<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>

class Solution {
public:
    bool canReorderDoubled(vector<int>& arr) {
        multiset<int>pos,neg;
        int cnt = 0;
        for(auto it: arr){
            if(it>0) pos.insert(it);
            else if(it<0) neg.insert(abs(it));
            else cnt++;
        }
        if(cnt%2) return false;
        if(pos.size()%2 || neg.size()%2 )return false;
        while(pos.size() ){
            auto start = *pos.begin();
            if(pos.find(2*start) == pos.end()) return false;
            pos.erase(pos.find(start));
            pos.erase(pos.find(2*start));
        }
        while(neg.size()){
            auto start = *neg.begin();
            if(neg.find(2*start) == neg.end()) return false;
            neg.erase(neg.find(start));
            neg.erase(neg.find(2*start));
        }
        return true;
    }
};
91800