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Formatted question description: https://leetcode.ca/all/1496.html

1496. Path Crossing (Easy)

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've 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 will only consist of characters in {'N', 'S', 'E', 'W}

Related Topics:
String

Solution 1.

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public boolean isPathCrossing(String path) {
          char[] directionChars = {'N', 'S', 'E', 'W'};
          int[][] directions = { {0, 1}, {0, -1}, {1, 0}, {-1, 0} };
          Map<Character, Integer> map = new HashMap<Character, Integer>();
          for (int i = 0; i < 4; i++)
              map.put(directionChars[i], i);
          Set<String> set = new HashSet<String>();
          int x = 0, y = 0;
          set.add(Arrays.toString(new int[]{0, 0}));
          int length = path.length();
          for (int i = 0; i < length; i++) {
              char c = path.charAt(i);
              int index = map.get(c);
              int[] direction = directions[index];
              x += direction[0];
              y += direction[1];
              int[] array = {x, y};
              if (!set.add(Arrays.toString(array)))
                  return true;
          }
          return false;
      }
  }
  
  ############
  
  class Solution {
      public boolean isPathCrossing(String path) {
          int i = 0, j = 0;
          Set<Integer> vis = new HashSet<>();
          vis.add(0);
          for (int k = 0, n = path.length(); k < n; ++k) {
              switch (path.charAt(k)) {
                  case 'N' -> --i;
                  case 'S' -> ++i;
                  case 'E' -> ++j;
                  case 'W' -> --j;
              }
              int t = i * 20000 + j;
              if (!vis.add(t)) {
                  return true;
              }
          }
          return false;
      }
  }
  

• // OJ: https://leetcode.com/problems/path-crossing/
  // Time: O(N)
  // Space: O(N)
  class Solution {
      long hash(int x, int y) { return (long)x * 10000 + y; }
  public:
      bool isPathCrossing(string path) {
          unordered_set<long> s;
          int x = 0, y = 0;
          s.insert(hash(x, y));
          for (char c : path) {
              if (c == 'N') ++y;
              else if (c == 'E') ++x;
              else if (c == 'S') --y;
              else --x;
              long key = hash(x, y);
              if (s.count(key)) return true;
              s.insert(key);
          }
          return false;
      }
  };
  

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

• func isPathCrossing(path string) bool {
  	i, j := 0, 0
  	vis := map[int]bool{0: true}
  	for _, c := range path {
  		switch c {
  		case 'N':
  			i--
  		case 'S':
  			i++
  		case 'E':
  			j++
  		case 'W':
  			j--
  		}
  		if vis[i*20000+j] {
  			return true
  		}
  		vis[i*20000+j] = true
  	}
  	return false
  }
  

• function isPathCrossing(path: string): boolean {
      let [i, j] = [0, 0];
      const vis: Set<number> = new Set();
      vis.add(0);
      for (const c of path) {
          if (c === 'N') {
              --i;
          } else if (c === 'S') {
              ++i;
          } else if (c === 'E') {
              ++j;
          } else if (c === 'W') {
              --j;
          }
          const t = i * 20000 + j;
          if (vis.has(t)) {
              return true;
          }
          vis.add(t);
      }
      return false;
  }
  
  

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