Problem Description

You are given a string path consisting of characters 'N', 'S', 'E', and 'W', where each character represents a directional movement:

• 'N' means move one unit north (up)
• 'S' means move one unit south (down)
• 'E' means move one unit east (right)
• 'W' means move one unit west (left)

You start at the origin point (0, 0) on a 2D coordinate plane and follow the movements specified by the characters in path in order.

The task is to determine if you ever visit the same coordinate twice during your journey. Return true if at any point you step on a location you've already visited (including potentially returning to the origin), and false if you never revisit any coordinate.

For example:

• If path = "NES", you move north to (0, 1), then east to (1, 1), then south to (1, 0). Since no coordinate is visited twice, the answer is false.
• If path = "NESWW", you move north to (0, 1), east to (1, 1), south to (1, 0), west to (0, 0), and west again to (-1, 0). Since you return to the origin (0, 0) on the fourth move, the answer is true.

Intuition

The key insight is that we need to track every position we've visited as we walk through the path. If we ever arrive at a position we've been to before, we know the path crosses itself.

Think of it like leaving footprints as you walk. Each time you take a step, you check if there's already a footprint at your new location. If there is, you've crossed your own path.

To implement this, we can:

1. Keep track of our current position using coordinates (i, j) where i represents the vertical position and j represents the horizontal position
2. Store all visited positions in a set for fast lookup - checking if a position exists in a set is an O(1) operation
3. Start by adding the origin (0, 0) to our visited set since that's where we begin
4. For each movement in the path, update our coordinates accordingly and check if the new position already exists in our visited set

The coordinate updates follow a simple pattern:

• Moving north decreases the vertical coordinate by 1: i -= 1
• Moving south increases the vertical coordinate by 1: i += 1
• Moving east increases the horizontal coordinate by 1: j += 1
• Moving west decreases the horizontal coordinate by 1: j -= 1

If at any point our new position (i, j) is already in the visited set, we immediately return true because we've crossed our path. If we complete the entire journey without revisiting any position, we return false.

This approach is efficient because we only need to traverse the path once, and each position check takes constant time using the set data structure.

Solution Approach

The implementation uses a hash set to track visited coordinates and simulates the path traversal step by step.

Data Structure:

• Use a set vis to store visited coordinates as tuples (i, j)
• Initialize vis with the starting position (0, 0)

Algorithm Steps:

1. Initialize position and visited set:

   i = j = 0  # Starting at origin
   vis = {(0, 0)}  # Mark origin as visited

2. Process each movement in the path:

   • Iterate through each character c in the path string
   • Use a match-case statement (Python 3.10+) to update coordinates based on direction:

     match c:
         case 'N': i -= 1  # Move north (up)
         case 'S': i += 1  # Move south (down)
         case 'E': j += 1  # Move east (right)
         case 'W': j -= 1  # Move west (left)

3. Check for path crossing:

   • After each move, check if the new position (i, j) already exists in vis
   • If it exists, we've crossed our path, so return True
   • Otherwise, add the new position to vis:

     if (i, j) in vis:
         return True
     vis.add((i, j))

4. Return result:

   • If we complete the entire path without finding any duplicate positions, return False

Time Complexity: O(n) where n is the length of the path string. We visit each character once, and set operations (lookup and insertion) are O(1) on average.

Space Complexity: O(n) in the worst case, as we might need to store up to n + 1 unique positions (including the origin) if the path never crosses itself.

The solution elegantly handles the problem by maintaining a single set of visited positions and checking for duplicates in real-time as we traverse the path, eliminating the need for any complex geometric calculations.

Example Walkthrough

Let's trace through the path "NESW" step by step to see if we cross our own path:

Initial Setup:

• Starting position: (0, 0)
• Visited set: {(0, 0)}

Step 1: Process 'N' (North)

• Current position: (0, 0)
• Move north: i = 0 - 1 = -1, j = 0
• New position: (-1, 0)
• Check if (-1, 0) is in visited set: No
• Add (-1, 0) to visited set
• Visited set: {(0, 0), (-1, 0)}

Step 2: Process 'E' (East)

• Current position: (-1, 0)
• Move east: i = -1, j = 0 + 1 = 1
• New position: (-1, 1)
• Check if (-1, 1) is in visited set: No
• Add (-1, 1) to visited set
• Visited set: {(0, 0), (-1, 0), (-1, 1)}

