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Merge Similar Items

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.

Solution Explanation: Merging Similar Items

The problem asks to merge two lists of items, where each item is represented by a value and a weight. Items with the same value should have their weights summed. The result should be a sorted list of unique items with their combined weights.

Approach

The most efficient approach involves using a hash table (or a similar data structure like a dictionary in Python or a map in C++/Java) to store the total weight for each unique item value. This allows for fast lookups and updates as we iterate through both input lists.

Here's a breakdown of the steps:

  1. Initialization: Create a hash table (or array if the value range is known and small). The keys will be the item values, and the values will be their cumulative weights.

  2. Iteration: Iterate through both items1 and items2. For each item [value, weight]:

    • Check if the value exists as a key in the hash table.
    • If it exists, add the weight to the current weight associated with that value.
    • If it doesn't exist, add the value as a key to the hash table with the weight as its value.

  3. Result Construction: Create a list to store the result. Iterate through the hash table (or array) and add each [value, weight] pair to the result list. Sort the list by value in ascending order.

  4. Return: Return the sorted result list.

Time and Space Complexity Analysis

  • Time Complexity: The algorithm iterates through items1 and items2 once each, taking O(n + m) time, where n is the length of items1 and m is the length of items2. The sorting step takes O((n+m) log(n+m)) time in the worst case, but since the number of unique items is likely much smaller than n+m, it often performs better. The overall time complexity is dominated by the iteration and sorting, resulting in O((n+m)log(n+m)). If we use an array instead of a hash table and the range of values is small, the sorting will dominate, resulting in O((n+m)log(n+m)).

  • Space Complexity: The hash table (or array) requires space proportional to the number of unique item values, which is at most O(n + m) in the worst case (all values are unique). The result list also takes O(n + m) space in the worst case. Therefore, the space complexity is O(n + m).

Code Examples (Python, Java, C++)

The provided code examples in the question demonstrate this approach effectively. Note that the Python solution uses the Counter object from the collections module which provides a convenient way to count items. The other solutions use arrays (or similar) to accomplish the same. The choice of using an array instead of a hash table in some solutions is a tradeoff between efficiency and memory usage, optimal when the range of item values is small and known in advance.