How do you solve Flatten Nested List Iterator on LeetCode?

Flatten Nested List Iterator (LeetCode #341) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding nested structures is crucial for this problem.

What is the brute force solution for Flatten Nested List Iterator?

The brute force approach for Flatten Nested List Iterator is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach involves recursively traversing the nested list and collecting all integers into a flat list. This method is straightforward but inefficient as it requires multiple passes through the data. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Flatten Nested List Iterator?

Think about how to manage depth in nested structures. Practice writing both recursive and iterative solutions. Be prepared to explain your thought process clearly.

What are common mistakes when solving Flatten Nested List Iterator?

Not handling the case where the list is empty. Confusing the order of operations when flattening the list.

#341

Flatten Nested List Iterator

Medium

StackTreeDepth-First SearchDesignQueueIteratorStackRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses an explicit stack to manage the nested structure without needing to create a separate flattened list upfront. This approach allows us to iterate through the elements as needed, maintaining a pointer to the current position.

⚙️

Algorithm

3 steps

1Step 1: Initialize a stack with the nested list.
2Step 2: While the stack is not empty, check the top element.
3Step 3: If the top element is an integer, return it. If it's a list, pop it and push its elements onto the stack.

solution.py16 lines

1# Full working Python code
2class NestedIterator:
3    def __init__(self, nestedList):
4        self.stack = list(reversed(nestedList))
5
6    def next(self):
7        return self.stack.pop().getInteger()
8
9    def hasNext(self):
10        while self.stack:
11            top = self.stack[-1]
12            if top.isInteger():
13                return True
14            self.stack.pop()
15            self.stack.extend(reversed(top.getList()))
16        return False

ℹ

Complexity note: The time complexity is O(n) because each element is processed once. The space complexity is O(n) due to the stack used to store elements during the flattening process.

1Understanding nested structures is crucial for this problem.
2Using a stack allows us to manage depth without recursion.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.