Last updated on March 7th, 2025 at 08:13 pm

Here, we see a Simplify Path LeetCode Solution. This Leetcode problem is solved using different approaches in many programming languages, such as C++, Java, JavaScript, Python, etc.

List of all LeetCode Solution

Topics

Stack, String

Companies

Facebook, Microsoft

Level of Question

Medium

Simplify Path LeetCode Solution

1. Problem Statement

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.

2. Coding Pattern Used in Solution

The coding pattern used in this code is “Stack-based Path Simplification”. The stack is used to simulate the behavior of navigating through a file system, where .. means moving up one directory, and . or empty strings are ignored.

3. Code Implementation in Different Languages

3.1 Simplify Path C++

class Solution {
public:
    string simplifyPath(string path) {
        string res, s;
        stack<string>stk;
        stringstream ss(path);
        while(getline(ss, s, '/')) {
            if (s == "" || s == ".") continue;
            if (s == ".." && !stk.empty()) stk.pop();
            else if (s != "..") stk.push(s);
        }
        while(!stk.empty()){
            res = "/"+ stk.top() + res;
            stk.pop();
        }
        return res.empty() ? "/" : res;        
    }
};

3.2 Simplify Path Java

class Solution {
    public String simplifyPath(String path) {
	String[] dir = path.split("/");
	String[] stack = new String[dir.length];
	int ptr = 0;
	for(int i = 0; i < dir.length; i++){
		if(dir[i].equals(".") || dir[i].equals("")){
			continue;
		}else if(dir[i].equals("..")){
			if(ptr > 0) ptr--;
		}else{
			stack[ptr] = dir[i];
			ptr++;
		}
	}
	StringBuilder result = new StringBuilder();
	for(int i = 0; i < ptr; i++){
		result.append("/");
		result.append(stack[i]);
	}
	return result.length() == 0 ? "/" : result.toString();    
    }
}

3.3 Simplify Path JavaScript

var simplifyPath = function(path) {
    let stack = [];
    path = path.split('/');
    
    for (let i=0;i<path.length;i++) {
        if (path[i]=='.' || path[i]=='') continue;
        if (path[i]=='..') stack.pop();
        else stack.push(path[i]);
    }
    
    return '/'+stack.join('/');    
};

3.4 Simplify Path Python

class Solution(object):
    def simplifyPath(self, path):
        stack = []
        for token in path.split('/'):
            if token in ('', '.'):
                pass
            elif token == '..':
                if stack: stack.pop()
            else:
                stack.append(token)
        return '/' + '/'.join(stack)

4. Time and Space Complexity

	Time Complexity	Space Complexity
C++	O(n)	O(n)
Java	O(n)	O(n)
JavaScript	O(n)	O(n)
Python	O(n)	O(n)

• Time Complexity: The time complexity is O(n) for all implementations because:
  • Splitting the path into tokens takes O(n), where n is the length of the input string.
  • Each token is processed once in the loop, which is also O(n).
  • Constructing the final result from the stack is O(n) in the worst case.
• Space Complexity: The space complexity is O(n) for all implementations because:
  • The stack can hold at most n tokens in the worst case.
  • The final result string or array also requires O(n) space.
• The logic is consistent across all implementations, with minor differences in syntax and language-specific features.