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, 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

    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)