Check if a sequence of path visits any coordinate twice or not

Given string str of length N only consisting of characters ‘N’, ‘S’, ‘E’, or ‘W’, each representing moving one unit North, South, East, or West, respectively. A man starts at the origin (0, 0) on a 2D plane and walks according to the directions in the string. The task is to check whether the man crosses any coordinate in his walk, which he has already visited or not. If found to be true, then print “Crossed”. Otherwise, print “Not Crossed”.

Examples:

Input: str = “NESW”
Output: Crossed
Explanation:
The path for the man is given by:
(0, 0) -> (0, 1) -> (1, 1) -> (0, 1) -> (0, 0).
Since, the coordinate (0, 0) is visited again.
Therefore, print Crossed as he has crossed the path.

Input: str = “SESS”
Output: Not Crossed
Explanation:
The path for the man is given by:
(0, 0) -> (0, -1) -> (1, -1) -> (1, -2) -> (1, -3).
From the above path, the man never visited the same coordinate again.
Therefore, print Not Crossed.

Approach: The idea is to maintain a Set for the co-ordinates encountered to check whether already visited co-ordinates occurs or not. Below are the steps:

  1. Initialize two variables X and Y and initiate the values as zero. These variables will represent x and y coordinates.
  2. Insert X and Y in the set.
  3. Iterate a loop over the range [0, N), and increment the X or Y coordinate as per the below condition:
    • If the character is ‘N’, then increment Y by 1.
    • If the character is ‘S’, then decrement Y by 1.
    • If the character is ‘E’, then increment X by 1.
    • If the character is ‘W’, then decrement X by 1.
  4. Insert the updated X and Y coordinates in the above step for each character in the set.
  5. While inserting the co-ordinates in the above step if any present co-ordinates are encountered then print Crossed.
  6. Otherwise, print Not Crossed as all the points of co-ordinates are unique.

Below is the implementation of the above approach

C++

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// C++ program for the above appproach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if the man crosses
// previous visited coordinate or not
bool isCrossed(string path)
{
    if (path.size() == 0)
        return false;
 
    // Stores the count of
    // crossed vertex
    bool ans = false;
 
    // Stores (x, y) coordinates
    set<pair<int, int> > set;
 
    // The coordinates for the origin
    int x = 0, y = 0;
    set.insert({ x, y });
 
    // Iterate over the string
    for (int i = 0; i < path.size(); i++) {
 
        // Condition to increment X or
        // Y co-ordinates respectively
        if (path[i] == 'N')
            set.insert({ x, y++ });
 
        if (path[i] == 'S')
            set.insert({ x, y-- });
 
        if (path[i] == 'E')
            set.insert({ x++, y });
 
        if (path[i] == 'W')
            set.insert({ x--, y });
 
        // Check if (x, y) is already
        // visited
        if (set.find({ x, y })
            != set.end()) {
 
            ans = true;
            break;
        }
    }
 
    // Print the result
    if (ans)
        cout << "Crossed";
    else
        cout << "Not Crossed";
}
 
// Driver Code
int main()
{
    // Given string
    string path = "NESW";
 
    // Function Call
    isCrossed(path);
 
    return 0;
}

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Java

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// Java program for
// the above appproach
import java.awt.Point;
import java.util.HashSet;
class GFG{
 
// Function to check if the man crosses
// previous visited coordinate or not
static void isCrossed(String path)
{
  if (path.length() == 0)
    return;
 
  // Stores the count of
  // crossed vertex
  boolean ans = false;
 
  // Stores (x, y) coordinates
  HashSet<Point> set =
          new HashSet<Point>();
 
  // The coordinates for the origin
  int x = 0, y = 0;
  set.add(new Point(x, y));
 
  // Iterate over the String
  for (int i = 0; i < path.length(); i++)
  {
    // Condition to increment X or
    // Y co-ordinates respectively
    if (path.charAt(i) == 'N')
      set.add(new Point(x, y++));
 
    if (path.charAt(i) == 'S')
      set.add(new Point(x, y--));
 
    if (path.charAt(i) == 'E')
      set.add(new Point(x++, y));
 
    if (path.charAt(i) == 'W')
      set.add(new Point(x--, y));
 
    // Check if (x, y) is already
    // visited
    if (set.contains(new Point(x, y)))
    {
      ans = true;
      break;
    }
  }
 
  // Print the result
  if (ans)
    System.out.print("Crossed");
  else
    System.out.print("Not Crossed");
}
 
// Driver Code
public static void main(String[] args)
{
  // Given String
  String path = "NESW";
 
  // Function Call
  isCrossed(path);
}
}
 
// This code is contributed by Princi Singh

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Python3

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# Python3 program for the above appproach
 
# Function to check if the man crosses
# previous visited coordinate or not
def isCrossed(path):
 
    if (len(path) == 0):
        return bool(False)
 
    # Stores the count of
    # crossed vertex
    ans = bool(False)
 
    # Stores (x, y) coordinates
    Set = set()
 
    # The coordinates for the origin
    x, y = 0, 0
    Set.add((x, y))
 
    # Iterate over the string
    for i in range(len(path)):
 
        # Condition to increment X or
        # Y co-ordinates respectively
        if (path[i] == 'N'):
            Set.add((x, y))
            y = y + 1
 
        if (path[i] == 'S'):
            Set.add((x, y))
            y = y - 1
             
        if (path[i] == 'E'):
            Set.add((x, y))
            x = x + 1
 
        if (path[i] == 'W'):
            Set.add((x, y))
            x = x - 1
             
        # Check if (x, y) is already visited
        if (x, y) in Set:
            ans = bool(True)
            break
 
    # Print the result
    if (ans):
        print("Crossed")
    else:
        print("Not Crossed")
 
# Driver code
 
# Given string
path = "NESW"
 
# Function call
isCrossed(path)
 
# This code is contributed by divyeshrabadiya07

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C#

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// C# program for the above appproach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to check if the man crosses
// previous visited coordinate or not
static void isCrossed(String path)
{
    if (path.Length == 0)
        return;
 
    // Stores the count of
    // crossed vertex
    bool ans = false;
 
    // Stores (x, y) coordinates
    HashSet<KeyValuePair<int, int>> mySet =
    new HashSet<KeyValuePair<int, int>>();
 
    // The coordinates for the origin
    int x = 0, y = 0;
    mySet.Add(new KeyValuePair<int, int>(x, y));
 
    // Iterate over the String
    for(int i = 0; i < path.Length; i++)
    {
         
        // Condition to increment X or
        // Y co-ordinates respectively
        if (path[i] == 'N')
            mySet.Add(
                new KeyValuePair<int, int>(x, y++));
 
        if (path[i] == 'S')
            mySet.Add(
                new KeyValuePair<int, int>(x, y--));
 
        if (path[i] == 'E')
            mySet.Add(
                new KeyValuePair<int, int>(x++, y));
 
        if (path[i] == 'W')
            mySet.Add(
                new KeyValuePair<int, int>(x--, y));
 
        // Check if (x, y) is already
        // visited
        if (mySet.Contains(
                new KeyValuePair<int, int>(x, y)))
        {
            ans = true;
            break;
        }
    }
 
    // Print the result
    if (ans)
        Console.Write("Crossed");
    else
        Console.Write("Not Crossed");
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given String
    String path = "NESW";
 
    // Function call
    isCrossed(path);
}
}
 
// This code is contributed by akhilsaini

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Output: 

Crossed







 

Time Complexity: O(N*log N)
Auxiliary Space: O(N)

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