LeetCode problem#1598
Crawler Log Folder
Solved in 1 languageJava
Solution
1class Solution {
2 public int minOperations(String[] logs) {
3 int level = 0;
4
5 for (int i = 0; i < logs.length; i++) {
6 if (logs[i].equals("./")) {
7 // do nothing
8 }
9 else if (logs[i].equals("../")) {
10 if (level > 0) {
11 level--;
12 }
13 }
14 else {
15 // going down a folder
16 level++;
17 }
18 }
19
20 return level;
21 }
22}How it works
The given Java code is a solution to a Leetcode problem that requires you to track your current directory level after processing a series of file path operations defined in an array of strings called logs. The operations in logs can be thought of as commands to navigate through a file system.
Problem Breakdown
Each string in the logs array represents a command that modifies the current directory level. The commands can be:
"./": This command refers to the current directory, meaning you do not change your level."../": This command refers to moving up to the parent directory. If you are not already at the top level (level 0), this command decreases the current level by one. If you are on the top level (level 0), it does not change the level.- Any other string (which we can assume refers to a valid directory name): This command indicates going down into that directory, which increases the level by one.
Solution Explanation
Now, let's walk through the provided code in detail:
1class Solution {
2 public int minOperations(String[] logs) {
3 int level = 0; // Initialize current directory level
4
5 for (int i = 0; i < logs.length; i++) { // Iterate over each command in logs
6 if (logs[i].equals("./")) {
7 // do nothing
8 }
9 else if (logs[i].equals("../")) {
10 if (level > 0) {
11 level--; // Move up a level only if we are not already at level 0
12 }
13 }
14 else {
15 // Going down a folder
16 level++; // Move down a level for any directory name
17 }
18 }
19
20 return level; // Return the final level
21 }
22}Step-by-step Explanation
-
Initialization:
int level = 0;: This variablelevelis initialized to0, representing the topmost directory level.
-
Iterating through logs:
- The code utilizes a
forloop to go through each element in thelogsarray.
- The code utilizes a
-
Command Processing:
- For
"./": If the command is to refer to the current directory, there's no action taken (// do nothing). - For
"../": If the command is to go up one level:- It checks if
levelis greater than0to ensure we’re not at the root level (level0). If so, it decrements the level.
- It checks if
- For any other command (which we assume are names of directories): This indicates moving into that directory, thus increasing the level by
1.
- For
-
Returning the Level:
- After processing all commands, the final value of
levelrepresents the depth within the directory structure, which is returned as the result.
- After processing all commands, the final value of
Summary
This solution effectively simulates navigating through a file system based on relative movements given by the commands in the logs array. The key points in the logic ensure that it accurately manages the level based on the defined commands, and it returns the number of levels below the root directory that you end up at upon completion of all commands. The overall time complexity is O(n), where n is the length of the logs array, as it processes each command exactly once.
Written by AI from the code above, so double-check anything important.
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