Merge Similar Items

For this problem, we need to keep track of the total weight for each unique value from both items1 and items2. We can accomplish this using a dictionary to store the weights for each unique value. Then we convert the dictionary to the final result list.

Here are the steps:

  1. Initialize an empty dictionary weights to store the weight for each unique value.
  2. Iterate over each item in items1 and items2, and for each item, add its weight to the corresponding value in the dictionary.
  3. After the loops, convert the dictionary to a list of lists where each list contains a value and its total weight. Sort this list by the value.

Here is the Python code for this:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13

class Solution:
    def mergeSimilarItems(self, items1: List[List[int]], items2: List[List[int]]) -> List[List[int]]:
        weights = {}
        for value, weight in items1 + items2:
            if value in weights:
                weights[value] += weight
            else:
                weights[value] = weight

        result = [[value, weight] for value, weight in weights.items()]
        result.sort(key=lambda x: x[0])

        return result

In this code, we first iterate over the items in both items1 and items2. We use the + operator to concatenate these two lists into a single list that we can iterate over. For each item, we check if its value is already a key in the weights dictionary. If it is, we add the item’s weight to the current total weight for that value. If it’s not, we add a new entry in the dictionary for that value with the item’s weight.

After the loops, we convert the weights dictionary to a list of lists and sort it by the value of the items. This is done by using a list comprehension to create a new list where each element is a list containing a value and its total weight. We then sort this list by the values using the sort method with a key function that returns the value of each item.

This solution has a time complexity of O(n log n), where n is the total number of items in both items1 and items2, because we need to sort the final result. The space complexity is O(n) for storing the weights in the dictionary.

Identifying Problem Isomorphism

“Merge Similar Items” can be mapped to “Intersection of Two Arrays II”.

In both problems, we are dealing with two different collections and the aim is to combine them in some way based on a common attribute (the item value in “Merge Similar Items” and the array element in “Intersection of Two Arrays II”).

However, the handling of the common elements differs between the two problems. In “Merge Similar Items”, we add up the weights of common elements, while in “Intersection of Two Arrays II”, we need to find all common elements, which requires a somewhat different approach.

“Intersection of Two Arrays II” is simpler because it only needs to keep track of common elements without any additional operation (like summing weights), and there’s no need to handle elements that exist in one array but not the other.

“Merge Similar Items” is more complex due to the additional operations on the weights and handling elements that exist in one item list but not the other.

As an approximate isomorphism, the concepts of iterating through two collections and tracking common elements underlie both problems.

Similar to https://leetcode.com/problems/merge-two-2d-arrays-by-summing-values/