort of 3600 (less th an the effort obtained by the direct move)an the effort

The results after the optimal sequence of exchanges are

50 20 30 80 70

50 80 30 20 70

20 80 30 80 70

As this takes an effort of 3600, the output is 3600.

30 20 40 80 70 60

There are 6 parcels, and the heaviest should be at position 3.   Hence the final order needs to be 20 30 80 40 60 70.   If we look at the initial position, we see that 20 and 30 need to be exchanged (effort 600), 40 and 80 need to be exchanged (effort 3200) and 60 and 70 need to be exchanged (effort 4200).   Hence the total effort is 600+3200+4200=8000.  

If we use the same approach as in Example 1, we get the following efforts

(600)    20 30 40 80 70 60

(3200) 20 30 80 40 70 60    

(1200) 60 30 80 40 70 20

(1400) 60 30 80 40 20 70

(1200) 20 30 80 40 60 70

A total effort of 7600 is obtained rather than an effort of 8000, which is the output.

It is the sports event of the year for the residents of Sportsville.   Their team had finally made it to the finals of the Bowls League Cup.

They have booked tickets for the city contingent for the same row, and the size of the contingent (N) is smaller than the number of seats booked(S).Unfortunately, there was rain the previous night and some of the seats are still wet. Some of the contingent love Bowls so much and are excited enough not to mind sitting on a wet chair. There are k of these. However, others want to sit on a dry seat so that they can enjoy the match more.

The contingent wants to minimize the distance between the first and last person in the row so that they can still conduct Mexican Waves, and other forms of support for their team.

Because they want to sit together, any block of 15 or more contiguous unoccupied seats between the first person sitting and the last person sitting is unacceptable.

There are M blocks of seats, starting with a dry block, with alternating wet and dry blocks.   The number of seats in each block is known.

Given S (the number of seats in the row), N (the size of the contingent),k (the number of the contingent who are willing to sit in a wet seat), and the distribution of wet and dry blocks, write a program to find the minimum distance between the first and the last member of the contingent in the row.

The first line contains four comma separated numbers representing S, N, k and M respectively.

The second line is a set of M comma separated numbers representing the number of seats in each block of seats.   The first block is dry, and the remaining blocks alternate between wet and dry.

One integer representing the minimum distance between the first and last member of the row.   If it is impossible to seat all the members according to their preferences,and with the unoccupied seat restriction,   the result should be 0.

S,N,k < 1000,   M < 30

Input  

3,10,30,5,30,22

S = 100, and there are 100 seats in the row.   N=50, and there are 50 members in the contingent. k=5, and 5 people (out of the 50) do not mind sitting on wet seats.   M=6, and there are 6 blocks of seats.   The number of seats in each block is 3,10,30,5,30 and 22, with the first block of 3 seats being dry, the next 10 being wet and so on.  

One possible positioning to achieve the minimum distance is to place the a set of 30 people in seats 14 to 43 (the dry block), the 5 people who do not mind sitting on wet seats in the wet block 44 to 48, and the remaining 15 people (of the 50) in the seats 49 to 63.   There is no unoccupied seat between the first person and the last person, and so this is acceptable.The distance between the last allocated seat (63) and the first allocated seat (14), is 49.   This is the output.

3,7,10,10,20,10,20,20

S = 100, and there are 100 seats in the row.   N=50, and there are 50 members in the contingent. k=5, and 5 people (out of the 50) do not mind sitting on wet seats.   M=8, and there are 8 blocks of seats.  

One possible positioning is to have a set of 10 people sit in the dry block 11 – 20, the 5 people who will accept wet seats in seats 21 – 25 (in the wet block 21 – 30), another 20 people in the dry block 31 – 50, leave the wet block 51-60 empty, and seat the remaining 15 people in seats 61 – 75 (in the dry block 61-80.   There is a block of 5 unoccupied seats (26-30) between the first person and the last person.   As this is not more than 15, this is acceptable. The distance from the last allocated seat (75) and the first allocated seat (11) is 64.   This is the result.

A cylindrical water cistern was built in an apartment complex in Aquatown. The bottom rests on concrete and is not accessible. It has a height h and a radius r.

A cylindrical water cistern was built in an apartment complex in Aquatown.

The bottom rests on concrete and is not accessible. It has a height h and a radius r,

A mathematical bug is sitting on the cistern at point A, and has established a coordinate system to cover the entire accessible area. The bug is sitting a distance s from the top of the cistern, and the nearest point at the top is B.

