LeetCode #1496 — EASY

Path Crossing

Build confidence with an intuition-first walkthrough focused on hash map fundamentals.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map

Example 1

"NES"

Example 2

"NESWW"

Step 02

Core Insight

What unlocks the optimal approach

Simulate the process while keeping track of visited points.
Use a set to store previously visited points.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1496: Path Crossing
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;
    }
}

// Accepted solution for LeetCode #1496: Path Crossing
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
}

# Accepted solution for LeetCode #1496: Path Crossing
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

// Accepted solution for LeetCode #1496: Path Crossing
struct Solution;
use std::collections::HashSet;

impl Solution {
    fn is_path_crossing(path: String) -> bool {
        let mut hs: HashSet<(i32, i32)> = HashSet::new();
        hs.insert((0, 0));
        let mut x = 0;
        let mut y = 0;
        for c in path.chars() {
            match c {
                'N' => {
                    y += 1;
                }
                'S' => {
                    y -= 1;
                }
                'E' => {
                    x += 1;
                }
                _ => {
                    x -= 1;
                }
            }
            if !hs.insert((x, y)) {
                return true;
            }
        }
        false
    }
}

#[test]
fn test() {
    let path = "NES".to_string();
    let res = false;
    assert_eq!(Solution::is_path_crossing(path), res);
    let path = "NESWW".to_string();
    let res = true;
    assert_eq!(Solution::is_path_crossing(path), res);
}

// Accepted solution for LeetCode #1496: Path Crossing
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.