What is the time complexity of Simplify Path?

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

What data structures are used to solve Simplify Path?

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

What are the key insights for solving Simplify Path?

Insight 1: Using a stack is crucial for keeping track of the directory structure and easily navigating up the hierarchy when encountering '..'. Insight 2: Splitting the path by '/' allows us to process each directory component individually.

Which companies ask LeetCode 71. Simplify Path in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Capital One and 17 more companies.

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Simplify Path - LeetCode 71 Solution

Simplify Path - Complete Solution Guide

Simplify Path is LeetCode problem 71, 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 an absolute path for a Unix-style file system, which always begins with a slash '/' . Your task is to transform this absolute path into its simplified canonical path . The rules of a Unix-style file system are as follows: A single period '.' represents the current directory. A double period '..' represents the previous/parent directory. Multiple consecutive slashes such as '//' and '///' are treated as a single slash '/' . Any sequence of periods that does not match the rules above

Detailed Explanation

The problem asks us to simplify a given absolute Unix-style file system path. The simplified path must adhere to specific rules: it should start with a single slash, directories should be separated by exactly one slash, and it shouldn't end with a slash unless it represents the root directory. Single periods ('.') represent the current directory and should be ignored. Double periods ('..') represent the parent directory, so we need to go up one level in the directory structure. Multiple consecutive slashes should be treated as a single slash. The input is a string representing the absolute path, and the output is the simplified canonical path.

Solution Approach

The solution utilizes a stack to maintain the directory hierarchy. The input path is first split into components using '/' as a delimiter. Each component is then processed: If it's '..', we pop an element from the stack (if the stack is not empty) to go up one level. If it's a directory name (not empty and not '.' or '..'), we push it onto the stack. Finally, we construct the simplified path by joining the elements in the stack with '/' and prepending a leading '/'. Special handling is required to ensure the output is '/' if the stack is empty (root directory).

Step-by-Step Algorithm

Step 1: Split the input path string into a list of components using the '/' character as a delimiter.
Step 2: Initialize an empty stack to store the directory names.
Step 3: Iterate through each component in the list.
Step 4: If the component is '..', check if the stack is not empty. If it's not empty, pop the top element from the stack (representing moving to the parent directory).
Step 5: If the component is not empty and not equal to '.' and not equal to '..', push the component onto the stack (representing a valid directory).
Step 6: After processing all components, construct the simplified path by joining the elements in the stack with '/' as a delimiter.
Step 7: Prepend a '/' to the resulting string to ensure it's an absolute path.
Step 8: If the stack is empty after processing (meaning we navigated back to the root directory or started at the root), return "/". Otherwise, return the constructed path.

Key Insights

Insight 1: Using a stack is crucial for keeping track of the directory structure and easily navigating up the hierarchy when encountering '..'.
Insight 2: Splitting the path by '/' allows us to process each directory component individually.
Insight 3: Handling edge cases like consecutive slashes, single periods, double periods, and ensuring the final path format is correct is essential for a robust solution.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Stack.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Capital One, Citadel, Gojek, Grab, Harness, Meta, Microsoft, OpenAI, Patreon, Roku, Snowflake, Tesla, TikTok, Tinkoff, Upstart, Visa, Yahoo, Yandex.

Simplify Path | LeetCode 71

Medium

String Stack

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

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public String simplifyPath(String path) {
        Stack<String> stack = new Stack<>();
        String[] components = path.split("/");

        for (String component : components) {
            if (component.equals("..")) {
                if (!stack.isEmpty()) {
                    stack.pop();
                }
            } else if (!component.isEmpty() && !component.equals(".")) {
                stack.push(component);
            }
        }

        return "/" + String.join("/", stack);
    }
}

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Problem Info

Description

Problem 71: Simplify Path - Detailed explanation and solution approaches.

Topics

String Stack

Companies

AdobeAmazonAppleBloombergCapital OneCitadelGojekGrab

Stats

Difficulty:Medium

Languages:4

Companies:22

Problem #:71

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

8 steps

1
Step 1: Split the input path string into a list of components using the '/' character as a delimiter.
2
Step 2: Initialize an empty stack to store the directory names.
3
Step 3: Iterate through each component in the list.
4
Step 4: If the component is '..', check if the stack is not empty. If it's not empty, pop the top element from the stack (representing moving to the parent directory).
5
Step 5: If the component is not empty and not equal to '.' and not equal to '..', push the component onto the stack (representing a valid directory).
6
Step 6: After processing all components, construct the simplified path by joining the elements in the stack with '/' as a delimiter.
7
Step 7: Prepend a '/' to the resulting string to ensure it's an absolute path.
8
Step 8: If the stack is empty after processing (meaning we navigated back to the root directory or started at the root), return "/". Otherwise, return the constructed path.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the linear iteration through the path and its components. The space complexity arises from the potential for the stack to hold all directory names in the input path if there are no ".." operations to remove elements from the stack.

📖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

In-place modification (less efficient)

Instead of using a stack, the path can be modified in-place. This involves iterating through the path and manipulating a pointer to track the current position of the simplified path. This approach requires more complex pointer manipulation and is generally less readable.

Time: O(n)Space: O(1) (excluding the input string)

Trade-offs: While the space complexity is better, the code is significantly more complex and harder to understand. The stack-based approach is more readable and easier to maintain for the same overall time complexity.

✨

Key Insights

1Insight 1: Using a stack is crucial for keeping track of the directory structure and easily navigating up the hierarchy when encountering '..'.
2Insight 2: Splitting the path by '/' allows us to process each directory component individually.
3Insight 3: Handling edge cases like consecutive slashes, single periods, double periods, and ensuring the final path format is correct is essential for a robust solution.

📚

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: Path starts with multiple slashes (e.g., '///home//foo'). The `split()` function handles this by creating empty strings in the resulting array, which are ignored because the condition `component` and `component != '.'` filters them out.
Edge case 2: Path contains single periods (e.g., '/./home/./foo'). These are ignored because the condition `component != '.'` filters them out during processing.
Edge case 3: Path contains double periods that navigate above the root directory (e.g., '/../'). The stack's conditional `if stack:` prevents popping from an empty stack, so the path remains at the root, and the algorithm correctly outputs '/'.
Edge case 4: Trailing slash (e.g., '/home/'). The final path construction ` '/' + '/'.join(stack)` handles it by not adding a trailing slash unless the stack is empty (root directory case).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the stack is empty after processing all components (e.g., '/../'). In this case, the output should be '/' (root directory).

✗Common mistake 2: Incorrectly handling multiple consecutive slashes. The splitting logic automatically handles this, but forgetting to filter out empty strings after the split can lead to errors.

✗Common mistake 3: Not correctly dealing with leading or trailing slashes in the input. Leading slashes are required in the output, while trailing slashes should be removed (unless the path is just '/').

✗Common mistake 4: Not freeing allocated memory in C implementation, leading to memory leaks.

✅

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: Navigating file systems in operating systems. This algorithm is directly applicable to simplifying paths entered by users or generated by programs.

🏦Real-world scenario 2: Web server URL normalization. Web servers often need to normalize URLs to ensure consistent behavior, and this algorithm can be used to simplify URL paths that contain redundant components.

🎮Real-world scenario 3: Software development tools that analyze or manipulate file paths, such as build systems or IDEs.

💡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

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

⚠️ Watch Out

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

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: August 11, 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 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 "Simplify Path"?

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?