<p>Given an integer array <code>nums</code>, return <em>the number of <strong>reverse pairs</strong> in the array</em>.</p>

<p>A <strong>reverse pair</strong> is a pair <code>(i, j)</code> where:</p>

<ul>
	<li><code>0 &lt;= i &lt; j &lt; nums.length</code> and</li>
	<li><code>nums[i] &gt; 2 * nums[j]</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,2,3,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The reverse pairs are:
(1, 4) --&gt; nums[1] = 3, nums[4] = 1, 3 &gt; 2 * 1
(3, 4) --&gt; nums[3] = 3, nums[4] = 1, 3 &gt; 2 * 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,4,3,5,1]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The reverse pairs are:
(1, 4) --&gt; nums[1] = 4, nums[4] = 1, 4 &gt; 2 * 1
(2, 4) --&gt; nums[2] = 3, nums[4] = 1, 3 &gt; 2 * 1
(3, 4) --&gt; nums[3] = 5, nums[4] = 1, 5 &gt; 2 * 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution {
public:
    int getSum(vector<int> &tree, int idx) {
        int sum = 0;
        for(int i = idx+1; i>0; i-=(i&(-i))) sum += tree[i];
        return sum;
    }
    void update(vector<int> &tree, int idx, int n, int del) {
        for(int i = idx+1; i<=n; i+=(i&(-i))) tree[i] += del;
    }
    int reversePairs(vector<int>& nums) {
        int n = nums.size();
        vector<pair<long long,int>> data, queries;
        for(int i = 0; i<n; i++) {
            data.push_back({nums[i], i});
            queries.push_back({2ll*nums[i], i});
        }
        sort(data.begin(), data.end(), greater<pair<long long,int>>());
        sort(queries.begin(), queries.end(), greater<pair<long long,int>>());
        int ans = 0;
        int idx = 0;
        vector<int> tree(n+1, 0);
        for(int i = 0; i<queries.size(); i++) {
            long long limit = queries[i].first;
            int qIdx = queries[i].second;
            while(idx < data.size() && data[idx].first > limit) {
                update(tree, data[idx].second, n, 1);
                idx++;
            }
            int res = getSum(tree, qIdx-1);
            ans += res;
        }
        return ans;
    }
};
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