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Calculate the Square of Euclidean Distance Traveled based on given conditions

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Given an array commands[], consisting of signed integers denoting distance and direction to be travelled along with the coordinates, and array obstacles[] denoting the coordinates which cannot be accessed, the task is to find the square of the square of maximum Euclidean distance that can be travelled started from the origin (0, 0) and facing north, following the commands specified in the sequence as in the commands[] array, of the following three types: 

  • -2: Turn left by 90 degrees.
  • -1: Turn right by 90 degrees.
  • 1<= X <= 9: Move forward by X units.

Examples: 

Input: commands[] = {4, -1, 4, -2, 4}, obstacles[] = {{ 2, 4 }} 
Output: 65 
Explanation: 
Step 1: (0, 0) -> (0, 4) 
Step 2: (0, 4) -> (1, 4) 
Step 3 and 4: 
Obstacles 
Step 5: (1, 4) -> (1, 8)

Input: commands[] = {4, -1, 3}, obstacles[] = {} 
Output: 25 

Approach: Follow the steps below to solve the problem: 

  1. Initially, the robot is at (0, 0) facing north.
  2. Assign variables to keep track of the current position and direction of the robot after each step.
  3. Store the coordinates of obstacles in a HashMap.
  4. Make 2 arrays(dx[], dy[]) and store all possible movements in x and y coordinates according to the change in direction.
  5. If direction change is encountered, change the present direction referring to the 2 arrays.
  6. Otherwise, keep on moving in the same direction until a direction change is encountered if no obstacles occur in between.
  7. Finally, calculate the square of the x and y coordinates.

Below is the implementation of the above approach: 

C++




// C++ program to implement
// the above approach
#include<bits/stdc++.h>
using namespace std;
 
void robotSim(vector<int> commands,
              vector<pair<int, int>> obstacles)
{
     
    // Possible movements in x coordinate.
    vector<int> dx = { 0, 1, 0, -1 };
 
    // Possible movements in y coordinate.
    vector<int> dy = { 1, 0, -1, 0 };
 
    int x = 0, y = 0;
 
    int di = 0;
 
    // Put all obstacles into hashmap.
    map<pair<int, int>, int>obstacleSet;
     
    for(auto i:obstacles)
        obstacleSet[i] = 1;
 
    // Maximum distance
    int ans = 0;
 
    // Iterate commands.
    for(int cmd : commands)
    {
         
        // Left direction
        if (cmd == -2)
            di = (di-1) % 4;
         
        // Right direction
        else if (cmd == -1)
            di = (di + 1) % 4;   
             
        // If no direction changes
        else
        {
            for(int i = 0; i < cmd; i++)
            {
                 
                // Checking for obstacles.
                if (obstacleSet.find({x + dx[di],
                                      y + dy[di]}) ==
                                      obstacleSet.end())
                {
                     
                    // Update x coordinate
                    x += dx[di];
                     
                    // Update y coordinate
                    y += dy[di];
                     
                    // Updating for max distance
                    ans = max(ans, x * x + y * y);
                }
            }
        }
    }
    cout << ans << endl;
}
 
// Driver Code
int main()
{
    vector<int> commands = { 4, -1, 4, -2, 4 };
    vector<pair<int, int>> obstacles = { { 2, 4 } };
     
    robotSim(commands, obstacles);
}
 
// This code is contributed by grand_master


Java




// Java program to implement
// the above approach
import java.util.*;
import java.awt.Point;
 
class GFG{
     
static void robotSim(List<Integer> commands,
                     List<Point> obstacles)
{
     
    // Possible movements in x coordinate.
    List<Integer> dx = Arrays.asList(0, 1, 0, -1);
  
    // Possible movements in y coordinate.
    List<Integer> dy = Arrays.asList(1, 0, -1, 0);
  
    int x = 0, y = 0;
    int di = 0;
  
    // Put all obstacles into hashmap.
    HashMap<Point, Integer> obstacleSet = new HashMap<>();
     
    for(Point i : obstacles)
        obstacleSet.put(i, 1);
  
    // Maximum distance
    int ans = 0;
  
    // Iterate commands.
    for(Integer cmd : commands)
    {
         
        // Left direction
        if (cmd == -2)
            di = (di - 1) % 4;
          
        // Right direction
        else if (cmd == -1)
            di = (di + 1) % 4;   
              
        // If no direction changes
        else
        {
            for(int i = 0; i < cmd; i++)
            {
                 
                // Checking for obstacles.
                if (!obstacleSet.containsKey(
                    new Point(x + dx.get(di),
                              y + dy.get(di))))
                {
                     
                    // Update x coordinate
                    x += dx.get(di);
                      
                    // Update y coordinate
                    y += dy.get(di);
                      
                    // Updating for max distance
                    ans = Math.max(ans, x * x + y * y);
                }
            }
        }
    }
    System.out.println(ans);
}
 
