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Problem 1598 Crawler Log Folder

Problem Statement

Link - Problem 1598

Question

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 ith step.

The file system starts in the main folder, then the operations in logs are performed.

Return the minimum number of operations needed to go back to the main folder after the change folder operations.

Example 1

Input: logs = ["d1/","d2/","../","d21/","./"]
Output: 2
Explanation: Use this change folder operation "../" 2 times and go back to the main folder.

Example 2

Input: logs = ["d1/","d2/","./","d3/","../","d31/"]
Output: 3

Example 3

Input: logs = ["d1/","../","../","../"]
Output: 0

Constraints

- `1 <= logs.length <= 10^3`
- `2 <= logs[i].length <= 10`
- logs[i] contains lowercase English letters, digits, '.', and '/'.
- logs[i] follows the format described in the statement.
- Folder names consist of lowercase English letters and digits.

Solution

class Solution {
public:
    int minOperations(vector<string>& logs) {
        int count = 0;
        std::ios::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);
        for(string it:logs)
        {
            if(it == "../")
            {
                count--;
                if(count<0)
                    count = 0;
            }
            else if(it != "./")
                count++;
        }
        return count;
    }
};

Complexity Analysis

• Time Complexity : O(N)
• Space Complexity : O(1)

Explanation

1. Intuition

- Since the directory is Linear in the given problem, we can keep track of the count of the operations.
- If the operation is `../` then we decrement the count and if the count is less than 0, we reset it to 0.
- If the operation is not `./` then we increment the count.
- Finally, we return the count.

2. Implementation

- Initialize the `count` to 0.
- Iterate over the `logs` and check if the operation is `../` then decrement the `count` and reset it to 0 if it is less than 0.
- If the operation is not `./` then increment the `count`.
- Finally, return the `count`.