Path crossing

We will start with the origin (0, 0) and for each move we will update the current position. If the current position is already in the set of visited positions, then we have a path crossing.

class Solution:
    def isPathCrossing(self, path: str) -> bool:
        x, y = 0, 0
        seen = set([(x, y)])
        for p in path:
            if p == 'N': x += 1
            elif p == 'S': x -= 1
            elif p == 'E': y += 1
            else: y -= 1
            if (x, y) in seen:
                return True
            seen.add((x, y))
        return False

Time complexity: O(n)
Space complexity: O(n)