<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:
//https://www.desgard.com/algo/docs/part3/ch02/2-segment-tree-combat/
//https://leetcode.com/problems/reverse-pairs/?envType=problem-list-v2&envId=pi1zgo2m
vector<int>segtree;
void build(int n){
    segtree.resize(4*n,0);
}
int query(int ind,int low,int high,int l,int h){
    if(h<low || l>high){
        return 0;
    }
    if(low>=l && high<=h){
        return segtree[ind];
    }
    int mid=(low+high)/2;
    return query(2*ind+1,low,mid,l,h)+query(2*ind+2,mid+1,high,l,h);
}
void update(int ind,int low,int high,int changeind){
    if(low==high){
        segtree[ind]++;
        return ;
    }
    int mid=(low+high)/2;
    if(changeind>=low && changeind<=mid){
        update(2*ind+1,low,mid,changeind);
    }
    else{
        update(2*ind+2,mid+1,high,changeind);
    }
    segtree[ind]=segtree[2*ind+1]+segtree[2*ind+2];
}
    int reversePairs(vector<int>& nums) {
        //cout<<pow(2,31)-1<<endl;
        vector<int>v;
       
        for(int i=0;i<nums.size();i++){
            v.push_back(nums[i]);
        }
        sort(v.begin(),v.end());
        map<int,int>mp;
        int st=1;
        for(int i=0;i<v.size();i++){
            if(mp.find(v[i])==mp.end()){
            mp[v[i]]=st;
            st++;
            }
        }
       
        // 0 - nums[i]/2 
        build(50001);
        int ans=0;
        int n=nums.size();
        set<int>vis;
        for(int i=n-1;i>=0;i--){
            int thresh=nums[i]/2;
            if(nums[i]%2!=0 && nums[i]>0){
            thresh++;
            }
            auto it=vis.lower_bound(thresh);
            if(it!=vis.begin()){
                int limit=mp[*(--it)];//limit according to compressed values
                ans=ans+query(0,0,50000,0,limit);
            }
            // else{
            //     if(v1[i]>0){
            //      k=query(0,0,50000,0,nums[i]/2); 
            //     }
            //     else{
            //      k=query(0,0,50000,0,nums[i]/2+1);     
            //     } 
            // }
            //cout<<i<<" "<<nums[i]<<" "<<k<<endl;
            vis.insert(nums[i]);
            update(0,0,50000,mp[nums[i]]);
        }
        return ans;
    }
    //3/2=1  iske equal
    //-3/2=-1 isse kamm
    //4/2=2 isse kamm 
};
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