An automobile company manufactures both a two wheeler (TW) and a four wheeler (FW)

An automobile company manufactures both a two wheeler (TW) and a four wheeler (FW)

Problem Statement – An automobile company manufactures both a two wheeler (TW) and a four wheeler (FW). A company manager wants to make the production of both types of vehicle according to the given data below:

  • 1st data, Total number of vehicle (two-wheeler + four-wheeler)=v
  • 2nd data, Total number of wheels = W

The task is to find how many two-wheelers as well as four-wheelers need to manufacture as per the given data.

Example :

Input :

  • 200  -> Value of V
  • 540   -> Value of W

Output :

  • TW =130 FW=70

Explanation:

130+70 = 200 vehicles

(70*4)+(130*2)= 540 wheels

Constraints :

  • 2<=W
  • W%2=0
  • V<W

Print “INVALID INPUT” , if inputs did not meet the constraints.

The input format for testing 

The candidate has to write the code to accept two positive numbers separated by a new line.

  • First Input line – Accept value of V.
  • Second Input line- Accept value for W.

The output format for testing 

  • Written program code should generate two outputs, each separated by a single space character(see the example)
  • Additional messages in the output will result in the failure of test case

C++ Coading

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

 
int main () 
{
  int v, w;
  cin >> v >> w;
  float x = ((4 * v) - w) / 2;
  if ((w & 1) || w < 2 || w <= v)
    {
      cout << "INVALID INPUT";
      return 0;
    }
  cout << "TW=" << x << " " << "FW=" << v - x;

}

Java Coading

import java.util.*;
public class Solution
{
    public static void main(String[] args)
    {
             Scanner sc=new Scanner(System.in);
             int v=sc.nextInt();
             int w=sc.nextInt();
             float res=((4*v)-w)/2;
             if(w>=2 && (w%2==0) && v<w ) 
               System.out.println("TW= "+(int)(res)+" FW= "+(int)(v-res));
             else
                System.out.println("INVALID INPUT");
    }
}

Python Coading

v=int(input())
w=int(input())
if (w&1)==1 or w<2 or w<=v:
    print("INVALID INPUT")
else:
    x=((4*v) -w)//2
    print("TW={0} FW={1}".format(x,v-x))

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