1400 - 1499
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:
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.
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 Falseobject 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
}
}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;
}
};