Introduction

In this blog post, we will explore how to solve the LeetCode problem "2426" 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

You are given two 0-indexed integer arrays nums1 and nums2, each of size n, and an integer diff. Find the number of pairs (i, j) such that: 0 <= i < j <= n - 1 and nums1[i] - nums1[j] <= nums2[i] - nums2[j] + diff. Return the number of pairs that satisfy the conditions. Example 1: Input: nums1 = [3,2,5], nums2 = [2,2,1], diff = 1 Output: 3 Explanation: There are 3 pairs that satisfy the conditions: 1. i = 0, j = 1: 3 - 2 <= 2 - 2 + 1. Since i < j and 1 <= 1, this pair satisfies the conditions. 2. i = 0, j = 2: 3 - 5 <= 2 - 1 + 1. Since i < j and -2 <= 2, this pair satisfies the conditions. 3. i = 1, j = 2: 2 - 5 <= 2 - 1 + 1. Since i < j and -3 <= 2, this pair satisfies the conditions. Therefore, we return 3. Example 2: Input: nums1 = [3,-1], nums2 = [-2,2], diff = -1 Output: 0 Explanation: Since there does not exist any pair that satisfies the conditions, we return 0. Constraints: n == nums1.length == nums2.length 2 <= n <= 105 -104 <= nums1[i], nums2[i] <= 104 -104 <= diff <= 104

Explanation

Here's a breakdown of the solution approach, followed by the Python code:

• Transform the Inequality: The core idea is to rearrange the given inequality nums1[i] - nums1[j] <= nums2[i] - nums2[j] + diff to nums1[i] - nums2[i] <= nums1[j] - nums2[j] + diff. Let arr[i] = nums1[i] - nums2[i]. Then the inequality becomes arr[i] <= arr[j] + diff.

• Count Valid Pairs using Merge Sort: Iterate through all possible pairs (i, j) where i < j. This would be O(n^2) if done naively. A much better approach is to use a modified merge sort. During the merge step, for each element arr[i] in the left subarray, we efficiently count the number of elements arr[j] in the right subarray that satisfy arr[i] <= arr[j] + diff. This is done by iterating through the right sub-array using binary search to find the elements that satisfy the desired inequality.

• Merge Sort for Stable Counting: Standard merge sort algorithms can be adapted. The key is to add the logic of counting during the merge operation efficiently.

• Runtime and Storage Complexity: Time complexity is O(n log n) due to the modified merge sort. Storage complexity is O(n) due to the temporary array used in merge sort.

Code

    def count_pairs(nums1, nums2, diff):
    """
    Counts the number of pairs (i, j) such that i < j and nums1[i] - nums1[j] <= nums2[i] - nums2[j] + diff.

    Args:
        nums1: The first array of integers.
        nums2: The second array of integers.
        diff: The integer difference.

    Returns:
        The number of pairs that satisfy the conditions.
    """

    n = len(nums1)
    arr = [nums1[i] - nums2[i] for i in range(n)]
    count = 0

    def merge_sort(arr, low, high):
        nonlocal count
        if low < high:
            mid = (low + high) // 2
            merge_sort(arr, low, mid)
            merge_sort(arr, mid + 1, high)
            count += merge(arr, low, mid, high, diff)

    def merge(arr, low, mid, high, diff):
        count = 0
        i = low
        j = mid + 1
        temp = []

        while i <= mid and j <= high:
            if arr[i] <= arr[j] + diff:
                count += (high - j + 1)
                i += 1
            else:
                j += 1

        i = low
        j = mid + 1

        while i <= mid and j <= high:
            if arr[i] <= arr[j]:
                temp.append(arr[i])
                i += 1
            else:
                temp.append(arr[j])
                j += 1
        while i <= mid:
            temp.append(arr[i])
            i += 1
        while j <= high:
            temp.append(arr[j])
            j += 1

        for i in range(len(temp)):
            arr[low + i] = temp[i]
        return count

    merge_sort(arr, 0, n - 1)
    return count