How do you solve Path Crossing on LeetCode?

Path Crossing (LeetCode #1496) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a set allows for O(1) average time complexity for checking previously visited positions.

What is the brute force solution for Path Crossing?

The brute force approach for Path Crossing is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every position we visit and see if we've been there before. This is straightforward but inefficient, as we will compare each position with all previous positions. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Path Crossing?

Path Crossing uses the following concepts: Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Path Crossing?

Always clarify the problem requirements and constraints before jumping into coding. Think about edge cases, such as paths that immediately cross themselves. Practice explaining your thought process as you code, as communication is key in interviews.

What are common mistakes when solving Path Crossing?

Not using a data structure like a set to track visited positions, leading to inefficient checks. Confusing the coordinates when updating position.

#1496

Path Crossing

Easy

Hash TableStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of checking all previous positions, we can use a set to track visited coordinates efficiently. This allows us to check for crossings in constant time.

⚙️

Algorithm

3 steps

1Step 1: Initialize a set to store visited positions, starting with the origin (0, 0).
2Step 2: For each direction in the path, update the current position accordingly.
3Step 3: After moving, check if the current position is in the set. If yes, return true; otherwise, add it to the set.

solution.py22 lines

1# Full working Python code
2path = "NESWW"
3
4def isPathCrossing(path):
5    visited = set()
6    visited.add((0, 0))
7    x, y = 0, 0
8    for direction in path:
9        if direction == 'N':
10            y += 1
11        elif direction == 'S':
12            y -= 1
13        elif direction == 'E':
14            x += 1
15        elif direction == 'W':
16            x -= 1
17        if (x, y) in visited:
18            return True
19        visited.add((x, y))
20    return False
21
22print(isPathCrossing(path))

ℹ

Complexity note: The time complexity is linear because we only traverse the path once, and the space complexity is linear due to storing visited positions in a set.

1Using a set allows for O(1) average time complexity for checking previously visited positions.
2The problem can be visualized as navigating a grid, where each direction corresponds to a movement on the grid.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.