Shortest road trip

#1

how to find rest of 3 quadrants for shortest road trip?

#include
using namespace std;

int main() {

//Read the line 
char ch;
ch = cin.get();

int x = 0,y=0;

while(ch!='\n'){
    
    switch(ch){
        case 'N':y++;break;
        case 'S':y--;break;
        case 'E':x++;break;
        case 'W':x--;break;
    }
    
    ch = cin.get();
}

cout<<"Net Displacement is "<<x<<" and "<<y <<endl;

if(x>=0 && y>=0){
    //East Part 
    for(int i=0;i<x;i++){
        cout<<'E';
    }
    
    //North Part
    for(int i=0;i<y;i++){
        cout<<'N';
    }
    cout<<endl;
    
}
//Rest of 3 quadrants???

}

#2

Hi Nitin,
Here, you need to understand the question first. We need to sort the directions and output the net displacement & directions (N,E,S,W) which will give you the shortest path to reach the final point.

So, here in switch case, you are analyzing directions from the input according to which we are finding the final coordinates to reach the end point.
After that, we need to print the coordinates x and y in terms of directions too.

There are 4 quadrants in which the coordinates can lay down:-

  1. North-East - you figured out yourself
  2. North-West
  3. South-West
  4. South-East

You need to find the other 3 quadrants by putting the those directions on places.

I suggest you try the code yourself first, I am sure you will be able to do it on your own.

If you have any other doubts, please feel free to ask again.

1 Like
#3

Hi rahul,
Thanks, I solved.

#4

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