<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>

class Solution:
    def checkIfExist(self, arr: List[int]) -> bool:
        rev = sorted(arr , reverse=True)
        i = 0
        j = len(arr) - 1
        b = False
        while i < j:
            if i != j :
                if rev[i] == 2 * rev[j] or rev[j] == 2 * rev[i]:
                    b = True
                    break
                else:
                    j -= 1
                    if j <= i:
                        j = len(arr) - 1
                        i += 1
            else:
                j = len(arr) - 1
                i += 1
        return b
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