<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>

void merge(vector<int>& nums,int start,int mid,int end,int &count){
    int s1 = mid-start+1;
    int s2 = end-mid;
    vector<long int> l(s1,0);
    vector<long int> r(s2,0);
    for(int i=0;i<s1;i++){
        l[i]=nums[start+i];
    }
    for(int i=0;i<s2;i++){
        r[i]=nums[mid+1+i];
    }
    int i=0;
    int j=0;
    while(i<s1 && j<s2){
        if(l[i] > 2*r[j]){
            count = count + s1-i;
            j++;
        }
        else{
            i++;
        }
    }
    // while(i<s1){
    //     i++;
    //     if(i<s1){
    //     count=count+s2;
    //     }
    // }
    i=0;
    j=0;
    int k=start;
    while(i<s1 && j<s2){
        if(l[i] <= r[j]){
            nums[k]=l[i];
            i++;
        }
        else{
            nums[k]=r[j];
            j++;
        }
        k++;
    }
    while(i<s1){
        nums[k]=l[i];
        i++;
        k++;
    }
    while(j<s2){
        nums[k]=r[j];
        j++;
        k++;
    }
}
void mergesort(vector<int> &nums,int start,int end,int &count){
    if(start >= end){return;}
    int mid = (start+end)/2;
    mergesort(nums,start,mid,count);
    mergesort(nums,mid+1,end,count);
    merge(nums,start,mid,end,count);
}
class Solution {
public:
    int reversePairs(vector<int>& nums) {
        int n=nums.size();
        int count = 0;
        mergesort(nums,0,n-1,count);
        return count;
    }
};
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