<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) {
        // 0, 1
        // 2, 3
        // 4, 5
        // 6, 7
        vector<int> neg;
        vector<int> pos;
        unordered_map<int,int> pos_count;
        unordered_map<int,int> neg_count;
        for (int i : arr) {
            if (i >= 0) {
                pos.push_back(i);
                pos_count[i]++;
            } else {
                neg.push_back(i);
                neg_count[i]++;
            }
        }
        if (pos.size() % 2 != 0 || neg.size() % 2 != 0) {
            return false;
        }
        sort(pos.begin(), pos.end());

        sort(neg.begin(), neg.end(), greater<int>());
        return process(pos_count, pos) && process(neg_count, neg);
    }
    
    bool process(unordered_map<int,int>& count, vector<int>& vec) {
        for (int i = 0; i < vec.size(); i++) {
            int x = vec[i];
            if (count.find(x) != count.end() && count[x] > 0) {
                int find = x * 2;
                if (count.find(find) == count.end()) {
                //    cout << "cannot find: " << find << endl;
                    return false;
                }
                count[find]--;
                if (count[find] == 0)  {
                    count.erase(find);
                }
                count[x]--;
                if (count[x] == 0) {
                    count.erase(x);
                }
            }
        }
        // cout << count.size() << endl;
        // for (auto& p : count) {
        //     cout << p.first << "," << p.second << endl;
        // }
        return count.size() == 0;
    }
};
68000