How do you solve Merge Two 2D Arrays by Summing Values on LeetCode?

Merge Two 2D Arrays by Summing Values (LeetCode #2570) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using a hash map allows for efficient summation of values without nested loops.

What is the brute force solution for Merge Two 2D Arrays by Summing Values?

The brute force approach for Merge Two 2D Arrays by Summing Values is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves iterating through both arrays and summing values for each unique id. This is straightforward but inefficient, as it checks every id against every other id. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Merge Two 2D Arrays by Summing Values?

Merge Two 2D Arrays by Summing Values uses the following concepts: Array, Hash Table, Two Pointers. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Merge Two 2D Arrays by Summing Values?

Always consider edge cases, such as empty arrays or arrays with non-overlapping ids. Explain your thought process clearly while coding; it shows your understanding. Practice writing clean and efficient code, as readability is key in interviews.

What are common mistakes when solving Merge Two 2D Arrays by Summing Values?

Forgetting to handle ids that exist in only one array. Not using a hash map, leading to inefficient solutions.

#2570

Merge Two 2D Arrays by Summing Values

Easy

ArrayHash TableTwo PointersHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

Using a hash map allows us to efficiently sum values for each id in a single pass through both arrays. This reduces the time complexity significantly by avoiding nested loops.

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Algorithm

4 steps

1Step 1: Create a hash map to store the sum of values for each id.
2Step 2: Iterate through nums1 and add each id and its value to the hash map.
3Step 3: Iterate through nums2 and update the hash map by adding values for existing ids or creating new entries.
4Step 4: Convert the hash map to a list and sort it by id.

solution.py9 lines

1from collections import defaultdict
2
3def mergeArrays(nums1, nums2):
4    id_map = defaultdict(int)
5    for id1, val1 in nums1:
6        id_map[id1] += val1
7    for id2, val2 in nums2:
8        id_map[id2] += val2
9    return sorted(id_map.items())

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step after inserting elements into the hash map, while the space complexity is O(n) for storing the sums.

1Using a hash map allows for efficient summation of values without nested loops.
2Sorting the final results is necessary to meet the output requirements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.