Introduction

In this blog post, we will explore how to solve the LeetCode problem "2363" 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, items1 and items2, representing two sets of items. Each array items has the following properties: items[i] = [valuei, weighti] where valuei represents the value and weighti represents the weight of the ith item. The value of each item in items is unique. Return a 2D integer array ret where ret[i] = [valuei, weighti], with weighti being the sum of weights of all items with value valuei. Note: ret should be returned in ascending order by value. Example 1: Input: items1 = [[1,1],[4,5],[3,8]], items2 = [[3,1],[1,5]] Output: [[1,6],[3,9],[4,5]] Explanation: The item with value = 1 occurs in items1 with weight = 1 and in items2 with weight = 5, total weight = 1 + 5 = 6. The item with value = 3 occurs in items1 with weight = 8 and in items2 with weight = 1, total weight = 8 + 1 = 9. The item with value = 4 occurs in items1 with weight = 5, total weight = 5. Therefore, we return [[1,6],[3,9],[4,5]]. Example 2: Input: items1 = [[1,1],[3,2],[2,3]], items2 = [[2,1],[3,2],[1,3]] Output: [[1,4],[2,4],[3,4]] Explanation: The item with value = 1 occurs in items1 with weight = 1 and in items2 with weight = 3, total weight = 1 + 3 = 4. The item with value = 2 occurs in items1 with weight = 3 and in items2 with weight = 1, total weight = 3 + 1 = 4. The item with value = 3 occurs in items1 with weight = 2 and in items2 with weight = 2, total weight = 2 + 2 = 4. Therefore, we return [[1,4],[2,4],[3,4]]. Example 3: Input: items1 = [[1,3],[2,2]], items2 = [[7,1],[2,2],[1,4]] Output: [[1,7],[2,4],[7,1]] Explanation: The item with value = 1 occurs in items1 with weight = 3 and in items2 with weight = 4, total weight = 3 + 4 = 7. The item with value = 2 occurs in items1 with weight = 2 and in items2 with weight = 2, total weight = 2 + 2 = 4. The item with value = 7 occurs in items2 with weight = 1, total weight = 1. Therefore, we return [[1,7],[2,4],[7,1]]. Constraints: 1 <= items1.length, items2.length <= 1000 items1[i].length == items2[i].length == 2 1 <= valuei, weighti <= 1000 Each valuei in items1 is unique. Each valuei in items2 is unique.

Explanation

Here's a breakdown of the solution:

• Hashing for Efficiency: Use a dictionary (hash map) to store the value as the key and the accumulated weight as the value. This allows for O(1) average time complexity for lookups and updates.
• Iterate and Accumulate: Iterate through both items1 and items2. For each item, either add its weight to the existing weight in the dictionary or insert a new entry if the value is not yet present.

• Sort and Return: Convert the dictionary into a list of lists (2D array) and sort it based on the value in ascending order before returning it.

• Runtime Complexity: O(n + m + k log k) where n is the length of items1, m is the length of items2, and k is the number of unique values.

• Storage Complexity: O(k) where k is the number of unique values (to store the dictionary).

Code

    def merge_items(items1, items2):
    """
    Merges two lists of items, summing the weights of items with the same value.

    Args:
        items1: A list of lists, where each inner list is [value, weight].
        items2: A list of lists, where each inner list is [value, weight].

    Returns:
        A list of lists, where each inner list is [value, total_weight], sorted by value.
    """

    item_weights = {}

    for value, weight in items1:
        item_weights[value] = item_weights.get(value, 0) + weight

    for value, weight in items2:
        item_weights[value] = item_weights.get(value, 0) + weight

    result = []
    for value, weight in item_weights.items():
        result.append([value, weight])

    result.sort()  # Sort by value

    return result