// Driver Code
public static void main(String[] args)
{
    List<Integer> commands = Arrays.asList(
        4, -1, 4, -2, 4);
    List<Point> obstacles = new ArrayList<>();
    obstacles.add(new Point(2, 4));
      
    robotSim(commands, obstacles);
}
}
 
// This code is contributed by divyesh072019


Python3




# Python3 Program to implement
# the above approach
def robotSim(commands, obstacles):
 
    # Possible movements in x coordinate.
    dx = [0, 1, 0, -1]
 
    # Possible movements in y coordinate.
    dy = [1, 0, -1, 0]
 
    # Initialise position to (0, 0).
    x, y = 0, 0
 
    # Initial direction is north.
    di = 0
 
    # Put all obstacles into hashmap.
    obstacleSet = set(map(tuple, obstacles))
 
    # maximum distance
    ans = 0
 
    # Iterate commands.
    for cmd in commands:
         
        # Left direction
        if cmd == -2:
            di = (di-1) % 4
         
        # Right direction
        elif cmd == -1:
            di = (di + 1) % 4
         
        # If no direction changes
        else:
            for i in range(cmd):
                # Checking for obstacles.
                if (x + dx[di], y + dy[di]) \
                not in obstacleSet:
                     
                    # Update x coordinate
                    x += dx[di]
                     
                    # Update y coordinate
                    y += dy[di]
                     
                    # Updating for max distance
                    ans = max(ans, x * x + y * y)
    print(ans)
 
 
# Driver Code
if __name__ == "__main__":
    commands = [4, -1, 4, -2, 4]
    obstacles = [[2, 4]]
    robotSim(commands, obstacles)


C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;  
class GFG {
     
  static void robotSim(List<int> commands,
                       List<Tuple<int, int>> obstacles)
  {
 
    // Possible movements in x coordinate.
    List<int> dx = new List<int>(new int[]{ 0, 1, 0, -1 });
 
    // Possible movements in y coordinate.
    List<int> dy = new List<int>(new int[]{ 1, 0, -1, 0 });   
    int x = 0, y = 0;    
    int di = 0;
 
    // Put all obstacles into hashmap.
    Dictionary<Tuple<int, int>, int> obstacleSet =
      new Dictionary<Tuple<int, int>, int>(); 
 
    foreach(Tuple<int, int> i in obstacles)
      obstacleSet[i] = 1;
 
    // Maximum distance
    int ans = 0;
 
    // Iterate commands.
    foreach(int cmd in commands)
    {
 
      // Left direction
      if (cmd == -2)
        di = (di - 1) % 4;
 
      // Right direction
      else if (cmd == -1)
        di = (di + 1) % 4;   
 
      // If no direction changes
      else
      {
        for(int i = 0; i < cmd; i++)
        {
 
          // Checking for obstacles.
          if (!obstacleSet.ContainsKey(new Tuple<int,int>(x + dx[di],
                                                          y + dy[di])))
          {
 
            // Update x coordinate
            x += dx[di];
 
            // Update y coordinate
            y += dy[di];
 
            // Updating for max distance
            ans = Math.Max(ans, x * x + y * y);
          }
        }
      }
    }
    Console.WriteLine(ans);
  }
 
  // Driver code
  static void Main()
  {
    List<int> commands = new List<int>(new int[]{ 4, -1, 4, -2, 4 });
    List<Tuple<int, int>> obstacles = new List<Tuple<int, int>>();
    obstacles.Add(new Tuple<int, int>(2,4));   
    robotSim(commands, obstacles);
  }
}
 
// This code is contributed by divyeshrabadiya07.


Javascript




// JavaScript program to implement
// the above approach
 
 
function robotSim(commands, obstacles)
{
    // Possible movements in x coordinate.
    let dx = [0, 1, 0, -1];
 
    // Possible movements in y coordinate.
    let dy = [1, 0, -1, 0];
 
    let x = 0, y = 0;
 
    let di = 0;
 
    // Put all obstacles into hashmap.
    let obstacleSet = new Map();
     
    obstacles.forEach(i=>{
        obstacleSet.set([i].join(), 1);
    })
     
    // Maximum distance
    let ans = 0;
 
    // Iterate commands.
    for(let j = 0; j < commands.length; j++)
    {
        let cmd = commands[j];
        // Left direction
        if (cmd == -2)
            di = (di-1) % 4;
         
        // Right direction
        else if (cmd == -1)
            di = (di + 1) % 4;   
             
        // If no direction changes
        else
        {
            for(let i = 0; i < cmd; i++)
            {
                // Checking for obstacles.
                if (!obstacleSet.has([x + dx[di], y + dy[di]].join()))
                {
                    // Update x coordinate
                    x = x + dx[di];
                     
                    // Update y coordinate
                    y = y + dy[di];
 
                    // Updating for max distance
                    ans = Math.max(ans, x * x + y * y);
                }
            }
        }
    }
    console.log(ans);
}
 
// Driver Code
let commands = [4, -1, 4, -2, 4];
let obstacles = [[2, 4]];
 
robotSim(commands, obstacles);
 
// This code is contributed by Gautam goel (gautamgoel962)


Output: 

65

 

Time Complexity: O(N) 
Auxiliary Space: O(N + M), where M is the length of the obstacle array.

 



Last Updated : 15 Jun, 2022
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