What is the time complexity of Mini Parser?

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

What data structures are used to solve Mini Parser?

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

What are the key insights for solving Mini Parser?

Insight 1: The nested structure suggests using a stack to keep track of the currently open lists and their parent lists. Insight 2: The algorithm needs to differentiate between integers and nested lists based on the starting character of the input string or when encountering a '[' character.

Which companies ask LeetCode 385. Mini Parser in interviews?

This problem is frequently asked at Airbnb.

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Mini Parser - LeetCode 385 Solution

Mini Parser - Complete Solution Guide

Mini Parser is LeetCode problem 385, 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

Given a string s represents the serialization of a nested list, implement a parser to deserialize it and return the deserialized NestedInteger . Each element is either an integer or a list whose elements may also be integers or other lists. Example 1: Input: s = "324" Output: 324 Explanation: You should return a NestedInteger object which contains a single integer 324. Example 2: Input: s = "[123,[456,[789]]]" Output: [123,[456,[789]]] Explanation: Return a NestedInteger object containing a nest

Detailed Explanation

The problem requires you to implement a parser that can deserialize a string representation of a nested list into a `NestedInteger` object. A `NestedInteger` can either hold a single integer or a nested list of other `NestedInteger` objects. The input string `s` will consist of digits, square brackets ('[' and ']'), commas (','), and optionally a negative sign ('-'). The goal is to build the nested list structure based on the string's format and return the root `NestedInteger`.

Solution Approach

The solution employs a stack-based approach to parse the string and build the nested list structure. It iterates through the string character by character. When it encounters an opening bracket '[', it creates a new `NestedInteger` (representing a list) and pushes it onto the stack. When it finds a comma ',' or a closing bracket ']', it checks if there's a number that has been accumulated. If there is, it creates a new `NestedInteger` with that number and adds it to the `NestedInteger` at the top of the stack (the currently open list). When it encounters a closing bracket ']', it means a list is complete. It pops the completed list from the stack and, if the stack is not empty, adds the completed list to the `NestedInteger` on top of the stack. If the input string represents a single integer (doesn't start with '['), it directly creates a `NestedInteger` with that integer value and returns it.

Step-by-Step Algorithm

Step 1: Check if the input string starts with '['. If not, it represents a single integer; convert it to an integer and return a `NestedInteger` object holding that integer.
Step 2: Initialize an empty stack to store `NestedInteger` objects (representing lists).
Step 3: Initialize an empty string `current_num_str` to accumulate digits for forming integers.
Step 4: Iterate through the input string character by character.
Step 5: If the character is '[': Create a new `NestedInteger` (representing a list) and push it onto the stack.
Step 6: If the character is ',': If `current_num_str` is not empty, convert it to an integer, create a `NestedInteger` with that integer, and add it to the `NestedInteger` at the top of the stack. Reset `current_num_str` to an empty string.
Step 7: If the character is ']': If `current_num_str` is not empty, convert it to an integer, create a `NestedInteger` with that integer, and add it to the `NestedInteger` at the top of the stack. Reset `current_num_str` to an empty string. If the stack size is greater than 1, pop the `NestedInteger` from the stack (representing the closed list) and add it to the `NestedInteger` at the top of the stack.
Step 8: If the character is a digit or '-': Append the character to `current_num_str`.
Step 9: After iterating through the entire string, the `NestedInteger` at the bottom of the stack (the first element) represents the deserialized nested list. Return it.

Key Insights

Insight 1: The nested structure suggests using a stack to keep track of the currently open lists and their parent lists.
Insight 2: The algorithm needs to differentiate between integers and nested lists based on the starting character of the input string or when encountering a '[' character.
Insight 3: Accumulating digit characters into a string before converting it to an integer is crucial for handling multi-digit numbers and negative signs.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(d)

Topics

This problem involves: String, Stack, Depth-First Search.

Companies

Asked at: Airbnb.

Mini Parser | LeetCode 385

Medium

String Stack Depth-First Search

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

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 {
 *     // Constructor initializes an empty nested list.
 *     public NestedInteger();
 *
 *     // Constructor initializes a single integer.
 *     public NestedInteger(int value);
 *
 *     // @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();
 *
 *     // Set this NestedInteger to hold a single integer.
 *     public void setInteger(int value);
 *
 *     // Set this NestedInteger to hold a nested list and adds a nested integer to it.
 *     public void add(NestedInteger ni);
 *
 *     // @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.List;
import java.util.Stack;

class Solution {
    public NestedInteger deserialize(String s) {
        if (!s.startsWith("[")) {
            return new NestedInteger(Integer.parseInt(s));
        }

        Stack<NestedInteger> stack = new Stack<>();
        String currentNumStr = "";

        for (char charr : s.toCharArray()) {
            if (charr == '[') {
                stack.push(new NestedInteger());
            } else if (charr == ',') {
                if (!currentNumStr.isEmpty()) {
                    stack.peek().add(new NestedInteger(Integer.parseInt(currentNumStr)));
                }
                currentNumStr = "";
            } else if (charr == ']') {
                if (!currentNumStr.isEmpty()) {
                    stack.peek().add(new NestedInteger(Integer.parseInt(currentNumStr)));
                }
                currentNumStr = "";

                if (stack.size() > 1) {
                    NestedInteger closedList = stack.pop();
                    stack.peek().add(closedList);
                }
            } else { // digit or '-'
                currentNumStr += charr;
            }
        }

        return stack.get(0);
    }
}

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Problem Info

Description

Problem 385: Mini Parser - Detailed explanation and solution approaches.

