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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.

• 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;
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};
return true;
}
return false;
}
}

############

class Solution {
public boolean isPathCrossing(String path) {
int i = 0, j = 0;
Set<Integer> vis = new HashSet<>();
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;
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
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();
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;
}