<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 n = arr.size();
        unordered_map<int, int> mp;
        vector<int> pos;
        vector<int> neg;
        int count=0;
        for (auto it : arr) {
            mp[it]++;
            if(it>0) pos.push_back(it);
            else neg.push_back(it);
        }
        if(mp[0]%2!=0) return  false;
        sort(pos.begin(),pos.end());
        sort(neg.begin(),neg.end(),greater<int>());
        if(pos.size()%2==1 || neg.size()%2==1) return false;
        for(auto it:pos){
            if(mp[it]>0 && mp[2*it]>0){
                mp[it]--;
                mp[2*it]--;
                count++;
            }
        }
            for(auto it:neg){
            if(mp[it]>0 && mp[2*it]>0){
                mp[it]--;
                mp[2*it]--;
                count++;
            }
        }
        return 2 * count == n;
    }
};
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