Generate all subarrays of an array using recursion medium

okay, let's dive into generating all subarrays of an array using recursion, with a comprehensive explanation and code example.

**understanding the problem**

a subarray is a contiguous (adjacent) sequence of elements within an array. for example, if we have the array `[1, 2, 3]`, its subarrays are:

* `[1]`
* `[2]`
* `[3]`
* `[1, 2]`
* `[2, 3]`
* `[1, 2, 3]`

the goal is to write a function that takes an array as input and returns a list (or prints) all possible subarrays.

**why recursion?**

recursion allows us to break down a problem into smaller, self-similar subproblems. in this case, we can think of generating subarrays as:

1. considering the first element and generating all subarrays starting from it.
2. recursively generating subarrays starting from the next element.

**recursive approach**

the core idea is to use two nested recursive functions, or a single recursive function with multiple arguments to manage the start and end positions of the subarrays:

* **outer recursion:** iterates through the starting index of the subarray.
* **inner recursion:** iterates through the ending index of the subarray, starting from the current starting index.

**python code example**

**explanation:**

1. **`generate_subarrays_recursive(arr)`:**
* this is the main function. it initializes an empty `result` list to store the subarrays.
* it calls the helper function `generate_from_start(0)` to start the recursive process.
* finally, it returns the `result` list containing all subarrays.

2. **`generate_from_start(start_index)`:**
* this function iterates through the possible starting indices of the subarrays.
* **base case:** `if start_index == len(arr): return` - if the `start_index` goes beyond the bounds of the array, it means we've considered all possible starting positions, so we stop the recursion.
* it calls the `generate_to_end` helper function to generate all subarrays starting from the current `start_ ...

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