<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 reversePairs(vector<int>& nums) {
        int ans=0,n=nums.size();
        vector<int>v=nums,Fenwick(n+1);
        map<int,int>mp;
        sort(v.begin(),v.end());
        for(int i=0;i<n;i++)
            mp[v[i]]=i+1;
        for(int i=n-1;i>=0;i--){
            ans+=sum(Fenwick,lower_bound(v.begin(), v.end(), nums[i] / 2.0) - v.begin());
            add(Fenwick,mp[nums[i]]);
        }
        return ans;
    }

    void add(vector<int>&Fenwick,int index){
        while(index < Fenwick.size()){
            Fenwick[index]++;
            index += (index & (-index));
        }
    }

    int sum(vector<int>&Fenwick,int index){
        int ans=0;
        while(index){
            ans+=Fenwick[index];
            index -= (index & (-index));
        }
        return ans;
    }

    int find(vector<int>&v,int target){
        int m,pos=0,l=0,r=v.size()-1;
        while(r>l){
            m=(l+r)/2;
            if(v[m] < target){
                l=m+1;
                pos=l;
            }
            else{
                r=m;
                pos=r;
            }
        }
        return pos;
    }
};
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