Step 3: Process 'S' (South)

• Current position: (-1, 1)
• Move south: i = -1 + 1 = 0, j = 1
• New position: (0, 1)
• Check if (0, 1) is in visited set: No
• Add (0, 1) to visited set
• Visited set: {(0, 0), (-1, 0), (-1, 1), (0, 1)}

Step 4: Process 'W' (West)

• Current position: (0, 1)
• Move west: i = 0, j = 1 - 1 = 0
• New position: (0, 0)
• Check if (0, 0) is in visited set: Yes!
• We've returned to the origin, so we've crossed our path
• Return True

The path forms a square, and we end up back at the starting position (0, 0), which means we've crossed our own path.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the length of the path string.

• We iterate through each character in the path string exactly once
• For each character, we perform constant time operations:
  • Pattern matching with the match-case statement: O(1)
  • Checking if a coordinate exists in the set: O(1) average case for hash set lookup
  • Adding a coordinate to the set: O(1) average case for hash set insertion
• Total time complexity: O(n) * O(1) = O(n)

Space Complexity: O(n) where n is the length of the path string.

• The vis set stores visited coordinates as tuples
• In the worst case, the path never crosses itself, meaning we visit n + 1 unique positions (including the starting position (0, 0))
• Each tuple (i, j) takes constant space
• Total space complexity: O(n + 1) = O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Forgetting to Initialize the Origin in the Visited Set

One of the most common mistakes is forgetting to add the starting position (0, 0) to the visited set before processing the path. This leads to incorrect results when the path returns to the origin.

Incorrect Implementation:

def isPathCrossing(self, path: str) -> bool:
    row = col = 0
    visited = set()  # ❌ Empty set - origin not included!
  
    for direction in path:
        if direction == 'N':
            row -= 1
        elif direction == 'S':
            row += 1
        elif direction == 'E':
            col += 1
        elif direction == 'W':
            col -= 1
      
        if (row, col) in visited:
            return True
        visited.add((row, col))
  
    return False

Problem: If the path is "NESW" (which forms a square returning to origin), this incorrect code would return False instead of True because (0, 0) was never added to the visited set initially.

Correct Solution:

visited = {(0, 0)}  # ✅ Include origin from the start

2. Checking for Crossing Before Moving

Another pitfall is checking if the current position has been visited before actually moving to the new position.

Incorrect Implementation:

def isPathCrossing(self, path: str) -> bool:
    row = col = 0
    visited = {(0, 0)}
  
    for direction in path:
        # ❌ Checking before moving
        if (row, col) in visited:
            return True
          
        if direction == 'N':
            row -= 1
        elif direction == 'S':
            row += 1
        elif direction == 'E':
            col += 1
        elif direction == 'W':
            col -= 1
      
        visited.add((row, col))
  
    return False

Problem: This always returns True on the first iteration because (0, 0) is already in visited, even though we haven't actually moved yet.

Correct Solution:

for direction in path:
    # ✅ Move first
    if direction == 'N':
        row -= 1
    # ... other directions
  
    # ✅ Then check the new position
    if (row, col) in visited:
        return True
    visited.add((row, col))

3. Using Mutable Objects as Set Elements

Using lists instead of tuples for coordinates will cause a runtime error since lists are mutable and cannot be added to sets.

Incorrect Implementation:

visited = {[0, 0]}  # ❌ TypeError: unhashable type: 'list'
# or later in the code:
visited.add([row, col])  # ❌ TypeError

Correct Solution:

visited = {(0, 0)}  # ✅ Use tuples (immutable)
visited.add((row, col))  # ✅ Tuples are hashable

4. Incorrect Coordinate System Interpretation

Mixing up the coordinate system directions can lead to wrong results. Be consistent with your interpretation:

Potential Confusion:

# Some might interpret coordinates differently:
if direction == 'N':
    row += 1  # ❌ If treating north as positive y
# vs
if direction == 'N':
    row -= 1  # ✅ If treating north as negative row index (array-style)

Solution: Pick one coordinate system and stick with it consistently. The implementation should match the problem's expected behavior. Generally in 2D arrays, moving "up" (north) decreases the row index.