<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> L;
          vector<int> R;
          for (int a : arr) {
            if (a >= 0) {
                L.push_back(a);
            } else {
                R.push_back(-a);
            }
          }
          return canArrange(L) && canArrange(R);
    }

    bool canArrange(vector<int>& arr) {
        int n = arr.size();
        if (n % 2 == 1) {
            return false;
        }
        sort(arr.begin(), arr.end());
        vector<bool> V(n);
        int idx = 0;
        for (int i = 0; i < n; i++) {
            if (!V[i]) {
                while (idx <= i || (idx < n && 2 * arr[i] != arr[idx])) {
                    idx++;
                }
                if (idx == n) {
                    return false;
                }
                V[idx++] = true;
            }
        }
        return true;
    }
};
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