How do you solve Group By on LeetCode?

Group By (LeetCode #2631) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a HashMap allows for efficient grouping of items.

What is the brute force solution for Group By?

The brute force approach for Group By is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves iterating through each element of the array and checking if the key already exists in the result object. If it does, we append the item to the corresponding array; if not, we create a new array for that key. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Group By?

Clarify the requirements and edge cases before coding. Think about the efficiency of your solution and explain your thought process. Practice writing clean and concise code.

What are common mistakes when solving Group By?

Not using the correct data structure for grouping (like an object or map). Failing to handle cases where the key already exists.

#2631

Group By

Medium

Hash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution uses a single pass through the array and a HashMap (or object) to store the grouped results. This approach is efficient because it reduces the number of lookups and checks needed, leading to a linear time complexity.

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Algorithm

5 steps

1Step 1: Initialize an empty object to hold the grouped results.
2Step 2: Loop through each item in the array.
3Step 3: For each item, call the provided function to get the key.
4Step 4: Use the key to either add the item to an existing array or create a new array.
5Step 5: Return the result object.

solution.py6 lines

1def group_by(array, fn):
2    result = {}
3    for item in array:
4        key = fn(item)
5        result.setdefault(key, []).append(item)
6    return result

ℹ

Complexity note: The time complexity is O(n) because we loop through the array once, and each key insertion into the object takes average O(1) time. The space complexity is O(n) because in the worst case, we store all items in the result object.

1Using a HashMap allows for efficient grouping of items.
2Understanding the difference between time complexity in brute force vs optimal solutions is crucial.

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