Decompress File Paths in C++

A comma separated list of strings can be compressed by finding common substrings and grouping those strings together into Token Lists

More formally, a Token List is a list where each item can be one of the following:

a string
a concatenation of Token Lists

Examples:

ab,ac => a{b,c}
ac,ad,bc,bd => {a,b}{c,d}
axc,axd,bxc,bxd => {a,b}x{c,d}

This technique is used to compress folder paths since a folder like

README.md
resources/
    js/
        app.js
        boot.js
        comp/
            a.js
            b.js

can be represented as a list of strings:

[
    "README.md",
    "resources/js/app.js",
    "resources/js/boot.js",
    "resources/js/comp/a.js",
    "resources/js/comp/b.js"
]

and further encoded into comma separated string

“README.md,resources/js/app.js,resources/js/boot.js,resources/js/comp/a.js,resources/js/comp/b.js”

and then we can compress them to the following Token List:

README.md,resources/js/{app,boot,comp/{a,b}}.js

Given a Token List representing a compressed folder, return the initial list of paths.

Example 1:

Input:
"resources/{js/{app.js,comp/{a.js,b.js},boot.js}},README.md"

Output: [
    "README.md",
    "resources/js/app.js",
    "resources/js/boot.js",
    "resources/js/comp/a.js",
    "resources/js/comp/b.js"
]

Example 2:

Input:
"my_{first,second}_folder/{a,b}{.js,.py}"

Output: [
    "my_first_folder/a.js,
    "my_first_folder/a.py,
    "my_first_folder/b.js,
    "my_first_folder/b.py,
    "my_second_folder/a.js,
    "my_second_folder/a.py,
    "my_second_folder/b.js,
    "my_second_folder/b.py
]

Note: The resulting array can be in any order

Understanding the Problem

The core challenge of this problem is to decompress a given compressed folder path represented as a Token List into its original list of paths. This involves parsing the Token List, identifying the common substrings, and expanding them into all possible combinations.

This problem is significant in scenarios where file paths or similar hierarchical data structures need to be efficiently stored and transmitted. Common applications include file systems, URL routing, and configuration management.

Potential pitfalls include correctly handling nested Token Lists and ensuring all combinations are generated without missing any paths.

Approach

To solve this problem, we need to parse the Token List and recursively expand the nested structures. Here’s a step-by-step approach:

Identify the base case where the Token List is a simple string without any nested structures.
For nested structures, recursively expand each part and concatenate the results.
Use a stack or recursion to handle the nested Token Lists.

Let’s start with a naive solution and then optimize it.

Naive Solution

The naive solution involves iterating through the string and expanding each Token List one by one. This approach is not optimal due to its high time complexity, especially for deeply nested structures.

Optimized Solution

An optimized solution involves using a stack to manage the nested Token Lists and efficiently expanding them. This reduces the time complexity by avoiding redundant operations.

Algorithm

Here’s a step-by-step breakdown of the optimized algorithm:

Initialize a stack to keep track of the current context (i.e., the current Token List being processed).
Iterate through the input string character by character.
When encountering a '{', push the current context onto the stack and start a new context.
When encountering a '}', pop the context from the stack and concatenate the results.
For each comma-separated value, recursively expand it and concatenate the results.
Return the final list of paths.

Code Implementation

Here is the C++ code for the optimized solution:


#include <iostream>
#include <vector>
#include <string>
#include <stack>
#include <sstream>

using namespace std;

// Helper function to split a string by a delimiter
vector<string> split(const string &s, char delimiter) {
    vector<string> tokens;
    string token;
    istringstream tokenStream(s);
    while (getline(tokenStream, token, delimiter)) {
        tokens.push_back(token);
    }
    return tokens;
}

// Recursive function to expand the token list
vector<string> expand(const string &s) {
    stack<vector<string>> stk;
    vector<string> current;
    string token;
    for (char c : s) {
        if (c == '{') {
            if (!token.empty()) {
                current.push_back(token);
                token.clear();
            }
            stk.push(current);
            current.clear();
        } else if (c == '}') {
            if (!token.empty()) {
                current.push_back(token);
                token.clear();
            }
            vector<string> prev = stk.top();
            stk.pop();
            vector<string> combined;
            for (const string &p : prev) {
                for (const string &q : current) {
                    combined.push_back(p + q);
                }
            }
            current = combined;
        } else if (c == ',') {
            if (!token.empty()) {
                current.push_back(token);
                token.clear();
            }
        } else {
            token += c;
        }
    }
    if (!token.empty()) {
        current.push_back(token);
    }
    return current;
}

int main() {
    string input1 = "resources/{js/{app.js,comp/{a.js,b.js},boot.js}},README.md";
    string input2 = "my_{first,second}_folder/{a,b}{.js,.py}";

    vector<string> result1 = expand(input1);
    vector<string> result2 = expand(input2);

    cout << "Output for input1:" << endl;
    for (const string &path : result1) {
        cout << path << endl;
    }

    cout << "Output for input2:" << endl;
    for (const string &path : result2) {
        cout << path << endl;
    }

    return 0;
}

Complexity Analysis

The time complexity of the optimized solution is O(n * m), where n is the length of the input string and m is the maximum number of nested structures. The space complexity is O(n) due to the use of the stack and intermediate storage.

Edge Cases

Potential edge cases include:

Empty input string
Input string without any nested structures
Deeply nested structures

Each algorithm handles these edge cases by ensuring the base case is correctly identified and the recursion or stack operations are properly managed.

Testing

To test the solution comprehensively, consider the following test cases:

Simple input without nested structures
Input with multiple levels of nested structures
Edge cases such as empty input and deeply nested structures

Use a variety of test cases to ensure the solution is robust and handles all scenarios correctly.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Break down the problem into smaller, manageable parts.
Use recursion or a stack to handle nested structures.
Test the solution with a variety of test cases to ensure robustness.

Practice solving similar problems and studying algorithms to improve problem-solving skills.

Conclusion

In this blog post, we discussed how to decompress file paths represented as Token Lists. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for efficient data storage and transmission in various applications.

Additional Resources

For further reading and practice, consider the following resources:

LeetCode - Practice coding problems
GeeksforGeeks - Tutorials and coding challenges
C++ Documentation - Official C++ documentation

Decompress File Paths in C++ with Time Complexity Analysis