<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:
        res = 0  # Initialize a variable to store the count of reverse pairs
        
        def mergeSort(left, right):
            nonlocal res  # Allow access to the 'res' variable within the nested function
            
            # Base case: If the subarray has less than two elements, return
            if left >= right: 
                return
            
            # Calculate the middle index to divide the array into two halves
            mid = (left + right) // 2
            
            # Recursively sort the left half
            mergeSort(left, mid)
            # Recursively sort the right half
            mergeSort(mid + 1, right)

            # Count reverse pairs between the two halves
            i, j = left, mid + 1
            while i <= mid and j <= right:
                # If nums[i] > 2 * nums[j], then all elements from i to mid form reverse pairs with nums[j]
                if nums[i] > 2 * nums[j]:
                    res += mid - i + 1  # Add the number of such pairs to the result
                    j += 1
                else:
                    i += 1
            
            # Merge the two sorted halves
            temp = []
            i, j = left, mid + 1
            while i <= mid and j <= right:
                # Add the smaller element from either half to the temp array
                if nums[i] <= nums[j]:
                    temp.append(nums[i])
                    i += 1
                else:
                    temp.append(nums[j])
                    j += 1
            
            # Add any remaining elements from the left half to the temp array
            while i <= mid:
                temp.append(nums[i])
                i += 1
            
            # Add any remaining elements from the right half to the temp array
            while j <= right:
                temp.append(nums[j])
                j += 1
            
            # Copy the sorted elements back into the original array
            nums[left:right+1] = temp[:]

        # Call the mergeSort function on the entire array
        mergeSort(0, len(nums) - 1)
        
        # Return the total count of reverse pairs
        return res

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