Check If N and Its Double Exist
<p>Given an array <code>arr</code> of integers, check if there exist two indices <code>i</code> and <code>j</code> such that :</p>

<ul>
	<li><code>i != j</code></li>
	<li><code>0 &lt;= i, j &lt; arr.length</code></li>
	<li><code>arr[i] == 2 * arr[j]</code></li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [10,2,5,3]
<strong>Output:</strong> true
<strong>Explanation:</strong> For i = 0 and j = 2, arr[i] == 10 == 2 * 5 == 2 * arr[j]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [3,1,7,11]
<strong>Output:</strong> false
<strong>Explanation:</strong> There is no i and j that satisfy the conditions.
</pre>

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

<ul>
	<li><code>2 &lt;= arr.length &lt;= 500</code></li>
	<li><code>-10<sup>3</sup> &lt;= arr[i] &lt;= 10<sup>3</sup></code></li>
</ul>
Array; Hash Table; Two Pointers; Binary Search; Sorting
function checkIfExist(arr: number[]): boolean {
    const s: Set<number> = new Set();
    for (const x of arr) {
        if (s.has(x * 2) || (x % 2 === 0 && s.has((x / 2) | 0))) {
            return true;
        }
        s.add(x);
    }
    return false;
}
