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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Simplify Path in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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About LeetCode

LeetCode is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews. 

LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either EasyMedium, or Hard.

LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:

  • Mathematics/Basic Logical Based Questions
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  • Dynamic Programming
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  • Greedy Algorithms
  • Breadth-First Search
  • Depth-First Search
  • Sorting & Searching
  • BST (Binary Search Tree)
  • Database
  • Linked List
  • Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the ProblemSimplify Path– LeetCode Problem

Simplify Path– LeetCode Problem

Problem:

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.

Constraints:

  • 1 <= path.length <= 3000
  • path consists of English letters, digits, period '.', slash '/' or '_'.
  • path is a valid absolute Unix path.
Simplify Path– LeetCode Solutions
Simplify Path in C++:
class Solution {
 public:
  string simplifyPath(string path) {
    string ans;
    istringstream iss(path);
    vector stack;

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

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

    return ans.empty() ? "/" : ans;
  }
};
Simplify Path in Java:
class Solution {
  public String simplifyPath(String path) {
    final String[] dirs = path.split("/");
    Stack stack = new Stack<>();

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

    return "/" + String.join("/", new ArrayList(stack));
  }
}
Simplify Path in Python:
class Solution:
  def simplifyPath(self, path: str) -> str:
    stack = []

    for str in path.split('/'):
      if str in ('', '.'):
        continue
      if str == '..':
        if stack:
          stack.pop()
      else:
        stack.append(str)

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

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