What is the time complexity of Flatten Nested List Iterator?

The optimal solution has a time complexity of O(1).

What is the space complexity of Flatten Nested List Iterator?

The space complexity is O(d*n).

What data structures are used to solve Flatten Nested List Iterator?

This problem uses Stack, Tree, Depth-First Search, Design, Queue, Iterator. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Flatten Nested List Iterator?

Insight 1: Using a stack data structure is crucial for implementing the depth-first traversal of the nested list. The stack helps keep track of the nested lists that are yet to be explored. Insight 2: The `hasNext()` method is more complex than it seems. It's not just about checking if the stack is empty. We need to continuously unpack nested lists on the top of the stack until we find an integer or the stack becomes empty.

Which companies ask LeetCode 341. Flatten Nested List Iterator in interviews?

This problem is frequently asked at Airbnb, LinkedIn, Mixpanel, Netflix, OpenAI and 5 more companies.

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Flatten Nested List Iterator - LeetCode 341 Solution

Flatten Nested List Iterator - Complete Solution Guide

Flatten Nested List Iterator is LeetCode problem 341, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a nested list of integers nestedList . Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it. Implement the NestedIterator class: NestedIterator(List<NestedInteger> nestedList) Initializes the iterator with the nested list nestedList . int next() Returns the next integer in the nested list. boolean hasNext() Returns true if there are still some integers in the nested list and false otherwise. Your code wi

Detailed Explanation

The problem asks us to implement an iterator that flattens a nested list of integers. Each element in the list can be either an integer or another nested list. The `NestedIterator` class should have a constructor that takes the nested list as input, a `next()` method that returns the next integer in the flattened list, and a `hasNext()` method that returns `true` if there are more integers in the flattened list and `false` otherwise. Essentially, we need to traverse the nested list in a depth-first manner and return the integers one by one.

Solution Approach

The provided solutions use a stack to implement a depth-first traversal of the nested list. The constructor initializes the stack with the elements of the nested list in reversed order. The `hasNext()` method peeks at the top of the stack. If it's an integer, it returns true. If it's a nested list, it unpacks the list by popping it from the stack and pushing its elements onto the stack in reverse order. This process continues until the stack is empty or an integer is found. The `next()` method simply pops an integer from the stack and returns it, assuming `hasNext()` has already confirmed that there is indeed an integer on top of the stack.

Step-by-Step Algorithm

Step 1: Initialize the `NestedIterator` with the given `nestedList`. This involves creating a stack and pushing the elements of the `nestedList` onto the stack in reverse order.
Step 2: Implement the `hasNext()` method. While the stack is not empty, peek at the top element. If it's an integer, return `true`.
Step 3: If the top element is a nested list, pop it from the stack.
Step 4: Push all the elements of the popped nested list onto the stack in reverse order. This ensures correct traversal order.
Step 5: Repeat steps 2-4 until an integer is found or the stack becomes empty. If the stack becomes empty, return `false`.
Step 6: Implement the `next()` method. Pop the integer from the stack (guaranteed to be an integer because `hasNext()` has already been called) and return its value.

Key Insights

Insight 1: Using a stack data structure is crucial for implementing the depth-first traversal of the nested list. The stack helps keep track of the nested lists that are yet to be explored.
Insight 2: The `hasNext()` method is more complex than it seems. It's not just about checking if the stack is empty. We need to continuously unpack nested lists on the top of the stack until we find an integer or the stack becomes empty.
Insight 3: The order in which elements are added to the stack is important. Since we are popping from the stack (LIFO), elements from the inner nested lists should be pushed in reverse order to maintain the correct flattened order.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(d*n)

Topics

This problem involves: Stack, Tree, Depth-First Search, Design, Queue, Iterator.

Companies

Asked at: Airbnb, LinkedIn, Mixpanel, Netflix, OpenAI, Tesla, Walmart Labs, Warnermedia, X, Yandex.

Flatten Nested List Iterator | LeetCode 341

Medium

Stack Tree Depth-First Search Design Queue Iterator

Time: O(1)Space: O(d*n)

Solution Code

Available in:

/**
 * // This is the interface that allows for creating nested lists.
 * // You should not implement it, or speculate about its implementation
 * public interface NestedInteger {
 *
 *     // @return true if this NestedInteger holds a single integer, rather than a nested list.
 *     public boolean isInteger();
 *
 *     // @return the single integer that this NestedInteger holds, if it holds a single integer
 *     // Return null if this NestedInteger holds a nested list
 *     public Integer getInteger();
 *
 *     // @return the nested list that this NestedInteger holds, if it holds a nested list
 *     // Return empty list if this NestedInteger holds a single integer
 *     public List<NestedInteger> getList();
 * }
 */
import java.util.Iterator;
import java.util.List;
import java.util.Stack;

public class NestedIterator implements Iterator<Integer> {

    private Stack<NestedInteger> stack;

    public NestedIterator(List<NestedInteger> nestedList) {
        stack = new Stack<>();
        for (int i = nestedList.size() - 1; i >= 0; i--) {
            stack.push(nestedList.get(i));
        }
    }

