Path Crossing
<p>Given a string <code>path</code>, where <code>path[i] = &#39;N&#39;</code>, <code>&#39;S&#39;</code>, <code>&#39;E&#39;</code> or <code>&#39;W&#39;</code>, each representing moving one unit north, south, east, or west, respectively. You start at the origin <code>(0, 0)</code> on a 2D plane and walk on the path specified by <code>path</code>.</p>

<p>Return <code>true</code> <em>if the path crosses itself at any point, that is, if at any time you are on a location you have previously visited</em>. Return <code>false</code> otherwise.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1496.Path%20Crossing/images/screen-shot-2020-06-10-at-123929-pm.png" style="width: 400px; height: 358px;" />
<pre>
<strong>Input:</strong> path = &quot;NES&quot;
<strong>Output:</strong> false 
<strong>Explanation:</strong> Notice that the path doesn&#39;t cross any point more than once.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1496.Path%20Crossing/images/screen-shot-2020-06-10-at-123843-pm.png" style="width: 400px; height: 339px;" />
<pre>
<strong>Input:</strong> path = &quot;NESWW&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> Notice that the path visits the origin twice.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= path.length &lt;= 10<sup>4</sup></code></li>
	<li><code>path[i]</code> is either <code>&#39;N&#39;</code>, <code>&#39;S&#39;</code>, <code>&#39;E&#39;</code>, or <code>&#39;W&#39;</code>.</li>
</ul>
Hash Table; String
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;
}
