It's really nice that you started by explaining how the directory works. Very clear and organize video!

appending a slash at the end of path and appending a slash when it returns makes code much easier to implement. I added many conditions and it is not easy to read. I learned a lot today too. Thank you so much!

So the built-in function .split() come up in my mind, which is a good way to convert string into list, this will help a lot. Let's still take Neetcode's example of the path. If path is "/../abc//./def/", then path.split('/') gonna be [' ', '..', 'abc', ' ', '.', 'def', ' ']. And this is pretty easy to understand the if conditions in Neetcode's code. class Solution: def simplifyPath(self, path: str) -> str: stack = [] newPath = path.split('/') for c in newPath: if c == '..': if stack: stack.pop() elif c != '' and c != '.': stack.append(c) return '/' + '/'.join(stack)

just had this question for google

Hero. Your videos really help with confidence.

Leetcode's solution is also good to understand was well: class Solution: def simplifyPath(self, path: str) -> str: # Initialize a stack stack = [] # Split the input string on "/" as the delimiter # and process each portion one by one for portion in path.split("/"): # If the current component is a "..", then # we pop an entry from the stack if it's non-empty if portion == "..": if stack: stack.pop() elif portion == "." or not portion: # A no-op for a "." or an empty string continue else: # Finally, a legitimate directory name, so we add it # to our stack stack.append(portion) # Stich together all the directory names together final_str = "/" + "/".join(stack) return final_str

Great explanation as always, I just listen to you explain the problem. By looking at the problem description I wasn't sure I understood all the cases.

I tried something like this without having to create cur variable. There are 2 conditions - one for double dot in which case we pop from the stack if stack is not empty. The second case is for single dot - we append into our stack if dir is not equal to single dot. class Solution: def simplifyPath(self, path: str) -> str: stack = [] path = path.replace('//','/') for dir in path.split('/'): if dir: if dir == '..': if stack: stack.pop() else: if dir != '.': stack.append(dir) return '/' + '/'.join(stack) Submission stats: Runtime: 35 ms, faster than 84.95% of Python3 online submissions for Simplify Path. Memory Usage: 13.9 MB, less than 37.47% of Python3 online submissions for Simplify Path.

Same but by just using split on path, makes it's easier to understand `path_arr = path.split("/") stack = [] for sec in path_arr: # section is not empty and section is not "." which means just current directory if sec != '' and sec != "." if sec == "..": # pop previous val in stack is sec is ".." if stack: stack.pop() else: stack.append(sec) res = "/" + "/".join(stack) return res`

It'd be much simpler if you split on / and loop over the list and build your stack

You always give the clearest explanations! Thanks a lot!

Thank you so much for this simple solution. My solution was too complicated and got stuck.

I did not need to see the code after listening to your explanation. I solved it just after listening to you.

If you go from right to left then you do not even need a stack. / and . can be handled inplace. '..' means skip one directory.

"join" traverses stack in the FIFO order, but stack is a LIFO order, so I think this property should not be called "stack" or the join method should not be used with stack.

class Solution: def simplifyPath(self, path: str) -> str: stack = [] path = path.split("/") for p in path: if stack and p == "..": stack.pop() if p.isalnum() or (p != "." and p != ".." and len(p) > 0): stack.append(p) return "/" + "/".join(stack)

I submitted the solution provided and it was a wrong answer for the test case: "/a//b////c/d//././/..". Therefore, I have modified the solution using while inside while loop and it was accepted. class Solution: def simplifyPath(self, path: str) -> str: stack = [] i = 0 while i < len(path): if path[i] == '/': i += 1 continue else: cur = '' while i < len(path) and path[i] != '/': cur += path[i] i += 1 if cur == '..': if stack: stack.pop() elif cur == '.' or cur == '': i += 1 continue else: stack.append(cur) return '/' + '/'.join(stack)

your explanation is so great

I wonder how you approach while solving the question for the first time, how do you split it into smaller part or do you immediately come with the solution at once

ur videos are just awesome

JS solution: var simplifyPath = function(path) { const segments=path.split("/"); const stack=[]; for (let segment of segments){ if (segment==="" || segment ===".") continue else if (segment==="..") stack.pop() else stack.push(segment) } return "/" + stack.join("/"); };

using two stacks was more intuitive to me for some reason

Great explanation

amazing explanation

Why did the "/" and the end of the for loop definition make the code easier?

A better simple way of the solving the question 1. replace all multiple occurring slashes to single slash 2. split the string by "/" 3. now perform the stack operations on the split string array class Solution: def simplifyPath(self, path: str) -> str: stack = [] while "//" in path: path = path.replace("//","/") splitted_string = path.split("/") #split the string with "/" and perform opertations over the stack for op in splitted_string: if op=="..": if stack: stack.pop() else: continue elif op=="." or op=="": continue else: stack.append(op) return "/"+"/".join(stack)

really neat solution. Thank you.

Isn't modifying input a bad practice in interviews? Context: you added a trailing "/" to the path

How can we write the last line of code in c++ instead of join

i think it will fail on this test case input : "/../"

smooth

/I/Love/You

iam shsit bro

If this fails in leetcode, try adding cur != '..' to line 10. This is for test cases like '/..'

i dont understand why you copy others solution and then try to explain them without understanding them one bit. time wasted.

simple solution stack=[] lis=path.split("/") for items in lis: if(items==".."): if(stack): stack.pop() elif(items!="" and items!="."): stack.append(items) return "/"+"/".join(stack)

I tried to do the join at the end in c# but it gave me the list in the wrong order, any idea why? public class Solution { public string SimplifyPath(string path) { Stack directories = new(); StringBuilder current = new(); // /home//foo// foreach(char character in (path + "/")){ if(character == '/'){ if(current.ToString() == ".."){ if(directories.Count > 0){ directories.Pop(); } } else if(current.ToString() != "" && current.ToString() != "."){ directories.Push(current.ToString()); } current = new StringBuilder(); } else{ current.Append(character); } } // fix wrong order for return "/" + String.Join( "/", directories); Stack answer = new(); foreach(var directory in directories){ Console.WriteLine($"directory={directory}"); answer.Push(directory); } return "/" + String.Join( "/", answer); } }

Great explanation