<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 BIT {
    int N; 
    vector<int> ft; 

public: 
    BIT(int N): N(N), ft(N) {}

    void update(int idx, int val) {
        for ( ; idx < N ; idx |= idx + 1) {
            ft[idx] += val; 
        }
    }

    int query(int idx) {
        int ret = 0; 
        for ( ; idx >= 0 ; idx = (idx & (idx + 1)) - 1) {
            ret += ft[idx]; 
        }
        return ret; 
    }
}; 

class Solution {
public:
    int reversePairs(vector<int>& nums) {
        int n = nums.size(); 
        
        vector<int> vec = nums; 
        sort(vec.begin(), vec.end()); 
        
        unordered_map<int, vector<int>> idx; 
        for (int i = 0 ; i < n ; i++) {
            idx[vec[i]].push_back(i); 
        }
        
        int res = 0; 
        BIT obj(n); 
        for (int i = n - 1 ; i >= 0 ; i--) {
            double val = nums[i] / 2.0; 
            auto it = lower_bound(vec.begin(), vec.end(), val); 
            if (it != vec.begin()) {
                int j = prev(it) - vec.begin(); 
                res += obj.query(j); 
            }
            int pos = idx[nums[i]].back(); 
            obj.update(pos, 1); 
            idx[nums[i]].pop_back(); 
        }

        return res; 
    }
};
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