<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) {
        vector<int> negs;

        for (int i = arr.size() - 1; i >= 0; i--){
            if (arr[i] < 0){
                negs.push_back(arr[i]);
                arr.erase(arr.begin() + i);
            }
        }

        int n = arr.size();
        int m = negs.size();

        sort(arr.begin(), arr.end());
        sort(negs.rbegin(), negs.rend());

        int l = 0, r = 1;

        vector<bool> visitedArr(n, false);
        vector<bool> visitedNeg(m, false);

        int pairs = 0;

        while ((l < r) && (r < n)){
            if (arr[r] == arr[l] * 2){
                pairs++;
                visitedArr[l] = true;
                visitedArr[r] = true;
                while (visitedArr[l]){
                    l++;
                }
                r = max(r+1, l+1);
            } else if (arr[r] < arr[l] * 2){
                r++;
            } else if (arr[r] > arr[l] * 2){
                l++;
                while (visitedArr[l]){
                    l++;
                }
                r = max(r, l+1);
            }
        }

        l = 0;
        r = 1;

        while ((l < r) && (r < m)){
            if (negs[r] == negs[l] * 2){
                pairs++;
                visitedNeg[l] = true;
                visitedNeg[r] = true;
                while (visitedNeg[l]){
                    l++;
                }
                r = max(r+1, l+1);
            } else if (negs[r] > negs[l] * 2){
                r++;
            } else if (negs[r] < negs[l] * 2){
                l++;
                while (visitedNeg[l]){
                    l++;
                }
                r = max(r, l+1);
            }
        }
        
        if (pairs != (n+m) / 2){
            return false;
        }
        return true;
    }
};
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