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Flatten Nested List Iterator

Medium

Flatten Nested List Iterator

Medium

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:

Can the nested lists contain null or empty lists? What should the iterator return in these cases?
What is the maximum depth of the nested list? Is there a limit to the number of nested levels?
Can the integers within the nested lists be any integer type (e.g., int, long), or are they guaranteed to be within a certain range?
What should `hasNext()` return if the input `nestedList` is initially empty?
Are the `NestedInteger` objects immutable, or can I modify them during the iteration process?

Brute Force Solution

Approach

The nested list iterator problem asks us to go through a list that can contain both numbers and other lists, which themselves can contain numbers and lists, and so on. The brute force way to handle this is to simply unpack everything into one flat list first. This involves exploring every single element at every level until only numbers remain.

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

Start with the outermost list.
Go through each item in the list.
If the item is a number, add it to a new, separate list.
If the item is another list, open it up and repeat the process of going through each item within that nested list.
Keep doing this opening and adding process until you've gone through every single item in every nested list, and you only have numbers left.
At the end, you will have one single list containing all the numbers, in the correct order.

Code Implementation

def flatten_nested_list(nested_list):
    flattened_list = []

    def helper(nested_list_segment):
        for item in nested_list_segment:
            # Check if the item is an integer
            if isinstance(item, int):
                flattened_list.append(item)

            # If the item is a list, recursively call the function
            elif isinstance(item, list):
                helper(item)

    # Initiate the recursive helper function.
    helper(nested_list)

    return flattened_list

Big(O) Analysis

Time Complexity

O(n) – The brute force approach flattens the entire nested list into a single list of integers. We need to visit each element in the nested list structure exactly once to extract its value. Therefore, the time complexity is directly proportional to the total number of elements (integers and lists) across all levels of nesting in the input. Let n be the total number of elements across all nested lists. The algorithm visits each element once, so the time complexity is O(n).

Space Complexity

O(N) – The brute force approach, as described, flattens the entire nested list into a new, separate list before iteration. This new list stores all the numbers from the nested lists. In the worst-case scenario, where N is the total number of elements across all nested lists, the flattened list will hold N numbers. Therefore, the auxiliary space required to store the flattened list grows linearly with the total number of elements in the input, resulting in a space complexity of O(N).

Optimal Solution

Approach

The efficient way to navigate a nested list is to think of it like exploring a maze. We keep track of where we are and use a stack to remember the path we've taken, so we can easily backtrack and explore other paths.

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

Start at the very beginning of the initial list.
If the current item is a single value, give that value out.
If the current item is a list, put it on the top of a stack to explore it later. Then, go to the first thing inside that new list.
If you reach the end of the current list, go back to the last list we were exploring (the one on top of the stack) and continue from there.
Keep doing this until the stack is empty, meaning you've explored every list completely.

Code Implementation

class NestedIterator:
    def __init__(self, nested_list):
        self.stack_of_lists = [nested_list]

    def next(self):
        return self.current_value

    def hasNext(self):
        while self.stack_of_lists:
            top_list = self.stack_of_lists[-1]
            if not top_list:
                self.stack_of_lists.pop()
                continue

            first_element = top_list.pop(0)

            if isinstance(first_element, list):
                # Push list to stack, go deeper
                self.stack_of_lists.append(first_element)

            else:
                # Found an integer, store and return
                self.current_value = first_element
                return True

        return False

Big(O) Analysis

Time Complexity

O(n) – Let n be the total number of integers within the nested list structure. The iterator visits each integer element exactly once to return it via the next() method. The hasNext() method, in the worst-case, might traverse a portion of the nested list to find the next integer. However, overall, each integer is still visited only once across all calls to hasNext() and next(). Therefore, the time complexity is proportional to the total number of integers, which is O(n).

Space Complexity

O(D) – The algorithm uses a stack to store the path of nested lists being explored. In the worst-case scenario, the nested list could have a maximum nesting depth of D. Therefore, the stack could grow to a maximum size of D, where D is the maximum depth of nesting in the input list. The auxiliary space used by the stack is thus proportional to D, resulting in O(D) space complexity.

