How do you solve Chunk Array on LeetCode?

Chunk Array (LeetCode #2677) 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: Understanding how to manipulate arrays and create subarrays is crucial.

What is the brute force solution for Chunk Array?

The brute force approach for Chunk Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves iterating through the array and manually creating subarrays of the specified size. This method is straightforward but inefficient as it repeatedly checks and creates new arrays. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Chunk Array?

Always clarify the problem constraints and edge cases before coding. Think about time and space complexity as you develop your solution. Practice explaining your thought process while coding, as this is often part of the interview.

What are common mistakes when solving Chunk Array?

Not handling the last chunk correctly when it has fewer elements. Using nested loops unnecessarily, leading to inefficient solutions.

#2677

Chunk Array

Easy

ArrayTwo Pointers

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 approach uses a single loop to create chunks, which is more efficient. We directly append elements to the current chunk and start a new chunk when the current one reaches the specified size.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty list to hold the chunks and a temporary list for the current chunk.
2Step 2: Loop through each element in the original array.
3Step 3: Append the current element to the current chunk.
4Step 4: If the current chunk reaches the specified size, append it to the chunks list and reset the current chunk.

solution.py11 lines

1def chunk_array(arr, size):
2    chunks = []
3    current_chunk = []
4    for item in arr:
5        current_chunk.append(item)
6        if len(current_chunk) == size:
7            chunks.append(current_chunk)
8            current_chunk = []
9    if current_chunk:
10        chunks.append(current_chunk)
11    return chunks

ℹ

Complexity note: The time complexity is O(n) because we traverse the array once, and the space complexity is O(n) due to the storage of the chunks.

1Understanding how to manipulate arrays and create subarrays is crucial.
2Recognizing when to use loops versus built-in functions can optimize performance.

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