Problem Description

You are given two 2D integer arrays, items1 and items2, representing two sets of items. Each array contains items where each item is represented as [value, weight]:

• value represents the unique identifier/value of the item
• weight represents the weight of that item
• All values within each array are unique

Your task is to merge these two arrays by combining items with the same value. When items from both arrays have the same value, you need to add their weights together.

The output should be a 2D array where:

• Each element is [value, total_weight]
• total_weight is the sum of all weights for items with that specific value across both input arrays
• The result must be sorted in ascending order by value

For example:

• If items1 has [1, 5] and items2 has [1, 3], the merged result would include [1, 8] (weights 5 + 3 = 8)
• If a value appears in only one array, include it with its original weight
• The final result should list all unique values with their combined weights, sorted by value

Intuition

The key insight is that we need to group items by their value and sum up their weights. This is a classic aggregation problem where we need to combine data from two sources based on a common key (the value).

Think of it like merging inventory from two warehouses - if both warehouses have the same product (identified by value), we want to know the total quantity (weight) across both locations.

A hash table (or dictionary/counter) is the natural data structure for this task because:

1. We need to quickly look up whether we've seen a particular value before
2. We need to accumulate weights for the same value
3. Hash tables provide O(1) average time complexity for both insertion and lookup

The solution becomes straightforward:

• Iterate through both arrays
• For each item [value, weight], add the weight to our running total for that value
• Using a Counter in Python makes this even simpler as it automatically handles missing keys by starting at 0

Once we have all the aggregated weights by value, we just need to convert our hash table back to a list of [value, weight] pairs and sort them by value. The sorted() function on dictionary items naturally sorts by the first element (key/value) in ascending order, which is exactly what we need.

The beauty of this approach is that we don't need to worry about which array an item comes from or whether it appears in one or both arrays - we simply process all items uniformly and let the hash table handle the aggregation.

Learn more about Sorting patterns.

Solution Approach

The implementation uses a hash table to efficiently merge and aggregate items from both arrays. Here's how the solution works step by step:

1. Initialize a Counter (Hash Table)

cnt = Counter()

We use Python's Counter from the collections module, which is a specialized dictionary that automatically initializes missing keys with 0. This eliminates the need to check if a key exists before adding to it.

2. Iterate Through Both Arrays

for v, w in chain(items1, items2):
    cnt[v] += w

The chain() function from itertools concatenates items1 and items2 into a single iterable sequence. This allows us to process all items in one loop without writing separate loops for each array.

For each item [v, w] (value and weight):

• We add the weight w to the running total for value v
• If v doesn't exist in the counter yet, it starts at 0 and then adds w
• If v already exists, we simply add w to the existing total

3. Sort and Return the Result

return sorted(cnt.items())

The cnt.items() method returns key-value pairs as tuples (value, total_weight). The sorted() function:

• Converts these tuples into a list
• Sorts them by the first element (value) in ascending order by default
• Returns a list of lists in the format [[value1, weight1], [value2, weight2], ...]

Time Complexity: O((m + n) log(m + n)) where m and n are the lengths of items1 and items2. The iteration takes O(m + n) time, and sorting takes O(k log k) where k is the number of unique values (at most m + n).

Space Complexity: O(m + n) for storing the counter with at most m + n unique values.

Example Walkthrough

Let's walk through a concrete example to illustrate how the solution works:

Given:

• items1 = [[1, 3], [2, 2], [4, 5]]
• items2 = [[1, 1], [3, 4], [4, 2]]

Step 1: Initialize the Counter We start with an empty counter: cnt = {}

Step 2: Process items from both arrays

Processing items1:

• [1, 3]: Add value 1 with weight 3 → cnt = {1: 3}
• [2, 2]: Add value 2 with weight 2 → cnt = {1: 3, 2: 2}
• [4, 5]: Add value 4 with weight 5 → cnt = {1: 3, 2: 2, 4: 5}

Processing items2:

• [1, 1]: Value 1 already exists, add weight 1 to existing 3 → cnt = {1: 4, 2: 2, 4: 5}
• [3, 4]: Add value 3 with weight 4 → cnt = {1: 4, 2: 2, 3: 4, 4: 5}
• [4, 2]: Value 4 already exists, add weight 2 to existing 5 → cnt = {1: 4, 2: 2, 3: 4, 4: 7}

