> For the complete documentation index, see [llms.txt](https://maksimdan.gitbook.io/interview-practice-problems/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://maksimdan.gitbook.io/interview-practice-problems/leetcode_sessions/71-simplify-path.md).

# 71 Simplify Path

Given an absolute path for a file (Unix-style), simplify it.

For example,\
**path**=`"/home/"`, =>`"/home"`\
**path**=`"/a/./b/../../c/"`, =>`"/c"`

**Corner Cases:**

* Did you consider the case where path = "/../"?
* In this case, you should return "/".
* Another corner case is the path might contain multiple slashes '/' together, such as "/home//foo/".
* In this case, you should ignore redundant slashes and return "/home/foo".

**The Idea:** This problem requires some Unix domain knowledge. `.` means current directory, so in terms of a path, it adds nothing and can be ignored. `..` means move up to parent directory, and `...` means move up recursively. Utilizing a stack is a natural solution because moving to a parent directory is equivalently to moving to the previous or most recent element in an array.

**Complexity:** O(n) time and space

```python
class Solution:
    def simplifyPath(self, path):
        """
        :type path: str
        :rtype: str
        """

        tokens = [token for token in path.split('/') if token != '' and token != '.']
        s = []

        for token in tokens:
            if token == '..':
                if s: # separated ifs for case where '...'
                    s.pop()
            else:
                s.append(token)

        return '/' + '/'.join(s)


obj = Solution()
print(obj.simplifyPath('/...')) # /...
print(obj.simplifyPath('/../')) # /
print(obj.simplifyPath('/')) # /
print(obj.simplifyPath('/a/./b/../../c/')) # /c
print(obj.simplifyPath('/home//foo/')) # /home/foo
```


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