For a point C on the curved surface, the nearest point D on the top is determined, and the distance CD is taken as t. The angle p (in degrees) subtended at the centre of the circle E by the arc BD is measured (in a counterclockwise manner). The coordinates of C are taken as the pair (t,p), with t being greater than 0 and less than h, and with p being between 0 and 359 (inclusive).

For a point on the top surface, F, the distance to the centre E is taken (a), and the counterclockwise angle (in degrees) between EF and EB is taken. The coordinates of the point F are then taken as (-a,q). The value of a is between 0 and r, and the value of q is between 0 and 359.

All coordinates are integers, and if the point is on the top surface of the cylinder, the first coordinate is negative, and if it is on the curved surface of the cylinder, the first coordinate is positive.

From its staring point A, the bug needs to go to its destination, which is a point (like C or F) either on the curved surface or the top surface. The coordinates of the destination are given. The bug would like to go by the shortest path to its destination.

The goal is to determine the length of the shortest path the bug can take.

The first line has three comma separated positive integers giving r (the radius), h (the height of the cylinder) and s (the distance from the top of the starting point of the bug)

The next line has two comma separated integers (d and g) giving the coordinates of the destination. If the first integer (d) is negative, it is on top surface of the cylinder, and else it is on the curved surface of the cylinder

The output is a single integer giving the shortest distance that the bug can travel. The computed distance must be rounded to the nearest integer

40<s<=h<10000

0<=g<=359

If d is negative, d > -r

If d is positive, d < h

100,500,200

The value of r is 100, and h is 500. The distance of the bug from the top surface is 200.

The coordinates of the destination are (200,180). As the first coordinate is 200, the destination is on the curved surface (like point C), and at the same distance from the top surface as the bug. As the second coordinate is 180, the destination is exactly on the other side of the cylinder at the same height as the bug, The distance is half the circumference of the cylinder, or 314. This is the output.

The coordinates of the destination are (-50,180). As the first coordinate is negative (-50), the point is on the top surface of the cylinder (like point F), and EF is 50. As the second coordinate is 180, BEF is a straight line. The distance travelled is AB + BE + EF = 200 + 100 + 50=350. This is the output.

In the theory of numbers, square free numbers have a special place.   A square free number is one that is not divisible by a perfect square (other than 1).

In the theory of numbers, square free numbers have a special place.   A square free number is one that is not divisible by a perfect square (other than 1).   Thus 72 is divisible by 36 (a perfect square), and is not a square free number, but 70 has factors 1, 2, 5, 7, 10, 14, 35 and 70.   As none of these are perfect squares (other than 1), 70 is a square free number.

For some algorithms, it is important to find out the square free numbers that divide a number.   Note that 1 is not considered a square free number.  

In this problem, you are asked to write a program to find the number of square free numbers that divide a given number.

The only line of the input is a single integer N which is divisible by no prime number larger than 19

One line containing an integer that gives the number of square free numbers (not including 1)

N    < 10^9

If we list the numbers that divide 20, they are

1, 2, 4, 5, 10, 20

1 is not a square free number, 4 is a perfect square, and 20 is divisible by 4, a perfect square.   2 and 5, being prime, are square free, and 10 is divisible by 1,2,5 and 10, none of which are perfect squares.   Hence the square free numbers that divide 20 are 2, 5, 10.   Hence the result is 3.

N=72.   The numbers that divide 72 are

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

1 is not considered square free.    4, 9 and 36 are perfect squares, and 8,12,18,24 and 72 are divisible by one of the.   Hence only 2, 3 and 6 are square free.   (It is easily seen that none of them are divisible by a perfect square).   The result is 3

Given N number of x's, perform logic equivalent of the above Java code and print the output

Scanner sc = new Scanner(System.in);

long sum = 0;

int N = sc.nextInt();

for (int i = 0; i < N; i++) {

final long x = sc.nextLong(); // read input

String str = Long.toString((long) Math.pow(1 << 1, x));

str = str.length() > 2 ? str.substring(str.length() - 2) : str;

sum += Integer.parseInt(str);

logger.debug(sum%100);

First line contains an integer N

Second line will contain N numbers delimited by space

Number that is the output of the given code by taking inputs as specified above

1<=N<=10^7

0<=x<=10^18

Given a n*n Array matrix (A) with A[0][0] element as the starting point and any one element as the destination.

