Simplify Path Leetcode Solution

In this post, we are going to solve the Simplify Path Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Simplify Path Leetcode Solution
Simplify Path Leetcode Solution


Given a string path, which is an absolute path (starting with a slash '/') to a file or directory in a Unix-style file system, convert it to the simplified canonical path.

In a Unix-style file system, a period '.' refers to the current directory, a double period '..' refers to the directory up a level, and any multiple consecutive slashes (i.e. '//') are treated as a single slash '/'. For this problem, any other format of periods such as '...' are treated as file/directory names.

The canonical path should have the following format:

  • The path starts with a single slash '/'.
  • Any two directories are separated by a single slash '/'.
  • The path does not end with a trailing '/'.
  • The path only contains the directories on the path from the root directory to the target file or directory (i.e., no period '.' or double period '..')

Return the simplified canonical path.

Example 1:

Input: path = "/home/"
Output: "/home"
Explanation: Note that there is no trailing slash after the last directory name.

Example 2:

Input: path = "/../"
Output: "/"
Explanation: Going one level up from the root directory is a no-op, as the root level is the highest level you can go.

Example 3:

Input: path = "/home//foo/"
Output: "/home/foo"
Explanation: In the canonical path, multiple consecutive slashes are replaced by a single one.


  • 1 <= path.length <= 3000
  • path consists of English letters, digits, period '.', slash '/' or '_'.
  • path is a valid absolute Unix path.

Now, lets see the leetcode solution of Simplify Path Leetcode Solution.

Simplify Path Leetcode Solution in Python

class Solution:
  def simplifyPath(self, path: str) -> str:
    stack = []

    for str in path.split('/'):
      if str in ('', '.'):
      if str == '..':
        if stack:

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

Simplify Path Leetcode Solution in CPP

class Solution {
  string simplifyPath(string path) {
    string ans;
    istringstream iss(path);
    vector<string> stack;

    for (string dir; getline(iss, dir, '/');) {
      if (dir.empty() || dir == ".")
      if (dir == "..") {
        if (!stack.empty())
      } else {

    for (const string& s : stack)
      ans += "/" + s;

    return ans.empty() ? "/" : ans;

Simplify Path Leetcode Solution in Java

class Solution {
  public String simplifyPath(String path) {
    final String[] dirs = path.split("/");
    Stack<String> stack = new Stack<>();

    for (final String dir : dirs) {
      if (dir.isEmpty() || dir.equals("."))
      if (dir.equals("..")) {
        if (!stack.isEmpty())
      } else {

    return "/" + String.join("/", stack);

Note: This problem Simplify Path is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

NEXT: Edit Distance Leetcode Solution

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