# Solving Leetcode Interviews in Seconds with AI: Reverse Pairs


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "493" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given an integer array nums, return the number of reverse pairs in the array. A reverse pair is a pair (i, j) where:  0 <= i < j < nums.length and nums[i] > 2 * nums[j].    Example 1:  Input: nums = [1,3,2,3,1] Output: 2 Explanation: The reverse pairs are: (1, 4) --> nums[1] = 3, nums[4] = 1, 3 > 2 * 1 (3, 4) --> nums[3] = 3, nums[4] = 1, 3 > 2 * 1  Example 2:  Input: nums = [2,4,3,5,1] Output: 3 Explanation: The reverse pairs are: (1, 4) --> nums[1] = 4, nums[4] = 1, 4 > 2 * 1 (2, 4) --> nums[2] = 3, nums[4] = 1, 3 > 2 * 1 (3, 4) --> nums[3] = 5, nums[4] = 1, 5 > 2 * 1    Constraints:  1 <= nums.length <= 5 * 104 -231 <= nums[i] <= 231 - 1  

	# Explanation
	Here's a solution to efficiently count reverse pairs in an integer array:

*   **Divide and Conquer:** The core idea is to use a modified merge sort. We recursively divide the array into two halves, count reverse pairs within each half, and then count reverse pairs that span across the two halves.
*   **Merge and Count:** During the merge step, we efficiently count the reverse pairs between the left and right subarrays. Because the subarrays are sorted, we can use a two-pointer approach to determine the number of elements in the left subarray that are greater than twice the elements in the right subarray.
*   **Sort for Recursion:** After counting, the merge step also sorts the combined subarray, ensuring that the recursive calls maintain the sorted property needed for the two-pointer counting approach.

*   **Time Complexity:** O(n log n), **Space Complexity:** O(n) due to the auxiliary space used in the merge operation.

	
	# Code
	```python
	def reversePairs(nums):
    def merge_sort(nums, start, end):
        if start >= end:
            return 0

        mid = (start + end) // 2
        count = merge_sort(nums, start, mid) + merge_sort(nums, mid + 1, end)

        # Count reverse pairs across the two halves
        j = mid + 1
        for i in range(start, mid + 1):
            while j <= end and nums[i] > 2 * nums[j]:
                j += 1
            count += j - (mid + 1)

        # Merge the two sorted halves
        left = nums[start:mid + 1]
        right = nums[mid + 1:end + 1]
        i, j = 0, 0
        k = start
        while i < len(left) and j < len(right):
            if left[i] <= right[j]:
                nums[k] = left[i]
                i += 1
            else:
                nums[k] = right[j]
                j += 1
            k += 1

        while i < len(left):
            nums[k] = left[i]
            i += 1
            k += 1

        while j < len(right):
            nums[k] = right[j]
            j += 1
            k += 1

        return count

    return merge_sort(nums, 0, len(nums) - 1)
	```