In the domain of algorithms and problem-solving, Leetcode offers a plethora of problems that challenge us to apply the best data structures and algorithms. One such intriguing problem is the “Crawler Log Folder” problem. In this problem, we have to process a series of operations in a list to determine the minimum number of operations required to go back to the main folder from a particular subfolder.

The problem statement is simple, but solving it requires an in-depth understanding of lists, stack data structure, and algorithmic optimization. This article aims to provide a comprehensive explanation of the problem, delve into the Pythonic solution, discuss the time and space complexities, and explore possible optimizations.

## Problem Statement

The original problem statement from Leetcode is as follows:

The Leetcode file system keeps a log each time some user performs a change folder operation.The operations are described below:

`"../"`

: Move to the parent folder of the current folder. (If you are already in the main folder, remain in the same folder).`"./"`

: Remain in the same folder.`"x/"`

: Move to the child folder named`x`

(This folder is guaranteed to always exist).

You are given a list of strings `logs`

where `logs[i]`

is the operation performed by the user at the `i-th`

step.

The file system starts in the main folder, return the minimum number of operations needed to go back to the main folder after the change folder operations.

### Example

```
Input: logs = ["d1/","d2/","../","d21/","./"]
Output: 2
```

## Understanding the Problem

In essence, the problem simulates a file system. We start in the main folder and move based on the operations provided in the list `logs`

. The aim is to find the minimum number of operations needed to return to the main folder. The operations are:

`"../"`

: Move up a folder. If already at the main folder, stay there.`"./"`

: Stay in the current folder.`"x/"`

: Move into a subfolder named`x`

.

## Algorithmic Approach

The problem is inherently sequential; each operation depends on the state resulting from the previous operation. As such, a stack-based approach is apt for solving this problem.Here is how we can solve this:

- Initialize a counter to keep track of the current folder level (depth).
- Traverse through each operation in the
`logs`

list. - Update the counter based on the operation:
- If the operation is
`"../"`

and we are not at the main folder, decrease the counter by 1. - If the operation is
`"./"`

, do nothing. - Otherwise, increase the counter by 1 (move into a subfolder).

- If the operation is
- At the end of the traversal, the counter will give the minimum number of operations needed to go back to the main folder, which is essentially the depth we are at.

## Python Code Implementation

Here is the Python code for this approach:

```
def minOperations(logs):
counter = 0
for log in logs:
if log == "../":
if counter > 0:
counter -= 1
elif log == "./":
continue
else:
counter += 1
return counter
```

## Code Explanation

- We initialize a
`counter`

to zero. This counter will serve as our measure of the depth of the current folder. - We iterate through each
`log`

in the`logs`

list. - Inside the loop, we apply the folder change operation and modify the
`counter`

accordingly:- If it is a
`"../"`

operation, we decrement the`counter`

only if we are not already in the main folder (`counter > 0`

). - If it is a
`"./"`

operation, we do nothing (`continue`

). - For all other logs, we increment the
`counter`

by 1.

- If it is a
- Finally, we return the
`counter`

, which is the minimum number of operations to go back to the main folder.

## Time and Space Complexity

The time complexity of this solution is O(n), where n is the length of the `logs`

list. This is because we traverse the list once, performing constant-time operations for each element.

The space complexity is O(1), as we only use a single integer variable `counter`

for storing the state. No additional data structures are required.

## Possible Optimizations and Variants

**Queue instead of Stack**: Since we are only interested in the final state, a queue-based approach will also yield the same result.**Using Enums for Operations**: For better code readability and future changes, you can use Enums to represent each folder operation instead of using raw strings.**Multi-level Folders**: An interesting variant of this problem could involve multi-level folders and would require a slightly more complex data structure to keep track of the current state.

## Conclusion

The “Crawler Log Folder” problem is an excellent exercise for practicing stack-based algorithms and offers an array of optimization possibilities. It is a straightforward but insightful problem that tests your understanding of basic data structures and algorithms.