<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 merge(vector<int>& nums,int l,int mid,int r){
        vector<int> a(mid-l+1);
        vector<int> b(r-mid);
        vector<int> c(r-l+1);
        for(int i=l;i<=mid;i++){
            a[i-l] = nums[i];
        }
        for(int i=mid+1;i<=r;i++){
            b[i-mid-1] = nums[i];
        }
        int j = 0,count = 0;
        for(int i=0;i<mid-l+1;i++){
            while(j<r-mid && a[i]>2*(long long)b[j]){
                j++;
            }
            count+= j;
        }
        int i=0,k=0;
        j=0;
        while(i<mid-l+1 && j<r-mid){
            if(a[i]<=b[j]){
                c[k++] = a[i];
                i++;
            }
            else{
                c[k++] = b[j];
                j++;
            }
        }
        while(i<mid-l+1){
            c[k++] = a[i];
            i++;
        }
        while(j<r-mid){
            c[k++] = b[j];
            j++;
        }
        for(int i=l;i<=r;i++){
            nums[i] = c[i-l];
        }
        return count;
    }
    int mergeSort(vector<int>& nums,int l,int r){
        if(r>l){
            int mid = l + (r-l)/2;
            int ans= 0;
            ans += mergeSort(nums,l,mid);
            ans += mergeSort(nums,mid+1,r);
            ans += merge(nums,l,mid,r);
            return ans;
        }
        return 0;
    }
    int reversePairs(vector<int>& nums) {
        return mergeSort(nums,0,nums.size()-1);
    }
};
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