2677. Chunk Array


Problem Description

The task is to take an input array arr and an integer size, and then divide or "chunk" the array into subarrays, where each subarray has a maximum length of size. The resulting array of subarrays should include all the elements of the original array in the same order, just segmented into chunks. If there are not enough elements at the end of arr to make up a full chunk of length size, then the last subarray will contain fewer than size elements.

Keep in mind that the array arr can be considered as being produced by a process like JSON.parse, implying that it is a valid JSON array which could contain any type of elements, not just numbers. Additionally, you should not rely on external libraries such as lodash, particularly avoiding the use of a function like _.chunk which essentially solves this problem.

Intuition

To chunk the array without using additional libraries, we can create a new array to hold our chunks, then iterate over the original array, slicing it into smaller arrays of length size. Here's the process:

  1. Initialize an empty array ans to store our chunks.
  2. Loop over the original array, using a counter i that starts at 0 and increments by size each time. This way, each iteration processes a chunk of the array.
  3. In each iteration, use the slice method to get a subarray of length size starting from the current index i.
  4. Push this subarray into our ans array.
  5. Continue the process until we've reached the end of the original array.
  6. The slice method will automatically handle the scenario where there aren't enough elements at the end of the array to form a complete chunk, resulting in the last subarray being the correct, potentially smaller size.
  7. Finally, return the ans array containing our chunks.

The intuition behind this approach is to systematically break down the original array into sections without needing complex logic or external libraries to assist us. The slice method available in TypeScript is quite handy as it easily allows extracting parts of an array based on index positions and does not modify the original array while doing so.

Solution Approach

The solution is simple in nature and relies on basic array manipulation techniques provided by TypeScript/JavaScript.

  • Algorithm: The solution involves a single pass through the input array, slicing out subarrays and collecting them into a result array.
  • Data Structures: Only one additional data structure is used, which is the result array that stores the chunks, referred to as ans in the code.
  • Patterns: The pattern used here is a common iteration pattern, where the index is incremented not by 1 but by the chunk size on each loop iteration.

Walking through the code:

  1. A result array ans is declared to store the chunks that will eventually be returned.

  2. A for-loop is used to iterate over the elements of the input array. The loop is controlled by index i, which starts at 0 and is incremented by size after each iteration to move to the next chunk.

    1for (let i = 0, n = arr.length; i < n; i += size) {
    2    ...
    3}
  3. Inside the loop body, the slice method is used to create a subarray consisting of elements from index i to i + size. The slice method is inclusive of the start index and exclusive of the end index.

    1ans.push(arr.slice(i, i + size));
  4. This subarray is then pushed into the ans array. The slice method will ensure that if the number of elements from index i is less than size, the subarray will include all elements up to the end of the array, capturing the potentially smaller last chunk if the array size is not perfectly divisible by size.

  5. The loop continues until i is greater than or equal to the length of arr, signifying that all elements have been included in the chunks.

  6. Once the loop is complete, ans, now containing all the chunked subarrays, is returned as the final result.

This approach is efficient because it only requires a single traversal of the original array and uses slice, which is an optimized native array method. There are no nested loops or redundant operations, making this approach ideal for chunking an array into subarrays of specified size.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

Which two pointer techniques do you use to check if a string is a palindrome?

Example Walkthrough

Let's use a small example to illustrate the solution approach. Suppose we have an input array arr represented as [1, 2, 3, 4, 5, 6, 7], and we want to chunk this array into subarrays of size 3.

Following the solution approach:

  1. We declare an empty result array ans to which we will add our chunks.

  2. We start a for-loop with a counter i initialized at 0, which will increment by the size after each iteration. Our size is 3, so i will take values 0, 3, and 6 during the loop.

    1for (let i = 0, n = arr.length; i < n; i += size) {
    2    ...
    3}
  3. On the first iteration (i = 0), we use the slice method to create a subarray from index 0 to index 3, which yields [1, 2, 3].

    1ans.push(arr.slice(0, 0 + 3)); // equivalent to arr.slice(0, 3)
  4. Then we push [1, 2, 3] into the ans array.

  5. Next iteration (i = 3), we use the slice method to create a subarray from index 3 to index 6, resulting in [4, 5, 6].

    1ans.push(arr.slice(3, 3 + 3)); // equivalent to arr.slice(3, 6)
  6. We push [4, 5, 6] into the ans array.

  7. On the final iteration (i = 6), we slice from index 6 to the end of the array, as slice naturally handles cases where the end index is beyond the array's length. This yields the subarray [7].

