# Solving Leetcode Interviews in Seconds with AI: Merge Similar Items


	# 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](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
	> 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
	```python
	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
	```