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This approach involves iterating over the array using a loop and extracting subarrays using slicing. The loop increments by the chunk size in each iteration, thus effectively slicing the array into chunks of the desired size.
Time Complexity: O(n), where n is the number of elements in the array.
Space Complexity: O(n), storing the entire chunked array requires space proportional to the input size.
1#include <vector>
std::vector<std::vector<int>> chunkArray(const std::vector<int>& arr, int size) {
std::vector<std::vector<int>> chunked;
for (int i = 0; i < arr.size(); i += size) {
std::vector<int> chunk;
for (int j = i; j < i + size && j < arr.size(); j++) {
chunk.push_back(arr[j]);
}
chunked.push_back(chunk);
}
return chunked;
}The chunkArray function iterates over arr with a step size of size. In each step, a vector chunk is populated with elements from arr starting from index i to i + size, and this chunk is added to the result.
This approach uses simple arithmetic operations to determine when to create a new subarray. Using the modulus operator allows checking if the number of currently collected elements is equal to the chunk size, upon which a new subarray is started.
Time Complexity: O(n), where n is the number of elements in the array.
Space Complexity: O(n), because the additional space needed is proportional to the input size.
1def chunk_array(arr, size):
2 chunked = []
3 chunk = []
4 for i in arr:
5 chunk.append(i)
6 if len(chunk) == size:
7 chunked.append(chunk)
8 chunk = []
9 if chunk:
10 chunked.append(chunk)
11 return chunkedThe method keeps appending elements to the current chunk list until its length reaches size, at which point the chunk is added to the chunked result list. If there is a leftover partial chunk at the end, it is also added to the result.