<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) {
        map<int,int> m;
        int n = arr.size();
        for(int i=0;i<n;i++){
            m[arr[i]]++;
        }
        if(m.find(0)!=m.end()){
            if(m[0]%2==1) return false;
            m.erase(0);
        }
        
        for(auto i : m){
            if(i.second==0) continue;
            cout<<i.first<<" "<<i.first%2;
            if(i.first<0) {
                if(i.first%2==-1) return false;
                else if(m[i.first/2]<m[i.first]) return false;
                else {
                    m[i.first/2] -= m[i.first];
                    m[i.first] = 0; 
                }
            }
            else {
                if(m[2*i.first]<m[i.first]) return false;
                m[2*i.first] -= m[i.first];
                m[i.first] = 0;    
            }
           
        }
        return true;
    }
};
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