    1ans.push(arr.slice(6, 6 + 3)); // equivalent to arr.slice(6, 9)
  8. We push the final chunk [7] into the ans array.

  9. Once the loop finishes, we have ans containing all the chunks: [[1, 2, 3], [4, 5, 6], [7]].

  10. This resulting array ans is then returned.

By following this process, we efficiently divide the original array into chunks of a given size, demonstrating the algorithm's effective use of TypeScript/JavaScript's native array methods.

Solution Implementation

1def chunk(array, size):
2    """Function to split an array into chunks of a specified size.
3  
4    Args:
5    array: A list of elements that needs to be split.
6    size: The size of each chunk.
7  
8    Returns:
9    A list of lists where each sublist is a chunk of the input array.
10    """
11  
12    chunks = []  # Initialize an empty list to hold the chunks
13
14    # Loop through the array, incrementing by 'size' on each iteration
15    for i in range(0, len(array), size):
16        # Slice the array from the current index 'i' up to 'i + size'
17        # and append this new chunk to the 'chunks' list
18        chunks.append(array[i:i + size])
19
20    return chunks  # Return the list of chunks
21
1import java.util.ArrayList;
2import java.util.List;
3
4// Function to split a list into chunks of a specified size
5public static List<List<Object>> chunk(List<Object> list, int size) {
6    List<List<Object>> chunks = new ArrayList<>(); // Initialize an empty list to hold the chunks
7  
8    // Loop through the list, incrementing by 'size' on each iteration
9    for (int i = 0; i < list.size(); i += size) {
10        // Calculate the end index for the current chunk, making sure it does not exceed the list size
11        int end = Math.min(i + size, list.size());
12      
13        // Create a sublist from the current index 'i' up to 'end' and add this new chunk to 'chunks'
14        chunks.add(new ArrayList<>(list.subList(i, end)));
15    }
16  
17    return chunks; // Return the list of chunks
18}
19
1#include <vector>
2
3// Function to split a vector into chunks of a specified size
4std::vector<std::vector<int>> chunk(const std::vector<int>& array, int size) {
5    std::vector<std::vector<int>> chunks; // Initialize an empty vector to hold the chunks
6
7    // Loop through the vector, incrementing by 'size' on each iteration
8    for (size_t i = 0; i < array.size(); i += size) {
9        // 'std::begin(array)' is the starting iterator of 'array'
10        // Move the start iterator 'i' positions forward
11        auto start_itr = std::next(std::begin(array), i);
12      
13        // Determine 'end_itr' ensuring we don't pass the end of the array
14        auto end_itr = std::next(start_itr, std::min(size, static_cast<int>(array.size()) - i));
15
16        // Construct a vector from the range [start_itr, end_itr) and add it to 'chunks'
17        chunks.emplace_back(start_itr, end_itr);
18    }
19
20    return chunks; // Return the vector of chunks
21}
22
1// Function to split an array into chunks of a specified size
2function chunk(array: any[], size: number): any[][] {
3    const chunks: any[][] = []; // Initialize an empty array to hold the chunks
4
5    // Loop through the array, incrementing by 'size' on each iteration
6    for (let i = 0, arrayLength = array.length; i < arrayLength; i += size) {
7        // Slice the array from the current index 'i' up to 'i + size'
8        // and push this new chunk into the 'chunks' array
9        chunks.push(array.slice(i, i + size));
10    }
11
12    return chunks; // Return the array of chunks
13}
14

Time and Space Complexity

Time Complexity

The time complexity of the code is O(n), where n is the total number of elements in the input array arr. The for loop iterates over the array in steps of size, and within each iteration, the .slice() method is called, which runs in O(k), where k is the size of the chunk being created. However, since k <= size and the steps of the loop are proportional to size, the overall number of operations depends linearly on n, thus resulting in linear time complexity.

Space Complexity

The space complexity of the code is also O(n). This is because the function creates a new array ans to store the chunks. In the worst case, when size is 1, the ans array will contain the same number of elements as the input array, spread across many sub-arrays, effectively duplicating all elements of the input. Therefore, the maximum space this function can take up is proportional to the number of items in the input array, hence O(n).


Fast Track Your Learning with Our Quick Skills Quiz:

What is the best way of checking if an element exists in a sorted array once in terms of time complexity? Select the best that applies.


Recommended Readings


Got a question? Ask the Monster Assistant anything you don't understand.

Still not clear? Ask in the Forum,  Discord or Submit the part you don't understand to our editors.


🪄