Problem Description

You need to implement an iterator that can traverse through a nested list structure where each element can be either:

• An integer value
• A list that contains more integers or nested lists

The nested list is represented using a NestedInteger interface that provides three methods:

• isInteger(): Returns True if the element holds a single integer, False if it holds a nested list
• getInteger(): Returns the integer value if the element is an integer
• getList(): Returns the nested list if the element is a list

Your task is to implement the NestedIterator class with three methods:

1. __init__(nestedList): Constructor that takes the nested list and initializes the iterator
2. next(): Returns the next integer in the flattened sequence
3. hasNext(): Returns True if there are more integers to iterate through, False otherwise

The iterator should flatten the nested structure and return integers one by one in the order they appear when traversing the nested list from left to right.

For example, given [[1,1],2,[1,1]], the iterator should return the integers in order: 1, 1, 2, 1, 1.

The solution approach uses a depth-first search (DFS) during initialization to flatten the entire nested list into a simple list of integers. It recursively traverses each element:

• If an element is an integer, it adds it directly to the flattened list
• If an element is a nested list, it recursively processes that list

The iterator then maintains an index pointer i to track the current position in the flattened list. The next() method increments this pointer and returns the current element, while hasNext() checks if there are more elements remaining by comparing the next position with the list length.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The nested list structure can be viewed as a graph where each list is a node that can contain other nodes (nested lists) or leaf values (integers). This creates a tree-like hierarchical structure.

Is it a tree?

• Yes: The nested list structure is indeed a tree. Each nested list can be seen as a node with children (either integers or other nested lists). There's a root (the main list), and we need to traverse all nodes to flatten the structure. There are no cycles - each element belongs to exactly one parent list.

DFS

• Yes: Since we identified this as a tree traversal problem, DFS is the appropriate algorithm. We need to visit each node (nested list) and recursively explore its children to extract all integers in the correct order.

Conclusion: The flowchart suggests using DFS (Depth-First Search) for flattening the nested list iterator. This makes perfect sense because:

1. We need to traverse a tree-like structure (nested lists)
2. We must visit every element to extract all integers
3. The order matters (left-to-right traversal)
4. DFS naturally handles the recursive nature of nested structures by going deep into each nested list before moving to the next sibling

The solution implements DFS in the constructor to pre-flatten the entire structure into a simple list, making the iterator operations (next() and hasNext()) very efficient with O(1) time complexity.

Intuition

When we look at this nested list structure, we're essentially dealing with a tree where each node can either be a leaf (integer) or a branch (nested list). The challenge is to visit every integer in the correct order, just like reading a book - we read each word sequentially, even though the book has a hierarchical structure of chapters, sections, and paragraphs.

The key insight is recognizing that we need to "flatten" this hierarchical structure into a linear sequence. Think of it like unpacking nested boxes - we open the main box, and for each item inside, if it's another box, we open that too, continuing until we find all the actual items (integers).

Since we need to access elements one by one through next() calls, we have two main strategies:

1. Lazy evaluation: Only flatten elements when needed during next() or hasNext() calls
2. Eager evaluation: Flatten everything upfront during initialization

The eager approach is simpler and more intuitive here. Why? Because:

• We're going to visit all elements anyway (the iterator will traverse the entire structure)
• Pre-flattening makes next() and hasNext() operations trivial - just array indexing
• We avoid the complexity of maintaining state about partially explored nested lists

The DFS pattern naturally fits this problem because it mirrors how we'd manually flatten the list: when we encounter a nested list, we immediately dive into it, process all its contents, then continue with the next element at the current level. This is exactly what DFS does - it goes as deep as possible before backtracking.

The recursive nature of DFS perfectly matches the recursive structure of nested lists. Each recursive call handles one level of nesting, and the base case (finding an integer) is where we actually collect the values into our flattened result.

Solution Approach

The solution implements an eager flattening strategy using DFS during the initialization phase. Let's walk through the implementation step by step:

1. Data Structure Setup

The NestedIterator class maintains two instance variables:

• self.nums: A flat list that will store all integers after flattening
• self.i: An index pointer initialized to -1 to track the current position in the iteration

2. DFS Flattening Algorithm

The core of the solution is the nested dfs function defined inside __init__:

def dfs(ls):
    for x in ls:
        if x.isInteger():
            self.nums.append(x.getInteger())
        else:
            dfs(x.getList())

This recursive function:

• Iterates through each element x in the current list ls
• For each element, checks if it's an integer using x.isInteger()
• If it's an integer: extracts the value with x.getInteger() and appends it to self.nums
• If it's a nested list: recursively calls dfs with x.getList() to process the nested structure

3. Initialization Process

def __init__(self, nestedList: [NestedInteger]):
    self.nums = []
    self.i = -1
    dfs(nestedList)

During initialization:

• Create an empty list self.nums to store the flattened integers
• Set the index pointer self.i to -1 (before the first element)
• Call dfs(nestedList) to flatten the entire structure upfront

4. Iterator Operations

Once the list is flattened, the iterator operations become trivial:

def next(self) -> int:
    self.i += 1
    return self.nums[self.i]

• Increment the index pointer
• Return the integer at the current position

def hasNext(self) -> bool:
    return self.i + 1 < len(self.nums)

• Check if the next position (self.i + 1) is within bounds of the flattened list

Time and Space Complexity:

• Initialization: O(N) time where N is the total number of integers, as we visit each element once
• next(): O(1) time - simple array access
• hasNext(): O(1) time - simple comparison
• Space: O(N) to store all integers in the flattened list, plus O(D) for the recursion stack where D is the maximum depth of nesting

This approach trades upfront computation cost for extremely efficient iteration operations, which is ideal when we expect to iterate through all elements.

