TCS CodeVita Questions

Approach

To solve this problem, we will first traverse the list and store the points in a list with an increment of 3 from 3 to 3*n-1. We will check if the x or y coordinate is the same as the previous point and z coordinate is the same as the previous point if so the bee will travel in an arc else the bee will always take the shortest route.

Algorithm

• We are storing the point in a list and traversing the list with an increment of 3 from index 3 to 3*n-1.
• To calculate distance, if the z-axis coordinate is the same as the previous point and any of the x or y coordinates is the same as the previous point, the bee will travel in an arc.
• Otherwise, the bee will take the shortest route.

Implementation in C++

#include <bits/stdc++.h>
using namespace std;
#define PIE 3.14
int sx,sy,sz;
float shortDist(int x,int y,int z)
{
    float dist;
    // checking if the Z-axis and any other one axis are the same.
    if(z == sz && ( y == sy || x == sx) && sz != 0) {
    //checking if the x axis of next co-ordinate is same
        if(x != sx){  
        dist = (2 * PIE * (abs(x-sx)))/6.0;
        }
        //checking if the y axis of next co-ordinate is same 
        else
        dist = (2 * PIE * (abs(y-sy)))/6.0;
    }
    //else meaning bee is moving to another face
    else{ 
    // finding eculidean distance between x and y and the abs distance of Z axis
    dist = (int)(sqrt(pow(x-sx,2) + pow(y-sy,2)) + abs(z-sz)); 
    }
    sx = x;sy = y;sz = z;
    return dist;
}
int main()
{
    int n,arr[50],x,y,z;
    float sum = 0.0;
    cin>>n;
    n = 3 * n;
    for(int i = 0;i < n;i++)
    {
        cin>>arr[i];
    }
    sx = arr[0];
    sy = arr[1];
    sz = arr[2];
    for(int i=3;i<n;i+=3)
    {
        sum += shortDist(arr[i],arr[i+1],arr[i+2]);
    }
    printf("%.2f",sum);
} 

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Input

3
1 2 10 2 1 10 1 1 9

Output

3.00

Implementation in Python

import math 
PIE=3.14
def shortDist( x,y,z,sx,sy,sz):
    dist = 0.0
    # if the x or y axis same as Z-axis
    if(z == sz and (y == sy or x==sx) and sz != 0): 
    # if the x axis of next co-ordinate is same 
        if(x != sx):    
            dist = (2 * PIE * (abs(x - sx)))/6.0
    # if the y axis of next co-ordinate is same
        else:  
            dist = (2 * PIE * (abs(y -sy)))/6.0    
    # else means bee is going to another face
    #finding eculidean distance
    else: 
        dist = int((math.sqrt(pow(x-sx,2) + pow(y-sy,2)) + abs(z - sz)))          
    #new starting cordinates
    sx = x
    sy = y
    sz = z  
    return dist,sx,sy,sz
n=int(input())
arr = [int(x) for x in input().split()]
#starting co-ordinate
sx = arr[0]
sy = arr[1]
sz = arr[2]
ans=0
for j in range(3,3*n,3):
    x,sx,sy,sz= shortDist(arr[j],arr[j+1],arr[j+2],sx,sy,sz)
    ans += x
print(round(ans,2))

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Input

3
2 2 10 2 1 10 2 1 8

Output

3.05

Implementation in Java

import java.util.*;
import java.lang.*;
import java.io.*;
class Main{
    final static float PI = 3.14f;
    static int sx, sy, sz;
    public static void main(String[] args) {
        int i, N, x, y, z;
        int arr[] = new int[50];
        float sum = 0.0f;
        Scanner sc = new Scanner(System.in);
        N = sc.nextInt();
        N = 3 * N;
        for(i = 0; i < N; i++){
            arr[i] = sc.nextInt();
        }
        sx = arr[0];
        sy = arr[1];
        sz = arr[2];
        for(i = 3; i < N ; i += 3){
            sum += shortDist(arr[i], arr[i+1], arr[i+2]);
        }
        System.out.printf("%.2f", sum);
    }
    private static float shortDist(int x, int y, int z) {
        float dis;
        if(z == sz && (y == sy || x == sx ) && sz != 0){
            if(x != sx){
                dis = (2 * PI * (Math.abs(x - sx))) / 6.0f;
            }else{
                dis = (2 * PI * (Math.abs(y - sy))) / 6.0f;
            }
        }else{
            dis = (int)(Math.sqrt(Math.pow(x - sx, 2) + Math.pow(y - sy, 2)) + Math.abs(z - sz));
        }
        sx = x;
        sy = y;
        sz = z;
        return dis;
    }
}

You can also try this code with Online Java Compiler

Run Code

Input

3
2 2 10 2 1 10 2 1 9

Output

2.05

Frequently Asked Questions

What is TCS CodeVita?

CodeVita is an online coding competition with three significant rounds: the Pre-Qualifier Round, the Qualifier Round, and the Grand Finale. 1. Pre-Qualifier Round: This will be an online coding competition in which participants must answer questions in 6 hours.

What is the eligibility to participate in TCS CodeVita?

Students from India's institutes who will complete their academic programs in 2022, 2023, 2024, and 2025 are eligible for the CodeVita competition.

What are the languages we can code in the exam?

Following are the languages we can use: C, C++, C#, Java, Perl, PHP, Python, and Ruby.

Conclusion

TCS organizes CodeVita every year, in which any graduate student can participate and give their all. It is used to encourage competitive programming and to help you improve your programming skills. In addition, top scorers are allowed to interview for job openings with TCS. In this blog, we have solved a sample question from TCS Codevita. 

Do you know TCS conducts one more placement exam to hire freshers - TCS NQT, refer to the blog TCS NQT Recruitment Process by Coding Ninjas, which has a stepwise explanation of the complete recruitment process.

You can also consider our Competitive Programming Course to give your career an edge over others!