Given a n*n Array matrix (A) with A[0][0] element as the starting point and any one element as the destination. Find the destination and print the route map. Rules: 1. Array Matrix with n*n elements such that n >=2 and n<=10. 2. Starting point A[0][0] value will be 'A'. 3. Destination value will be 'D' 4. There will be always 1 continuous route which can be straight or diagonal. 5. There are 4 types of hurdles and corresponding values : a. Stone denoted by 'S' b. Wall denoted by 'L' c. Water denoted by 'W' d. Thorn denoted by 'T' 6. Music provides mind peace. Which will be denoted by 'M'. It is not a hurdle. 7. The value of route will be 'R'.

First Line contains dimension N of Matrix A.

Next N Lines, each contains N values delimited by space

At every Step print the surrounded hurdles in ascending order of values. i.e. for every 'R' print the surrounding hurdles.

If there are no hurdled around step in the route, print 'NO HURDLES' for that step.

On reaching destination print 'DESTINATION'

Music 'M' is not a hurdle. It should not be included in output.

2 <= N <= 20

Example 1 Input 4 A S L D T R W R R M S R W R R M

DESTINATION

Frequently Asked Questions

In doubt? Don't worry! Check out our Frequently Asked Questions enclosed below.

Kindly refer to this blog for step by step registration process.

Yes, you can go back to the welcome page after starting to answer the coding problems.

No, you do not have to attempt the questions in any order.

After login, you can check the languages and their respective compiler/interpreter version under the 'Compilers & Interpreters' tab.

Yes, the timer will continue for Question 1.

Yes, before you submit your code, you will get an option - "I would like to provide attribution to the following sources". Select this option before submitting. You are free to refer to other material, but you need to declare the source code.

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tcs-codevita

Here are 14 public repositories matching this topic..., j33l / tcs-codevita-practice-problems.

TCS CodeVita 2020 solutions, https://practice.tcscodevita.com/IIT/index.jsp

• Updated Jul 29, 2021

rajeevranjancom / Placement_paper

Placement Paper

• Updated Oct 10, 2020

cmrcet-coding-club / codevita2020

• Updated Aug 17, 2020

amitvishw / TCSCodeVita

TCS CodeVita 2016 Submissions

• Updated Nov 27, 2017

Saptarshi-prog / Counting-Rock-Samples-Codevita

This repository contains Codevita solution of the "Counting Rock Samples".

• Updated Aug 14, 2020

Saptarshi-prog / Path-through-graph-TCS_Codevita-2020

This repository contains my solution to TCS Codevita Season 9 question of path through graph.

• Updated Sep 1, 2020

Saptarshi-prog / Consecutive-prime-sum-codevita

This repository contains my program for TCS code vita question of consecutive prime sums.

• Updated Aug 13, 2020

TharunBalaji2004 / tcs-codevita-practice

Repository for saving TCS CodeVita practice problems with solutions using C++/Java/Python 🚀

• Updated Nov 21, 2023

akatsuki-org / TCS-CodeVita-practice-problems

TCS CodeVita 2020 practice problem solutions, https://practice.tcscodevita.com/IIT/index.jsp

• Updated Aug 8, 2020

deshmukhmayur / tcs-codevita6

My solutions for TCS CodeVita6 Round 1.

• Updated Aug 11, 2017

open-for-everyone / Codevita

• Updated Aug 16, 2020

NamitNaik / CodeVita-Season-9

This repository contains codes in python for the problem statements that were asked in CodeVita Season 9

• Updated Jan 1, 2021

jainrocky / MOCKVITA-2020

• Updated Jul 18, 2020

11fenil11 / Competitive-Programming

This Repository will contain the solution of the coding problems from different Coding Platform sites like CodeForces, Hackerrank, Codechef and also Coding Competitions like Google Kickstart, Google Codejam, TCS codevita, etc.

• Updated Mar 27, 2021

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TCS Codevita 2020 Problem-A (Codekart) Code

For those of you requesting (that hardcore implementation) problem A code here it goes: It is quite self explanatory and took me 2 hrs to pass it during the contest :

what a fuck all question it was.

did it pass , mine codevita is in zone but by solving zone 1 ques i came uo with same approach ,i thought it will not pass because complexity is much please ans as soon as possible

here is the link

Yes it passed, with presentation error though.

do you have the problem statement, if yes please share