1496 Path Crossing

This is a problem that involves just simulating what’s stated in the problem. There’s no hidden trick, or optimization - simply simulate a point moving from $(0,0)$ to the directions given in the string. We use a set to store all visited points starting with $(0,0)$. If we come across any point that’s already visited, we return True, else False.

Code:

Python

class Solution:
    def isPathCrossing(self, path: str) -> bool:
        vis = set()
 
        dx, dy = 0, 0
        vis.add((dx,dy))
        for di in path:
            if di == 'N':   dy += 1
            elif di == 'E': dx -= 1
            elif di == 'W': dx += 1
            elif di == 'S': dy -= 1
 
            if (dx, dy) in vis: return True
            else: vis.add((dx,dy))
        
        
        return False

Big O Analysis

• Runtime

  The runtime complexity here is $O(n)$ as since we would be visiting all characters in the string at least once.

• Memory

  The memory usage is $O(n)$ since we are using a set to store visited points.

— A

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