Merge Similar Items

Easy

Asked by:

6 views

Topics:

ArraysDynamic Programming

You are given two 2D integer arrays, items1 and items2, representing two sets of items. Each array items has the following properties:

items[i] = [value_i, weight_i] where value_i represents the value and weight_i represents the weight of the i^th item.
The value of each item in items is unique.

Return a 2D integer array ret where ret[i] = [value_i, weight_i], with weight_i being the sum of weights of all items with value value_i.

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 <= value_i, weight_i <= 1000
Each value_i in items1 is unique.
Each value_i in items2 is unique.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the values within each item pair in the input arrays?
Can the input arrays be empty or null?
If there are no common items to merge, should I return an empty list or the original items arrays as is?
Is the order of the items within each pair significant? Should the output list be sorted in any particular way (e.g., by item value or total weight)?
Can the same item appear in multiple input pairs with potentially different weights? If so, should the final weight be the sum of all occurrences?

Brute Force Solution

Approach

The brute force approach to merging similar items involves checking every possible combination of items from two different lists. We'll go through each item in the first list and compare it to every item in the second list to see if they should be combined. If they can't be combined, we keep them separate.

Here's how the algorithm would work step-by-step:

Take the first item from the first list.
Compare it with every single item in the second list.
If you find an item in the second list that has the same type as the current item from the first list, combine them by adding their quantities together and put the new combined item in a new list.
If you go through the entire second list and don't find a match for the item from the first list, then just put the item from the first list into the new list as is.
Move to the second item in the first list and repeat the comparison process with every item in the second list.
Continue this process for every item in the first list.
After you've processed all items in the first list, add any remaining unique items that were originally only in the second list to the new combined list.
The new list now contains all the merged and unique items.

Code Implementation

def merge_similar_items_brute_force(items1, items2):
    merged_items = []

    for item1_index in range(len(items1)):
        item1_type = items1[item1_index][0]
        item1_quantity = items1[item1_index][1]
        found_match = False

        for item2_index in range(len(items2)):
            item2_type = items2[item2_index][0]
            item2_quantity = items2[item2_index][1]

            if item1_type == item2_type:
                # Combine quantities if types match
                merged_items.append([item1_type, item1_quantity + item2_quantity])
                found_match = True
                break

        # Add item1 if no match was found in items2
        if not found_match:
            merged_items.append([item1_type, item1_quantity])

    # Find and add items unique to items2
    for item2_index in range(len(items2)):
        item2_type = items2[item2_index][0]
        item2_quantity = items2[item2_index][1]
        is_unique = True

        # Check if the item from items2 already exists in merged_items
        for merged_item_index in range(len(merged_items)):
            if item2_type == merged_items[merged_item_index][0]:
                is_unique = False
                break

        # Append unique item2s to the merged list
        if is_unique:
            merged_items.append([item2_type, item2_quantity])

    merged_items.sort()

    return merged_items

Big(O) Analysis

Time Complexity

O(n*m) – Let n be the number of items in the first list and m be the number of items in the second list. The algorithm iterates through each of the n items in the first list. For each of these n items, it compares it with every one of the m items in the second list. Therefore, the total number of comparisons is n * m. Thus, the time complexity is O(n*m).

Space Complexity

O(N+M) – The provided approach creates a new list to store the merged items. In the worst case, where no items are similar, all items from the first list (of size N) and the unique items from the second list (of size M) will be added to this new list. Therefore, the auxiliary space used is proportional to the sum of the sizes of the two input lists, N+M. This leads to a space complexity of O(N+M).

Optimal Solution

Approach

The best way to solve this problem is to organize the information by item type. Then, we can easily combine the quantities for each item that is the same. Finally, we just need to present the combined list in an organized manner.

Here's how the algorithm would work step-by-step:

First, create a way to quickly look up the total quantity for each item type.
Go through the first list of items and their quantities, adding the quantities to your item lookup.
Do the same with the second list, adding their quantities to the same item lookup.
Now that you have the combined quantities, create a new list of items and their combined quantities.
Finally, organize this new list by item type in increasing order.

Code Implementation

def merge_similar_items(items1, items2):
    item_quantities = {}

    # Use a dictionary for fast quantity lookups by item.
    for item, quantity in items1:
        item_quantities[item] = quantity

    for item, quantity in items2:
        if item in item_quantities:
            item_quantities[item] += quantity
        else:
            item_quantities[item] = quantity

    # Combine items and quantities into a list of lists.
    merged_items = [[item, item_quantities[item]] for item in item_quantities]

    # Sort to match the problem description's requirements.
    merged_items.sort()

    return merged_items

Big(O) Analysis

Time Complexity

O(n log n) – Creating the item lookup, using a hash map or dictionary, takes O(n) time where n is the combined number of items across both input lists. Populating this lookup involves iterating through both lists once which is O(n). Sorting the final list of combined items, which can have at most n unique items, requires O(n log n) time using an efficient sorting algorithm. Therefore, the dominant operation is the sorting step, making the overall time complexity O(n log n).

Space Complexity

O(N) – The algorithm utilizes a lookup (e.g., a hash map) to store the total quantity for each item type. In the worst case, all items across both input lists are distinct, requiring the lookup to store N key-value pairs, where N is the total number of items in both lists. Additionally, a new list is created to hold the combined quantities, also potentially requiring space for N items. Therefore, the auxiliary space is O(N).

Edge Cases

Case	How to Handle
Both input arrays are empty	Return an empty list as there are no items to merge.
One input array is empty, the other is not	Return the non-empty array as it represents the merged items.
Input arrays contain items with the same value but different weights	The solution should sum the weights for the same value, representing a proper merge.
Input arrays contain extremely large weights that could cause integer overflow during summation	Use a larger integer type (e.g., long) or check for potential overflow before summation.
Input arrays are very large, impacting performance	Use a hash map to efficiently merge and sum weights, providing O(n+m) average time complexity.
Input arrays are sorted in reverse order	The hash map approach is unaffected by the input order, so it will still produce the correct result.
One of the value or weight is zero	The solution correctly sums the weights including zero values, properly handling zero weight adjustments.
All items in both input arrays have the same value, but different weights	The hash map will combine all the weights into a single entry for that value.

Both input arrays are empty

How to Handle:

Return an empty list as there are no items to merge.

One input array is empty, the other is not

How to Handle:

Return the non-empty array as it represents the merged items.

Input arrays contain items with the same value but different weights

How to Handle:

The solution should sum the weights for the same value, representing a proper merge.

Input arrays contain extremely large weights that could cause integer overflow during summation

How to Handle:

Use a larger integer type (e.g., long) or check for potential overflow before summation.

Input arrays are very large, impacting performance

How to Handle:

Use a hash map to efficiently merge and sum weights, providing O(n+m) average time complexity.

Input arrays are sorted in reverse order

How to Handle:

The hash map approach is unaffected by the input order, so it will still produce the correct result.

One of the value or weight is zero

How to Handle:

The solution correctly sums the weights including zero values, properly handling zero weight adjustments.

All items in both input arrays have the same value, but different weights

How to Handle:

The hash map will combine all the weights into a single entry for that value.