<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 mergeAndCount(vector<int>& nums,int left,int mid,int right){
        int lSize = mid-left+1;
        int rSize = right-mid;
        int n = right-left+1;
        vector<int> l(lSize),r(rSize),f(n);
        for(int i=0;i<lSize;i++){
            l[i] = nums[left+i];
        }
        for(int i=0;i<rSize;i++){
            r[i] = nums[mid+1+i];
        }
        int i=0,j=0,k=0;
        int count = 0;
        for(;i<lSize;i++){
            while(j<rSize && l[i] > 2LL*r[j]){
                j++;
            }
            count += j;
        }
        i = 0,j=0;
        while(i<lSize && j<rSize){
            if(l[i]<=r[j]){
                f[k] = l[i];
                i++;
            }
            else{
                f[k] = r[j];
                j++;
            }
            k++;
        }
        while(i<lSize){
            f[k++] = l[i++];
        }
        while(j<rSize){
            f[k++] = r[j++];
        }
        for(int i=0;i<n;i++){
            nums[left+i] = f[i];
        }
        
        return count;
    }
    int mergeSort(vector<int> &nums,int left,int right){
        int count = 0;
        if(left < right){
            int mid = left + (right-left)/2;
            count += mergeSort(nums,left,mid);
            count += mergeSort(nums,mid+1,right);
            count += mergeAndCount(nums,left,mid,right);
        }
        return count;
    }
    int reversePairs(vector<int>& nums) {
        int n = nums.size();
        return mergeSort(nums,0,n-1);
    }
};
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