<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:
    void merge(vector<int>&a,int lo,int mid,int hi){
        vector<long long int>temp(hi-lo+1);
        
        int left = lo;
        int right = mid+1;
        int k =0;
        while(left<=mid && right<=hi){
            if(a[left]>a[right]){
                temp[k++] = (a[right]);
                right++;
            }
            else{
                temp[k++] = (a[left]);
                left++;
            }
        }
        while(left<=mid){
            temp[k++] = (a[left]);
            left++;
        }
        while(right<=hi){
            temp[k++] = (a[right]);
            right++;
        }
        for(int i = lo, k = 0; i <= hi; i++, k++) {
            a[i] = temp[k];
        }

        
    }

int countPairs(vector<int> &arr, int low, int mid, int high) {
    int right = mid + 1;
    int cnt = 0;
    for (int i = low; i <= mid; i++) {
        while (right <= high && (long long)arr[i] > (long long)2*arr[right]) right++;
        cnt += (right - (mid + 1));
    }
    return cnt;
}
        int mergeSort(vector<int>&a,int lo,int hi){
    
        if(lo>=hi){ return 0; }

            int cnt =0;
            int mid = lo + (hi-lo)/2;
            cnt+= mergeSort(a,lo,mid);
            
            cnt+=mergeSort(a,mid+1,hi);

            cnt+=countPairs(a,lo,mid,hi);

            merge(a,lo,mid,hi);

        return cnt;
    }
    int reversePairs(vector<int>& nums) {
        int n = nums.size();
        return mergeSort(nums,0,n-1);
    }
};
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