1598. Crawler Log Folder.md

1598. Crawler Log Folder

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, 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

You are given a list of command/logs of navigating through folders. We need to find the minimum number of operations needed to go back to the main folder after the change folder operations.

Approach

There are three types of operations in the logs:

1. "../" move to parent folder of current folder
2. "./" stay on the same folder
3. And "x/" move down to a x sub folder.

As the second command does not add any value (there is no location change), we can ignore this one.

Third command move down to a sub folder and hence we need additional operations to go back to the Main. First command move upwards and hence reduce the operation needed to go back to the Main.

Algorithm

1. Declare operationNeeded = 0;
2. for each log in the logs:
   • if log is equal to "../"
     • if operationNeeded is greater than 0 (not on the Main folder)
       • Reduce required operationNeeded by one.
     • else do nothing
   • else if log is not equal to "./" (i,e. log equal to "x/")
     • Increase required operationNeeded by one.
3. Return operationNeeded

Explanation with example

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

log	operationNeeded
"d1/"	1
"d1/"	2
"../"	1
"d21/"	2
"./"	2

minimum number of operations needed = 2.

Complexity

• Time Complexity:

  • O(n) where n = size of the logs

• Space Complexity:

  • O(1)

Code

class Solution {
public:
    int minOperations(vector<string>& logs) {
        int operationNeeded = 0;

        for(string log:logs) {
            if(log == "../") {
                if(operationNeeded > 0) {
                    operationNeeded--;
                }
            } else if(log != "./") {
                operationNeeded++;
            }
        }
        return operationNeeded;
    }
};

Tags: C++, cpp, leetcode, leetcode 1598, Greedy, Implementation