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1496. Path Crossing

Description

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 <= 104
• path[i] is either 'N', 'S', 'E', or 'W'.

Solutions

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

• 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;
      }
  }
  

• class Solution {
  public:
      bool isPathCrossing(string path) {
          int i = 0, j = 0;
          unordered_set<int> s{ {0} };
          for (char& c : path) {
              if (c == 'N') {
                  --i;
              } else if (c == 'S') {
                  ++i;
              } else if (c == 'E') {
                  ++j;
              } else {
                  --j;
              }
              int t = i * 20000 + j;
              if (s.count(t)) {
                  return true;
              }
              s.insert(t);
          }
          return false;
      }
  };
  

• class Solution:
      def isPathCrossing(self, path: str) -> bool:
          i = j = 0
          vis = {(0, 0)}
          for c in path:
              match c:
                  case 'N':
                      i -= 1
                  case 'S':
                      i += 1
                  case 'E':
                      j += 1
                  case 'W':
                      j -= 1
              if (i, j) in vis:
                  return True
              vis.add((i, j))
          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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