<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>

#define ll long long

class Solution {
public:
    ll cnt=0;
    
    void f1(int l, int r, int mid, vector<ll> &arr){
        int n1 = mid-l+1;
        int n2 = r-mid;
        vector<int> v1;
        vector<int> v2;
        for(int i=0; i<n1; i++){
            v1.push_back(arr[l+i]);
        }
        for(int i=0; i<n2; i++){
            v2.push_back(arr[mid+i+1]);
        }
        
        int right=mid+1;
          for(int i=l;i<=mid;i++)//for the first array
          {
              while(right<=r && arr[i]>2*arr[right])
              {
                  right++;
              }
              cnt+=right-(mid+1);
          }
        
        int k=l;
        int pt1=0, pt2=0;
        while(pt1<n1 && pt2<n2){
            if(v1[pt1]<=v2[pt2]){
                arr[k++] = v1[pt1];
                pt1++;
            }
            else{
                arr[k++] = v2[pt2];
                pt2++;
            }
        }
        while(pt1<n1){
            arr[k++] = v1[pt1];
            pt1++;
        }
        while(pt2<n2){
            arr[k++] = v2[pt2];
            pt2++;
        }
        return;
    }
  
    void f(int l, int r, vector<ll> &arr){
        if(l==r) return;
        int mid = (l+r)/2;
        f(l, mid, arr);
        f(mid+1, r, arr);
        f1(l, r, mid, arr);
    }

    int reversePairs(vector<int>& nums) {
        cnt = 0;
        int n = nums.size();
        vector<ll> arr(n, 0);
        for(int i=0; i<n; i++) arr[i] = nums[i];
        f(0, n-1, arr);
        return cnt;
    }
};
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