<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) {
        int cnt=0;
        for(int i=0;i<arr.size();i++){
            if(arr[i]<0)cnt++;
        }
        if(cnt%2==1)return false;
        sort(arr.begin(),arr.end());
        map<int,int>mp;
        for(auto it:arr){
            if(mp.find(2*it)!=mp.end()){
                mp[2*it]--;
                if(mp[2*it]==0)mp.erase(2*it);
            }else if(mp.find(it/2)!=mp.end()  && 2*(it/2)==it){
                mp[it/2]--;
                if(mp[it/2]==0)mp.erase(it/2);
            }else mp[it]++;
        }
        
        return  mp.size()?false:true;
    }
};
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