Step 3: Convert to sorted list

• Extract items: [(1, 4), (2, 2), (3, 4), (4, 7)]
• Sort by value (already sorted in this case)
• Convert to list format: [[1, 4], [2, 2], [3, 4], [4, 7]]

Final Result: [[1, 4], [2, 2], [3, 4], [4, 7]]

Note how:

• Value 1 appeared in both arrays with weights 3 and 1, resulting in total weight 4
• Value 2 appeared only in items1 with weight 2
• Value 3 appeared only in items2 with weight 4
• Value 4 appeared in both arrays with weights 5 and 2, resulting in total weight 7

Solution Implementation

Time and Space Complexity

The time complexity is O((n + m) * log(n + m)) where n and m are the lengths of items1 and items2 respectively.

• Iterating through both lists using chain(items1, items2) takes O(n + m) time
• Building the Counter by adding values takes O(n + m) time
• The sorted() function on cnt.items() sorts at most n + m unique items, which takes O((n + m) * log(n + m)) time

The overall time complexity is dominated by the sorting step: O((n + m) * log(n + m))

The space complexity is O(n + m) where n and m are the lengths of items1 and items2 respectively.

• The Counter cnt stores at most n + m key-value pairs (in the worst case where all values are unique)
• The sorted() function returns a new list containing at most n + m items

Therefore, the space complexity is O(n + m).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Using a Regular Dictionary Instead of Counter

A common mistake is using a regular dictionary without properly handling missing keys:

Incorrect approach:

def mergeSimilarItems(self, items1, items2):
    weight_dict = {}
    for value, weight in chain(items1, items2):
        weight_dict[value] += weight  # KeyError if value doesn't exist!
    return sorted(weight_dict.items())

Why it fails: This will raise a KeyError when encountering a value for the first time since weight_dict[value] doesn't exist yet.

Solutions:

• Use Counter() as shown in the original solution (automatically handles missing keys)
• Or manually check for key existence:

weight_dict = {}
for value, weight in chain(items1, items2):
    if value in weight_dict:
        weight_dict[value] += weight
    else:
        weight_dict[value] = weight

• Or use defaultdict(int):

from collections import defaultdict
weight_dict = defaultdict(int)
for value, weight in chain(items1, items2):
    weight_dict[value] += weight

2. Forgetting to Return a List of Lists

The problem expects the output as a 2D array (list of lists), but sorted(cnt.items()) returns a list of tuples:

Potential issue:

return sorted(cnt.items())  # Returns [(1, 5), (2, 3), ...] instead of [[1, 5], [2, 3], ...]

Why it might matter: While Python often treats tuples and lists interchangeably, some test systems or strict type checkers might expect the exact format List[List[int]].

Solution if strict list format is required:

return [[value, weight] for value, weight in sorted(cnt.items())]

3. Incorrectly Handling the Sorting

Some might try to sort by weight instead of value, or forget to sort entirely:

Incorrect approaches:

# Sorting by weight (second element) instead of value
return sorted(cnt.items(), key=lambda x: x[1])

# Forgetting to sort at all
return list(cnt.items())

Solution: Always sort by value (the first element), which sorted() does by default when no key is specified.

4. Inefficient Merging with Nested Loops

Attempting to merge using nested loops to find matching values:

Inefficient approach:

def mergeSimilarItems(self, items1, items2):
    result = []
    processed = set()
  
    # Process items1
    for v1, w1 in items1:
        if v1 not in processed:
            total_weight = w1
            # Search for matching value in items2
            for v2, w2 in items2:
                if v1 == v2:
                    total_weight += w2
            result.append([v1, total_weight])
            processed.add(v1)
  
    # Add remaining items from items2
    for v2, w2 in items2:
        if v2 not in processed:
            result.append([v2, w2])
  
    return sorted(result)

Why it's problematic: This has O(m*n) time complexity for the merging phase alone, making it inefficient for large inputs.

Solution: Use a hash table approach as in the original solution for O(m+n) merging time.