    @Override
    public Integer next() {
        return stack.pop().getInteger();
    }

    @Override
    public boolean hasNext() {
        while (!stack.isEmpty()) {
            NestedInteger top = stack.peek();
            if (top.isInteger()) {
                return true;
            }

            stack.pop();
            List<NestedInteger> nested_list = top.getList();
            for (int i = nested_list.size() - 1; i >= 0; i--) {
                stack.push(nested_list.get(i));
            }
        }

        return false;
    }
}

/**
 * Your NestedIterator object will be instantiated and called as such:
 * NestedIterator i = new NestedIterator(nestedList);
 * while (i.hasNext()) v[f()] = i.next();
 */

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Problem Info

Description

Problem 341: Flatten Nested List Iterator - Detailed explanation and solution approaches.

Topics

Stack Tree Depth-First Search Design Queue Iterator

Companies

AirbnbLinkedInMixpanelNetflixOpenAITeslaWalmart LabsWarnermedia

Stats

Difficulty:Medium

Languages:4

Companies:10

Problem #:341

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize the `NestedIterator` with the given `nestedList`. This involves creating a stack and pushing the elements of the `nestedList` onto the stack in reverse order.
2
Step 2: Implement the `hasNext()` method. While the stack is not empty, peek at the top element. If it's an integer, return `true`.
3
Step 3: If the top element is a nested list, pop it from the stack.
4
Step 4: Push all the elements of the popped nested list onto the stack in reverse order. This ensures correct traversal order.
5
Step 5: Repeat steps 2-4 until an integer is found or the stack becomes empty. If the stack becomes empty, return `false`.
6
Step 6: Implement the `next()` method. Pop the integer from the stack (guaranteed to be an integer because `hasNext()` has already been called) and return its value.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(d*n)

📝

Why?

The time complexity for `next()` is O(1) because it simply pops the next integer from the stack. For `hasNext()`, in the worst-case, we might have to repeatedly unpack nested lists until we find an integer. The stack could potentially contain all elements of the nested list in the worst-case nested configuration. Therefore, space is dependent on the number of elements and depth of the nesting.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursion

Instead of using a stack, recursion can be used to flatten the list into a single list of integers during initialization. Then, a simple iterator can be built on top of this flattened list.

Time: O(n) for flattening the list, O(1) for `next()` and `hasNext()`.Space: O(n) for the flattened list.

Trade-offs: Using recursion for flattening simplifies the `next()` and `hasNext()` methods, making them O(1). However, it requires additional O(n) space to store the flattened list upfront and can potentially cause stack overflow issues if the nesting depth is very large. The stack-based approach is generally preferred due to its lower space usage in some scenarios and better control over the iteration process.

✨

Key Insights

1Insight 1: Using a stack data structure is crucial for implementing the depth-first traversal of the nested list. The stack helps keep track of the nested lists that are yet to be explored.
2Insight 2: The `hasNext()` method is more complex than it seems. It's not just about checking if the stack is empty. We need to continuously unpack nested lists on the top of the stack until we find an integer or the stack becomes empty.
3Insight 3: The order in which elements are added to the stack is important. Since we are popping from the stack (LIFO), elements from the inner nested lists should be pushed in reverse order to maintain the correct flattened order.

📚

Key Terms Explained

For Students

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Queue

FIFO structure - First In, First Out. Like a line at a store!

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty input list. The iterator should handle an empty input list gracefully, with `hasNext()` returning `false` immediately.
Edge case 2: Deeply nested lists. The stack should be able to handle deeply nested lists without causing a stack overflow (though practically limited by system memory).
Edge case 3: A list containing only empty lists: The algorithm should correctly return `false` for `hasNext()` in this case.
Edge case 4: Integer appearing as the first element, followed by a list. The stack correctly retrieves the integer first.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Pushing elements onto the stack in the wrong order. The elements of the nested list must be pushed in reverse order to ensure the correct flattened order when popping.

✗Common mistake 2: Not handling empty nested lists correctly. The algorithm should skip over empty lists without adding them to the stack.

✗Common mistake 3: Assuming `next()` is always called after `hasNext()`. Although the problem implies this, it is a good practice to add a check in `next()` to throw an exception if there is no next element, ensuring robustness.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Processing JSON data. JSON data can have nested structures, and an iterator similar to this can be used to traverse and extract information from a complex JSON object.

🏦Real-world scenario 2: Traversing file system directories. File systems have nested directory structures, and an iterator can be used to list all files within a directory and its subdirectories.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Stacks are perfect for problems involving matching (parentheses) or "next greater element" patterns. If you need to look back at recent elements, think stack.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Popping from an empty stack. Always check if the stack is non-empty before popping.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 9, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

What is a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Flatten Nested List Iterator"?

For this medium-level Stack problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether tree techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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