Array of Doubled Pairs
<p>Given an integer array of even length <code>arr</code>, return <code>true</code><em> if it is possible to reorder </em><code>arr</code><em> such that </em><code>arr[2 * i + 1] = 2 * arr[2 * i]</code><em> for every </em><code>0 &lt;= i &lt; len(arr) / 2</code><em>, or </em><code>false</code><em> otherwise</em>.</p>

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

<pre>
<strong>Input:</strong> arr = [3,1,3,6]
<strong>Output:</strong> false
</pre>

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

<pre>
<strong>Input:</strong> arr = [2,1,2,6]
<strong>Output:</strong> false
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [4,-2,2,-4]
<strong>Output:</strong> true
<strong>Explanation:</strong> We can take two groups, [-2,-4] and [2,4] to form [-2,-4,2,4] or [2,4,-2,-4].
</pre>

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

<ul>
	<li><code>2 &lt;= arr.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>arr.length</code> is even.</li>
	<li><code>-10<sup>5</sup> &lt;= arr[i] &lt;= 10<sup>5</sup></code></li>
</ul>
Greedy; Array; Hash Table; Sorting
class Solution {
public:
    bool canReorderDoubled(vector<int>& arr) {
        unordered_map<int, int> freq;
        for (int& v : arr) ++freq[v];
        if (freq[0] & 1) return false;
        vector<int> keys;
        for (auto& [k, _] : freq) keys.push_back(k);
        sort(keys.begin(), keys.end(), [](int a, int b) { return abs(a) < abs(b); });
        for (int& k : keys) {
            if (freq[k * 2] < freq[k]) return false;
            freq[k * 2] -= freq[k];
        }
        return true;
    }
};