Edge Cases

Case	How to Handle
Null or empty nestedList input	The constructor should handle a null or empty input list gracefully, potentially by initializing an empty internal data structure or throwing an exception if null input is not allowed.
NestedList containing only empty lists	The hasNext() method should correctly return false when all nested lists are empty, even if the top-level list is not empty.
Deeply nested lists	The implementation should avoid stack overflow errors when handling deeply nested lists, possibly by using an iterative approach or a more efficient data structure than recursion.
Nested lists with extremely large numbers of integers	Consider the memory usage and potential for integer overflow when handling a very large number of integers, possibly employing a stream processing approach if the entire flattened list cannot fit in memory.
Nested lists containing non-integer types	The solution needs to explicitly check if the nested element is an integer, or define clear behaviour, such as ignoring non-integer types or throwing an exception.
Consecutive integers after flattening	The iterator should correctly yield each integer one at a time even if multiple consecutive integers occur after flattening.
Nested list with only one integer	The next() and hasNext() methods should work correctly when the nested list ultimately flattens to a single integer.
List containing a very large number of nested lists, each with only one number.	The solution should maintain reasonable performance and memory usage even with a very wide but shallow nesting structure.

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:

Can the nested lists contain null or empty lists? What should the iterator return in these cases?
What is the maximum depth of the nested list? Is there a limit to the number of nested levels?
Can the integers within the nested lists be any integer type (e.g., int, long), or are they guaranteed to be within a certain range?
What should `hasNext()` return if the input `nestedList` is initially empty?
Are the `NestedInteger` objects immutable, or can I modify them during the iteration process?

Brute Force Solution

Approach

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

Start with the outermost list.
Go through each item in the list.
If the item is a number, add it to a new, separate list.
If the item is another list, open it up and repeat the process of going through each item within that nested list.
Keep doing this opening and adding process until you've gone through every single item in every nested list, and you only have numbers left.
At the end, you will have one single list containing all the numbers, in the correct order.

Code Implementation

def flatten_nested_list(nested_list):
    flattened_list = []

    def helper(nested_list_segment):
        for item in nested_list_segment:
            # Check if the item is an integer
            if isinstance(item, int):
                flattened_list.append(item)

            # If the item is a list, recursively call the function
            elif isinstance(item, list):
                helper(item)

    # Initiate the recursive helper function.
    helper(nested_list)

    return flattened_list

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start at the very beginning of the initial list.
If the current item is a single value, give that value out.
If the current item is a list, put it on the top of a stack to explore it later. Then, go to the first thing inside that new list.
If you reach the end of the current list, go back to the last list we were exploring (the one on top of the stack) and continue from there.
Keep doing this until the stack is empty, meaning you've explored every list completely.

Code Implementation

class NestedIterator:
    def __init__(self, nested_list):
        self.stack_of_lists = [nested_list]

    def next(self):
        return self.current_value

    def hasNext(self):
        while self.stack_of_lists:
            top_list = self.stack_of_lists[-1]
            if not top_list:
                self.stack_of_lists.pop()
                continue

            first_element = top_list.pop(0)

            if isinstance(first_element, list):
                # Push list to stack, go deeper
                self.stack_of_lists.append(first_element)

            else:
                # Found an integer, store and return
                self.current_value = first_element
                return True

        return False

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty nestedList input	The constructor should handle a null or empty input list gracefully, potentially by initializing an empty internal data structure or throwing an exception if null input is not allowed.
NestedList containing only empty lists	The hasNext() method should correctly return false when all nested lists are empty, even if the top-level list is not empty.
Deeply nested lists	The implementation should avoid stack overflow errors when handling deeply nested lists, possibly by using an iterative approach or a more efficient data structure than recursion.
Nested lists with extremely large numbers of integers	Consider the memory usage and potential for integer overflow when handling a very large number of integers, possibly employing a stream processing approach if the entire flattened list cannot fit in memory.
Nested lists containing non-integer types	The solution needs to explicitly check if the nested element is an integer, or define clear behaviour, such as ignoring non-integer types or throwing an exception.
Consecutive integers after flattening	The iterator should correctly yield each integer one at a time even if multiple consecutive integers occur after flattening.
Nested list with only one integer	The next() and hasNext() methods should work correctly when the nested list ultimately flattens to a single integer.
List containing a very large number of nested lists, each with only one number.	The solution should maintain reasonable performance and memory usage even with a very wide but shallow nesting structure.