<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>

long long merge(vector<long long>& arr, int start, int mid, int end) {
    int sizel = mid - start + 1;
    int sizer = end - mid;
    long long tot = 0;
    vector<long long> left(sizel), right(sizer);
    for (int i = 0; i < sizel; i++)
        left[i] = arr[start + i];
    for (int i = 0; i < sizer; i++)
        right[i] = arr[mid + i + 1];
    int i = 0, j = 0, k = start;
    while (i < sizel && j < sizer) {
        if (left[i] / 2 > right[j]) {
            tot += sizel - i;
            j++;
        } else {
            i++;
        } 
    }
    i = 0, j = 0 ;
    while (i < sizel && j < sizer) {
        if (left[i] > right[j]) {
            arr[k] = right[j];
            j++;
        } else {
            arr[k] = left[i];
            i++;
        }
        k++;
    }
    if (i == sizel) {
        while (j != sizer) {
            arr[k] = right[j];
            j++;
            k++;
        }
    } else {
        while (i != sizel) {
            arr[k] = left[i];
            k++;
            i++;
        }
    }
    return tot;
}
long long mergeSort(vector<long long>& arr, int begin, int end) {
    if (begin >= end)
        return 0;
    int mid = begin + (end - begin) / 2;
    long long tot = 0;
    tot += mergeSort(arr, begin, mid);
    tot += mergeSort(arr, mid + 1, end);
    tot += merge(arr, begin, mid, end);
    return tot;
}
class Solution {
public:
    int reversePairs(vector<int>& nums) {
        int n = nums.size(); 
        vector<long long> temp ;
        for(int i=0;i<n;i++) temp.push_back(nums[i]*2ll) ;
        return mergeSort(temp,0,n-1) ;
    }
};
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