Example Walkthrough

Let's walk through the solution with the example [[1,1],2,[1,1]].

Initial Structure Visualization:

[
  [1, 1],      // Element 0: nested list
  2,           // Element 1: integer
  [1, 1]       // Element 2: nested list
]

Step 1: Initialization

• Create empty nums = [] and set i = -1
• Call dfs([[1,1],2,[1,1]])

Step 2: DFS Traversal

Processing element 0: [1,1]

• isInteger() returns False → it's a list
• Recursively call dfs([1,1])
  • Process first 1: isInteger() returns True → append to nums = [1]
  • Process second 1: isInteger() returns True → append to nums = [1,1]

Processing element 1: 2

• isInteger() returns True → append to nums = [1,1,2]

Processing element 2: [1,1]

• isInteger() returns False → it's a list
• Recursively call dfs([1,1])
  • Process first 1: isInteger() returns True → append to nums = [1,1,2,1]
  • Process second 1: isInteger() returns True → append to nums = [1,1,2,1,1]

Step 3: Final State After Initialization

• nums = [1,1,2,1,1] (fully flattened)
• i = -1 (ready to start iteration)

Step 4: Using the Iterator

The iterator successfully returns all integers in order: 1, 1, 2, 1, 1.

Solution Implementation

Time and Space Complexity

Time Complexity:

• Constructor (__init__): O(n) where n is the total number of integers in the nested list structure. The DFS traversal visits each element exactly once, whether it's an integer or a nested list. For each integer element, we perform an O(1) append operation.
• next(): O(1) - Simply increments the index and returns the element at that position in the flattened list.
• hasNext(): O(1) - Just compares the current index with the length of the list.

Space Complexity:

• O(n + d) where n is the total number of integers in the nested structure and d is the maximum depth of nesting.
  • O(n) for storing all integers in self.nums list after flattening.
  • O(d) for the recursive call stack during DFS traversal, where d represents the maximum nesting depth.
  • In the worst case where the structure is deeply nested (like [[[[...[[1]]...]]]), the space complexity would be O(n) for both the storage and call stack combined.

Analysis: The approach flattens the entire nested structure upfront during initialization using DFS. This trades initialization time for constant-time iteration operations. All integers are extracted and stored in a simple list, making subsequent next() and hasNext() operations very efficient at O(1) each.

Common Pitfalls

1. Attempting Lazy Evaluation with Incorrect State Management

One common pitfall occurs when developers try to optimize by implementing lazy evaluation (flattening on-the-fly) but fail to properly manage the iterator state across nested levels.

Problematic Approach:

class NestedIterator:
    def __init__(self, nestedList):
        self.stack = nestedList[::-1]  # Reverse for stack operations
  
    def next(self):
        return self.stack.pop().getInteger()  # Assumes top is always an integer
  
    def hasNext(self):
        while self.stack:
            top = self.stack[-1]
            if top.isInteger():
                return True
            self.stack.pop()
            # PITFALL: Losing nested elements
            for item in top.getList()[::-1]:
                self.stack.append(item)
        return False

Issues:

• The next() method assumes the top element is always an integer
• State inconsistency between hasNext() and next() calls
• Complex stack manipulation prone to errors

Solution: Ensure hasNext() always leaves the stack with an integer on top if one exists:

class NestedIterator:
    def __init__(self, nestedList):
        self.stack = nestedList[::-1]
  
    def next(self):
        # hasNext() ensures top is an integer
        return self.stack.pop().getInteger()
  
    def hasNext(self):
        while self.stack:
            top = self.stack[-1]
            if top.isInteger():
                return True
            # Pop the list and expand it
            self.stack.pop()
            nested_list = top.getList()
            for item in nested_list[::-1]:
                self.stack.append(item)
        return False

2. Index Management Errors in Eager Flattening

When using eager flattening (as in the provided solution), a common mistake is incorrect index initialization or increment logic.

Problematic Pattern:

def __init__(self, nestedList):
    self.nums = []
    self.i = 0  # PITFALL: Starting at 0 instead of -1
    # ... flattening logic
  
def next(self):
    result = self.nums[self.i]  # Returns first element twice
    self.i += 1
    return result

Solution: Initialize index to -1 and increment before accessing:

def __init__(self, nestedList):
    self.nums = []
    self.i = -1  # Correct initialization
  
def next(self):
    self.i += 1  # Increment first
    return self.nums[self.i]

3. Memory Inefficiency with Deep Recursion

For deeply nested structures, the recursive DFS approach can cause stack overflow.

Problematic Scenario:

# Deep nesting like: [[[[[[1]]]]]]
def dfs(ls):
    for x in ls:
        if x.isInteger():
            self.nums.append(x.getInteger())
        else:
            dfs(x.getList())  # Can cause stack overflow

Solution: Use iterative approach with explicit stack:

def __init__(self, nestedList):
    self.nums = []
    self.i = -1
  
    # Iterative flattening using stack
    stack = list(reversed(nestedList))
    while stack:
        current = stack.pop()
        if current.isInteger():
            self.nums.append(current.getInteger())
        else:
            # Add nested items to stack in reverse order
            stack.extend(reversed(current.getList()))

4. Handling Empty Nested Lists

Failing to handle empty nested structures correctly can lead to unexpected behavior.

Problematic Assumption:

# Assuming nestedList always has elements
def __init__(self, nestedList):
    self.nums = []
    self.i = -1
    dfs(nestedList)
    # What if self.nums is still empty?

Solution: The current implementation handles this correctly, but ensure client code checks:

iterator = NestedIterator(nestedList)
if iterator.hasNext():  # Always check before calling next()
    value = iterator.next()