<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:
    def reversePairs(self, nums: List[int]) -> int:
        # clarify the input
        # what is maximum length of nums?
        # is num value positive?
        # brute force
        # use two for loop
        # for each number in nums
        # find next number satisfy condition and increase counter
        # this solution O(n^2) and O(1) space
        # since we want to find number with i < j
        # if we have a sorted array with all j > i, we can easily count
        # this bring me the idea of borrowing the merge sort algorithm
        # we split the nums into two half and sort each half
        # now we can compair a number on left half with number on right half
        # and update counter
        # we recursively running it
        # so the total time complexity is just like merge sort, which is O(nlogn)
        # space complexity is O(n)

        self.ans = 0

        def merge(A, B):
            return sorted(A + B)

        def merge_sort(i, j):
            if i >= j:
                return
            mid = (i + j) // 2
            merge_sort(i, mid)
            merge_sort(mid + 1, j)
            # we update count for our ans
            r = mid + 1
            for l in range(i, mid + 1):
                while r <= j and nums[l] > 2 * nums[r]:
                    r += 1
                self.ans += r - (mid + 1)
                
            # now we merge two sorted half
            nums[i: j + 1] = merge(nums[i: mid + 1], nums[mid + 1: j + 1])
        
        merge_sort(0, len(nums) - 1)
        return self.ans


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