Topics

String Stack Depth-First Search

Companies

Airbnb

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:385

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

9 steps

1
Step 1: Check if the input string starts with '['. If not, it represents a single integer; convert it to an integer and return a `NestedInteger` object holding that integer.
2
Step 2: Initialize an empty stack to store `NestedInteger` objects (representing lists).
3
Step 3: Initialize an empty string `current_num_str` to accumulate digits for forming integers.
4
Step 4: Iterate through the input string character by character.
5
Step 5: If the character is '[': Create a new `NestedInteger` (representing a list) and push it onto the stack.
6
Step 6: If the character is ',': If `current_num_str` is not empty, convert it to an integer, create a `NestedInteger` with that integer, and add it to the `NestedInteger` at the top of the stack. Reset `current_num_str` to an empty string.
7
Step 7: If the character is ']': If `current_num_str` is not empty, convert it to an integer, create a `NestedInteger` with that integer, and add it to the `NestedInteger` at the top of the stack. Reset `current_num_str` to an empty string. If the stack size is greater than 1, pop the `NestedInteger` from the stack (representing the closed list) and add it to the `NestedInteger` at the top of the stack.
8
Step 8: If the character is a digit or '-': Append the character to `current_num_str`.
9
Step 9: After iterating through the entire string, the `NestedInteger` at the bottom of the stack (the first element) represents the deserialized nested list. Return it.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(d)

📝

Why?

The time complexity is linear because each character in the input string is processed exactly once. The space complexity depends on the maximum depth of the nested lists, as the stack will hold one `NestedInteger` for each level of 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

Recursive Approach

A recursive approach can be used to parse the string. You would need to keep track of the current position in the string and recursively call the function to parse nested lists.

Time: O(n), where n is the length of the input string.Space: O(d), where d is the maximum depth of the nested list (due to recursion stack).

Trade-offs: The recursive approach can be more readable for some, but it might be less efficient due to the overhead of function calls and managing the current position in the string. The stack-based approach is generally more efficient because it avoids the overhead of recursion.

✨

Key Insights

1Insight 1: The nested structure suggests using a stack to keep track of the currently open lists and their parent lists.
2Insight 2: The algorithm needs to differentiate between integers and nested lists based on the starting character of the input string or when encountering a '[' character.
3Insight 3: Accumulating digit characters into a string before converting it to an integer is crucial for handling multi-digit numbers and negative signs.

📚

Key Terms Explained

For Students

Stack

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

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string. The provided solutions don't explicitly handle this. In C, the provided solution returns NestedIntegerInit(). Python, CPP, and Java solutions won't encounter an empty string because the constraints specify the string length is at least 1.
Edge case 2: Single integer input (e.g., "123"). The solutions correctly handle this case by creating a `NestedInteger` directly with the integer value.
Edge case 3: Nested lists with only integers (e.g., "[1,2,3]"). The solutions handle this correctly by creating individual `NestedInteger` objects for each integer and adding them to the list.
Edge case 4: Nested lists with nested lists (e.g., "[[1,2],[3,4]]"). The stack-based approach ensures correct handling of nested structures.
Edge case 5: Negative numbers (e.g., "-123"). The solutions correctly accumulate the '-' sign along with the digits into `current_num_str` and then convert it to an integer using `stoi` or `Integer.parseInt`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle negative numbers correctly. Forgetting to accumulate the '-' sign before converting to an integer will result in incorrect parsing.

✗Common mistake 2: Incorrectly handling commas. Commas indicate the end of a number, but the solutions also need to check if the `current_num_str` is empty to prevent parsing errors when there's consecutive commas.

✗Common mistake 3: Not resetting the `current_num_str` after parsing a number. This will lead to the next number being parsed incorrectly.

✗Common mistake 4: Forgetting to add the last number to the current list when the parsing ends with a ']'.

✅

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: Parsing JSON data. The problem is analogous to parsing JSON arrays, where nested structures are common.

🏦Real-world scenario 2: Deserializing complex data structures from strings in configuration files or network communication protocols. Many configuration files represent data using nested lists and dictionaries.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: June 26, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

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

@AlgoMasterJul 7

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

@InterviewProJun 30

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

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.

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 "Mini Parser"?

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

🤖 Need help with this problem?