Introduction

In this blog post, we will explore how to solve the LeetCode problem "2570" 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 2D integer arrays nums1 and nums2. nums1[i] = [idi, vali] indicate that the number with the id idi has a value equal to vali. nums2[i] = [idi, vali] indicate that the number with the id idi has a value equal to vali. Each array contains unique ids and is sorted in ascending order by id. Merge the two arrays into one array that is sorted in ascending order by id, respecting the following conditions: Only ids that appear in at least one of the two arrays should be included in the resulting array. Each id should be included only once and its value should be the sum of the values of this id in the two arrays. If the id does not exist in one of the two arrays, then assume its value in that array to be 0. Return the resulting array. The returned array must be sorted in ascending order by id. Example 1: Input: nums1 = [[1,2],[2,3],[4,5]], nums2 = [[1,4],[3,2],[4,1]] Output: [[1,6],[2,3],[3,2],[4,6]] Explanation: The resulting array contains the following: - id = 1, the value of this id is 2 + 4 = 6. - id = 2, the value of this id is 3. - id = 3, the value of this id is 2. - id = 4, the value of this id is 5 + 1 = 6. Example 2: Input: nums1 = [[2,4],[3,6],[5,5]], nums2 = [[1,3],[4,3]] Output: [[1,3],[2,4],[3,6],[4,3],[5,5]] Explanation: There are no common ids, so we just include each id with its value in the resulting list. Constraints: 1 <= nums1.length, nums2.length <= 200 nums1[i].length == nums2[j].length == 2 1 <= idi, vali <= 1000 Both arrays contain unique ids. Both arrays are in strictly ascending order by id.

Explanation

• Two Pointers: Utilize two pointers, one for each input array, to iterate through them simultaneously.
  • Comparison and Merging: Compare the IDs at the pointer positions. If the IDs are equal, sum their values and add the result to the merged array. If they are different, add the entry with the smaller ID to the merged array.
  • Handle Remaining Elements: After one array is exhausted, add the remaining elements from the other array to the merged array.

• Runtime Complexity: O(m + n), where m and n are the lengths of nums1 and nums2 respectively. Storage Complexity: O(m + n) in the worst case (no common IDs).

Code

    def mergeArrays(nums1, nums2):
    """
    Merges two sorted 2D arrays based on ID and sums values for common IDs.

    Args:
        nums1: The first 2D array (sorted by ID).
        nums2: The second 2D array (sorted by ID).

    Returns:
        A new 2D array containing the merged and summed data, sorted by ID.
    """
    merged_array = []
    i = 0
    j = 0

    while i < len(nums1) and j < len(nums2):
        id1, val1 = nums1[i]
        id2, val2 = nums2[j]

        if id1 == id2:
            merged_array.append([id1, val1 + val2])
            i += 1
            j += 1
        elif id1 < id2:
            merged_array.append([id1, val1])
            i += 1
        else:
            merged_array.append([id2, val2])
            j += 1

    # Add remaining elements from nums1, if any
    while i < len(nums1):
        merged_array.append(nums1[i])
        i += 1

    # Add remaining elements from nums2, if any
    while j < len(nums2):
        merged_array.append(nums2[j])
        j += 1

    return merged_array