Algorithm
Problem Name: 1233. Remove Sub-Folders from the Filesystem
Given a list of folders folder, return the folders after removing all sub-folders in those folders. You may return the answer in any order.
If a folder[i] is located within another folder[j], it is called a sub-folder of it.
The format of a path is one or more concatenated strings of the form: '/' followed by one or more lowercase English letters.
- For example,
"/leetcode"and"/leetcode/problems"are valid paths while an empty string and"/"are not.
Example 1:
Input: folder = ["/a","/a/b","/c/d","/c/d/e","/c/f"] Output: ["/a","/c/d","/c/f"] Explanation: Folders "/a/b" is a subfolder of "/a" and "/c/d/e" is inside of folder "/c/d" in our filesystem.
Example 2:
Input: folder = ["/a","/a/b/c","/a/b/d"] Output: ["/a"] Explanation: Folders "/a/b/c" and "/a/b/d" will be removed because they are subfolders of "/a".
Example 3:
Input: folder = ["/a/b/c","/a/b/ca","/a/b/d"] Output: ["/a/b/c","/a/b/ca","/a/b/d"]
Constraints:
1 <= folder.length <= 4 * 1042 <= folder[i].length <= 100folder[i]contains only lowercase letters and'/'.folder[i]always starts with the character'/'.- Each folder name is unique.
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public List removeSubfolders(String[] folder) {
Set set = new HashSet<>();
Arrays.sort(folder, new Comparator < String>(){
public int compare(String s1, String s2) {
return s1.length() - s2.length();
}
});
for (String fl : folder) {
String[] files = fl.split("/");
StringBuilder sb = new StringBuilder();
for (int i = 1; i < files.length; i++) {
sb.append("/").append(files[i]);
if (set.contains(sb.toString())) {
break;
}
}
if (sb.length() > 0) {
set.add(sb.toString());
}
}
return new ArrayList < >(set);
}
}
Input
folder = ["/a","/a/b","/c/d","/c/d/e","/c/f"]
Output
["/a","/c/d","/c/f"]
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def removeSubfolders(self, folder: List[str]) -> List[str]:
st = set(folder)
for f in folder:
if any(p in st for p in itertools.accumulate(f.split('/'), lambda x, y: x + '/' + y) if p and p != f):
st.discard(f)
return list(st)
Input
folder = ["/a","/a/b","/c/d","/c/d/e","/c/f"]
Output
["/a","/c/d","/c/f"]