1496 - Path Crossing (Easy)

Problem Link

https://leetcode.com/problems/path-crossing/

Problem Statement

Given a string path, where path[i] = 'N', 'S', 'E' or 'W', each representing moving one unit north, south, east, or west, respectively. You start at the origin (0, 0) on a 2D plane and walk on the path specified by path.

Return true if the path crosses itself at any point, that is, if at any time you are on a location you have previously visited. Return false otherwise.

Example 1:

Input: path = "NES"
Output: false 
Explanation: Notice that the path doesn't cross any point more than once.

Example 2:

Input: path = "NESWW"
Output: true
Explanation: Notice that the path visits the origin twice.

Constraints:

• $1 <= path.length <= 10^4$
• path[i] is either 'N', 'S', 'E', or 'W'.

Approach 1: Set

We can keep track of the current position on a 2D plane and check if path crosses itself by storing visited positions in a Set. If a visited position is encountered again, it means the path has crossed itself, then we return true. Otherwise, we returns false.

Python

Written by @jaffar

class Solution:
    def isPathCrossing(self, path: str) -> bool:
        x, y = 0, 0
        visited = {(x, y)}
        for move in path:
            if move == 'N': y += 1
            elif move == 'S': y -= 1
            elif move == 'E': x += 1
            elif move == 'W': x -= 1
            if (x, y) in visited:
                return True
            visited.add((x, y))
        return False

Scala

Written by @jit

object Solution {
  // Check if the coordinates traversed has duplicates
  def isPathCrossing(path: String): Boolean = {
    val coords = path.scanLeft((0, 0)) {
      case ((x, y), 'N') => (x, y + 1)
      case ((x, y), 'E') => (x + 1, y)
      case ((x, y), 'S') => (x, y - 1)
      case ((x, y), 'W') => (x - 1, y)
    }
    coords.distinct.size != coords.size
  }
}

C++

Written by @wingkwong

class Solution {
public:
    bool isPathCrossing(string path) {
        int x = 0, y = 0;
        set<pair<int, int>> s;
        s.insert({x, y});
        for (char c : path) {
            if (c == 'N') y += 1;
            else if (c == 'S') y -= 1;
            else if (c == 'E') x += 1;
            else x -= 1;
            if (s.find({x, y}) != s.end()) return true;
            s.insert({x, y});
        }
        return false;
    }
};