1598. Crawler Log Folder
Problem Descriptionβ
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.
Examplesβ
Example 1:
Input: logs = ["d1/","d2/","../","d21/","./"]
Output: 2
Explanation: Use this change folder operation "../" 2 times and go back to the main folder.
Constraintsβ
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 for 1598. Crawler Log Folderβ
- Solution
Implementationβ
function Solution(arr) { var minOperations = function(logs) { let cnt = 0; for (let log of logs) { if (log === "../") { if (cnt > 0) { cnt--; } } else if (log !== "./") { cnt++; } } return cnt; }; const input = logs = ["d1/","d2/","../","d21/","./"] const output = minOperations(input) return ( <div> <p> <b>Input: </b> {JSON.stringify(input)} </p> <p> <b>Output:</b> {output.toString()} </p> </div> ); }
Complexity Analysisβ
- Time Complexity:
- Space Complexity:
Code in Different Languagesβ
- TypeScript
- Python
- Java
- C++
function minOperations(logs: string[]): number {
let cnt = 0;
for (let log of logs) {
if (log === "../") {
if (cnt > 0) {
cnt--;
}
} else if (log !== "./") {
cnt++;
}
}
return cnt;
}
from typing import List
class Solution:
def minOperations(self, logs: List[str]) -> int:
cnt = 0
for log in logs:
if log == "../":
if cnt > 0:
cnt -= 1
elif log != "./":
cnt += 1
return cnt
import java.util.List;
public class Solution {
public int minOperations(List<String> logs) {
int cnt = 0;
for (String log : logs) {
if (log.equals("../")) {
if (cnt > 0) {
cnt--;
}
} else if (!log.equals("./")) {
cnt++;
}
}
return cnt;
}
}
class Solution {
public:
int minOperations(vector<string>& logs) {
int cnt=0;
for (int i = 0; i < logs.size(); i++) {
if (logs[i] == "../") {
if(cnt)
cnt--;
else continue;
} else if (logs[i] == "./") {
continue;
} else{
cnt++;
}
}
return cnt;
}
};
Referencesβ
-
LeetCode Problem: 1598. Crawler Log Folder
-
Solution Link: LeetCode Solution