Chunk Array

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Arrays

Given an array arr and a chunk size size, return a chunked array.

A chunked array contains the original elements in arr, but consists of subarrays each of length size. The length of the last subarray may be less than size if arr.length is not evenly divisible by size.

Please solve it without using lodash's _.chunk function.

Example 1:

Input: arr = [1,2,3,4,5], size = 1
Output: [[1],[2],[3],[4],[5]]
Explanation: The arr has been split into subarrays each with 1 element.

Example 2:

Input: arr = [1,9,6,3,2], size = 3
Output: [[1,9,6],[3,2]]
Explanation: The arr has been split into subarrays with 3 elements. However, only two elements are left for the 2nd subarray.

Example 3:

Input: arr = [8,5,3,2,6], size = 6
Output: [[8,5,3,2,6]]
Explanation: Size is greater than arr.length thus all elements are in the first subarray.

Example 4:

Input: arr = [], size = 1
Output: []
Explanation: There are no elements to be chunked so an empty array is returned.

Constraints:

arr is a string representing the array.
2 <= arr.length <= 10⁵
1 <= size <= arr.length + 1

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 is the data type of the elements in the input array `arr`, and what is the range of possible values?
Can the input array `arr` be null or empty? What should I return in those cases?
Is the chunk size `size` guaranteed to be a positive integer? What should I return if `size` is zero or negative?
Should the order of elements within each chunk be the same as in the original array?
Should I return a new array containing the chunks, or can I modify the original array in place (if that's even possible given the requirements)?

Brute Force Solution

Approach

The brute force method for chunking is like trying every single way to divide a long string of beads into smaller strands of a specific length. We explore all possible cut points to see which divisions work.

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

Start by grabbing the first few items to form the first group.
Check if this group is the right size.
If it is not the right size, either too small or too big, then try a different number of items for the first group.
Once you find a valid first group, move on and form the second group using the next few available items.
Repeat the process of checking the group size and adjusting the number of items until you've grouped all the original items.
Keep track of all the different ways you can successfully divide the items into valid groups.
Finally, select the best grouping based on your criteria (for example, the grouping with the fewest groups).

Code Implementation

def chunk_array_brute_force(input_list, chunk_size):    all_chunks = []
    def generate_chunks(current_index, current_chunks):        # Base case: If we've processed all elements,
        # add the current chunk combination to the result.        if current_index >= len(input_list):            all_chunks.append(current_chunks)
            return
        # Iterate through all possible chunk sizes
        # starting from the current index.        for next_index in range(current_index + 1, len(input_list) + 1):            new_chunk = input_list[current_index:next_index]
            # Recursively call generate_chunks to create
            # further chunk combinations.            generate_chunks(next_index, current_chunks + [new_chunk])
    generate_chunks(0, [])
    valid_chunks = []    # Filter out combinations with incorrect chunk sizes.    for chunks in all_chunks:        is_valid = True        for chunk in chunks:            if len(chunk) > chunk_size:                is_valid = False                break
        # Store only the chunk combinations with valid chunks.        if is_valid:            valid_chunks.append(chunks)
    # Return the first valid chunk combination found.    if valid_chunks:
        return valid_chunks[0]
    else:
        return []

Big(O) Analysis

Time Complexity

O(n^2) – The provided brute force approach explores all possible chunk sizes for the initial group, potentially iterating through lengths 1 to n. For each potential initial chunk size, it attempts to recursively chunk the remaining array. In the worst case, where many combinations are checked, each element might be visited multiple times across different chunking attempts. This exploration leads to a nested iterative behavior where the outer loop considers different starting chunk sizes and the inner processes the remaining array, resulting in approximately n * n/2 operations and therefore O(n^2) time complexity.

Space Complexity

O(N) – The algorithm keeps track of all the different ways the array can be successfully divided into valid groups. This implies storing multiple chunked arrays, potentially up to N, where N is the number of elements in the input array, in a data structure to compare them later to select the best one. Furthermore, storing the intermediate groups that are being constructed requires auxiliary space. Therefore, the auxiliary space used grows linearly with the input size N, leading to a space complexity of O(N).

Optimal Solution

Approach

To divide an array into smaller groups, the optimal strategy involves stepping through the array and creating new sub-arrays of the desired size. Each time a sub-array is filled, it is added to a main container, and the process is repeated until the entire original array is processed. This avoids unnecessary recalculations and efficiently builds the chunked array.

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

Start with an empty container to hold all the smaller groups.
Look at the first part of the main array, and decide how many elements should be in each smaller group.
Create a new, smaller group and start adding elements from the main array into it.
Keep adding elements until the smaller group is full.
Once the smaller group is full, put it into the main container.
Repeat the process by creating another smaller group and filling it with the next elements from the main array.
Keep doing this until there are no more elements left in the main array.
The main container now holds all the smaller groups, representing the chunked array.

Code Implementation

def chunk_array(input_list, chunk_size):
    # Initialize the list to store our chunks
    chunked_list = []

    current_index = 0
    while current_index < len(input_list):
        # Determine the end index for the current chunk.
        end_index = min(current_index + chunk_size, len(input_list))

        # Extract a chunk from the input list.
        current_chunk = input_list[current_index:end_index]

        # Add the chunk to the chunked list
        chunked_list.append(current_chunk)

        # Increment the current index to start the next chunk
        current_index += chunk_size

    return chunked_list

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n once. In each iteration, it adds the current element to a sub-array. When a sub-array is filled (based on the chunk size, which is constant), it's added to the result. The number of sub-arrays created depends on the chunk size and array length (n), but the total number of elements copied is still directly proportional to the input size n. Therefore, the time complexity is O(n).

Space Complexity

O(N) – The algorithm creates a main container to hold the smaller groups (chunks). In the worst-case scenario, where the chunk size is 1, the main container will hold N chunks, each potentially containing a single element from the original array. Therefore, the auxiliary space used is proportional to the number of elements in the input array, N. This results in a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty input array	Return an empty array because there are no elements to chunk.
Null input array	Throw an IllegalArgumentException (or equivalent language-specific exception) or return null, depending on requirements.
Chunk size is zero	Throw an IllegalArgumentException, as a chunk size of zero is invalid.
Chunk size is negative	Throw an IllegalArgumentException, as a chunk size cannot be negative.
Chunk size is larger than the input array's length	Return an array containing a single chunk equal to the entire input array.
Input array contains very large numbers	The solution should handle large numbers as array elements without integer overflow, assuming sufficient memory.
Chunk size equal to array length	Return an array containing a single chunk equal to the entire input array.
Large input array with a small chunk size (Scalability)	The solution should perform acceptably with a large input array and small chunk size using an iterative approach, avoiding excessive memory allocation or recursion.

Empty input array

How to Handle:

Return an empty array because there are no elements to chunk.

Null input array

How to Handle:

Throw an IllegalArgumentException (or equivalent language-specific exception) or return null, depending on requirements.

Chunk size is zero

How to Handle:

Throw an IllegalArgumentException, as a chunk size of zero is invalid.

Chunk size is negative

How to Handle:

Throw an IllegalArgumentException, as a chunk size cannot be negative.

Chunk size is larger than the input array's length

How to Handle:

Return an array containing a single chunk equal to the entire input array.

Input array contains very large numbers

How to Handle:

The solution should handle large numbers as array elements without integer overflow, assuming sufficient memory.

Chunk size equal to array length

How to Handle:

Return an array containing a single chunk equal to the entire input array.

Large input array with a small chunk size (Scalability)

How to Handle:

The solution should perform acceptably with a large input array and small chunk size using an iterative approach, avoiding excessive memory allocation or recursion.