Problem Description

You're given a file system that tracks folder navigation operations. The system starts in the main folder and processes a series of commands to move between folders.

There are three types of operations:

• "../" - Move to the parent folder (go up one level). If already at the main folder, stay there.
• "./" - Stay in the current folder (no movement).
• "x/" - Move into a child folder named x (go down one level into folder x).

Given an array logs containing these operations performed in sequence, determine how many steps it takes to return to the main folder after executing all operations.

For example, if you start at the main folder and execute:

1. "d1/" - go into folder d1 (depth = 1)
2. "d2/" - go into folder d2 (depth = 2)
3. "../" - go back to d1 (depth = 1)
4. "./" - stay in d1 (depth = 1)

You would need 1 operation to get back to the main folder.

The solution tracks the current depth from the main folder. Going into a child folder increases the depth by 1, going to parent folder decreases it by 1 (but never below 0), and staying in the same folder doesn't change the depth. The final depth value represents the minimum number of operations needed to return to the main folder.

Intuition

Think of the file system as a tree structure where the main folder is the root. When we navigate through folders, we're essentially moving up and down this tree. The key insight is that we only need to track our current depth in the tree - how many levels deep we are from the main folder.

Why does tracking depth work? Because going back to the main folder always requires moving up exactly as many levels as we've moved down. If we're 3 levels deep, we need exactly 3 "../" operations to get back to the main folder.

We can visualize this like climbing stairs in a building:

• Starting at ground floor (main folder) = depth 0
• Going into a child folder = going up one floor (depth + 1)
• Going to parent folder = going down one floor (depth - 1)
• Staying in same folder = staying on same floor (depth unchanged)

The beauty of this approach is that we don't need to remember the actual folder names or the exact path we took. We only care about how "deep" we are. Each non-special folder operation ("x/") takes us one level deeper, and each "../" brings us one level up (unless we're already at the main folder).

The constraint that we can't go above the main folder (depth can't be negative) is handled by using max(0, depth - 1) when processing "../". This ensures that if we're already at the main folder and try to go up, we stay at depth 0.

After processing all operations, our current depth tells us exactly how many "../" operations we'd need to perform to get back to the main folder, which is our answer.

Learn more about Stack patterns.

Solution Approach

The implementation uses a simple counter variable to track the current folder depth. Let's walk through the algorithm step by step:

1. Initialize a counter: Start with ans = 0 to represent that we begin at the main folder (depth 0).

2. Process each operation: Iterate through each operation v in the logs array.

3. Handle different operation types:

   • Parent folder operation ("../"):
     • Check if v == "../"
     • Decrease the depth by 1, but ensure it doesn't go below 0 using ans = max(0, ans - 1)
     • This handles the edge case where we're already at the main folder
   • Child folder operation (any folder name like "x/"):
     • Check if the operation doesn't start with "." using v[0] != "."
     • This condition cleverly filters out both "./" and "../" operations
     • If true, increment the counter: ans += 1
   • Stay operation ("./"):
     • This is implicitly handled - when v == "./", neither condition is met
     • The counter remains unchanged

4. Return the result: After processing all operations, ans contains the current depth, which equals the minimum number of operations needed to return to the main folder.

The algorithm has a time complexity of O(n) where n is the number of operations, as we iterate through the list once. The space complexity is O(1) as we only use a single integer variable to track the depth.

The elegance of this solution lies in its simplicity - by abstracting the problem to depth tracking, we avoid the complexity of maintaining actual folder paths or a stack data structure.

Example Walkthrough

Let's trace through a concrete example with logs = ["d1/", "d2/", "../", "d21/", "./"]

Initial State:

• We start at the main folder
• ans = 0 (depth counter initialized to 0)

Step 1: Process "d1/"

• This is a child folder operation (doesn't start with ".")
• We move into folder d1
• ans = 0 + 1 = 1
• Current position: main → d1 (depth 1)

Step 2: Process "d2/"

• This is a child folder operation (doesn't start with ".")
• We move into folder d2 (inside d1)
• ans = 1 + 1 = 2
• Current position: main → d1 → d2 (depth 2)

Step 3: Process "../"

• This is a parent folder operation
• We move back to d1
• ans = max(0, 2 - 1) = 1
• Current position: main → d1 (depth 1)

Step 4: Process "d21/"

• This is a child folder operation (doesn't start with ".")
• We move into folder d21 (inside d1)
• ans = 1 + 1 = 2
• Current position: main → d1 → d21 (depth 2)

Step 5: Process "./"

• This is a stay operation
• Neither condition is met (v != "../" and v[0] == ".")
• ans remains unchanged at 2
• Current position: main → d1 → d21 (depth 2)

Final Result:

• After all operations, ans = 2
• This means we need 2 operations ("../" twice) to return to the main folder
• Return value: 2

The algorithm correctly determines that from our final position (2 levels deep), we need exactly 2 parent folder operations to get back to the main folder.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the logs list. The algorithm iterates through each element in the logs list exactly once. Each operation inside the loop (string comparison, max operation, and increment/decrement) takes O(1) time. Therefore, the overall time complexity is linear with respect to the input size.

Space Complexity: O(1). The algorithm uses only a single integer variable ans to keep track of the current depth in the file system. No additional data structures are created that scale with the input size. The space usage remains constant regardless of how many log entries are processed.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrectly Handling Parent Directory at Main Folder

A common mistake is forgetting to check if we're already at the main folder when processing "../". Simply decrementing the depth counter without bounds checking would result in negative depth values.

Incorrect approach:

if operation == "../":
    current_depth -= 1  # Wrong! Can go negative

Correct approach:

if operation == "../":
    current_depth = max(0, current_depth - 1)  # Ensures depth never goes below 0

2. Misidentifying Folder Operations

Another pitfall is incorrectly identifying which operations represent moving into a child folder. Some might try to explicitly check for NOT being "./" or "../", which can be error-prone.

Incorrect approach:

if operation != "./" and operation != "../":
    current_depth += 1

Better approach:

elif operation[0] != ".":  # Cleverly filters out both "./" and "../"
    current_depth += 1

Or use the explicit conditions as shown in the solution for better readability.

3. String Comparison Without Considering Format

Assuming the operations might not have the trailing slash or have different formats can lead to bugs. The problem guarantees the format, but defensive programming might lead to over-complicated checks.

Over-complicated approach:

if operation.startswith("..") or operation == ".." or operation == "../":
    # Handle parent directory

Clean approach (assuming guaranteed format):

if operation == "../":
    # Handle parent directory

4. Using Complex Data Structures Unnecessarily

Some might think they need to maintain the actual folder path using a stack or list, which adds unnecessary complexity and memory usage.

Over-engineered approach:

folder_stack = []
for operation in logs:
    if operation == "../":
        if folder_stack:
            folder_stack.pop()
    elif operation != "./":
        folder_stack.append(operation)
return len(folder_stack)

Optimal approach: Simply track the depth with a single integer counter, as shown in the solution.

5. Off-by-One Errors in Final Result

Some might mistakenly think they need to add 1 to the final depth or perform additional calculations. The depth directly represents the number of "../" operations needed to return to the main folder.

Incorrect interpretation:

return current_depth + 1  # Wrong! The depth IS the answer

Correct interpretation:

return current_depth  # The depth equals the steps needed to go back