<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 lowerBoundExclusive(vector<int> &a,double target)
    {
        int n=a.size();
        int ans=-1;
        int l=0,r=n-1;
        while(l<=r)
        {
            int mid=l+(r-l)/2;
            if(a[mid]<target)
            {
                ans=mid;
                l=mid+1;
            }
            else r=mid-1;
        }
        return ans;
    }

    int find(vector<int> &a,vector<int> b)
    {
        int ans=0;
        int n=a.size();
        int m=b.size();
        for(int i=0;i<n;i++)
        {
            ans+=(lowerBoundExclusive(b,a[i]/2.0)+1);
        }
        return ans;
    }


    void mergeSort(vector<int> &a,int l,int r,int &count)
    {
        if(l<r)
        {
            int mid=l+(r-l)/2;
            mergeSort(a,l,mid,count);
            mergeSort(a,mid+1,r,count);
            merge(a,l,mid,r,count);
        }
    }
    void merge(vector<int> &a,int l,int mid,int r,int &count)
    {
       
        
        int n=mid-l+1;
        int m=r-mid;
        vector<int> left(n),right(m);
        for(int i=0;i<n;i++)
            left[i]=a[l+i];
        for(int j=0;j<m;j++)
            right[j]=a[mid+1+j];

        //Here we have both sorted arrays so count the number of 
        //largest numbers from first array to second array
        count+=find(left,right);

        int i=0;
        int j=0;
        while(i<n && j<m)
        {
            if(left[i]<right[j])
                a[l++]=left[i++];
            else a[l++]=right[j++];
        }
        while(i<n)
            a[l++]=left[i++];
        while(j<m)
            a[l++]=right[j++];
    }



    int reversePairs(vector<int>& a) {
        int n=a.size();
        int count=0;
        mergeSort(a,0,n-1,count);
        for(auto it:a)
            cout<<it<<" ";
        return count;
    }
};
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