Simplify Path - Stack - Leetcode 71 - Python

[@yaoyao6605]

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

[@licokr]

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!

[@danielsun716]

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)

[@anirudh1202]

This is a simpler solution

[@DesiqnedBy]

This made the solution make alot more sense, thank you!

[@edwardteach2]

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

[@dhruvpatel9708]

Personally I find leetcode solution more easy to understand

[@ladydimitrescu1155]

Honestly this is much more straightforward, thanks!

[@littletiger1228]

so much better

[@pranavbhatia8736]

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.

[@kaneknight4606]

Hero. Your videos really help with confidence.

[@NeetCode]

Thanks, appreciate the kind words 🙂

[@oijgg3p]

@@NeetCode I am doing leetcode like crazy, and I consider you as a mentor. Can't wait to reach one day and thank you after getting into google.

[@devnull711]

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.

[@sahilsharma2518]

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`

[@geekydanish5990]

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)

[@karthik1627]

Nice alternative solution 👌👌

[@andreivilla9009]

This is the exact same solution I came up with, I think it's neater and simpler to understand

[@batmunkhn6666]

just had this question for google

[@rahatsshowcase8614]

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("/"); };

[@jagrutitiwari2551]

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

[@ehabteima]

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

[@dreamakash]

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

[@aakankshajaiswal9770]

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)

[@curesnow6493]

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

[@mallepallihimasagar955]

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)

[@VladPopov7]

"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.

[@kywei7485]

You always give the clearest explanations! Thanks a lot!

[@ozgeylmaz8685]

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

[@hoyinli7462]

ur videos are just awesome

[@kcprxkln]

using two stacks was more intuitive to me for some reason

[@ajith4249]

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)

[@ShivamKumar-qv6em]

Great explanation

[@rajchavan2886]

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

[@tripham4884]

really neat solution. Thank you.

[@anikaboyd3613]

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

[@praneeth9002]

It helps to add the final scanned directory name to add into the stack

[@willowsongbird]

@@praneeth9002 how exactly is it helping?

[@edwardteach2]

For this specific test case: path("/a//b////c/d//././/..") < -- realize there is a ".." at the end of this path, which means we need to pop from the top of their stack If we didn't include the "/" in our for loop, we'll get the wrong answer. I don't like this problem since it's very tricky. But oh well. We needed the "/" at the end of our for loop [for c in path + "/"] in order to pop the 'd'. Thus our answer would be '/a/b/c', not 'a/b/c/d' (if you removed the '/' in the for loop) Best way to explain it is to debug it, then you'll understand.

[@lavanya_m01]

@@willowsongbird A simple example is take this case "/abc/def". If you don't add "/" to the end, you wouldn't add "def" to the stack, so ur ans would be "/abc", but if we add "/" to the input, our output will be "/abc/def" which is correct.

[@karthikh8993]

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

[@anuragchakraborty7607]

You have to use another stack , fill the new stack by popping elements of current stack . And pop each element from the new stack and make a string ans , ans += "/" + newstack->top() ;

[@arpitjain5179]

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

[@CodeMonkeyy]

Yes I agree. It fails on this test input

[@arjunreddy2647]

/I/Love/You

[@wallyyu5426]

smooth

[@amansayer4943]

iam shsit bro

[@alokesh985]

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

[@lavanya_m01]

it's not needed.. it'll be handled by the first if statement where we check if cur == '..'

[@vatsaljain7029]

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

[@haroldaltamirano3958]

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); } }

[@davidpetro1780]

amazing explanation

[@thanirmalai]

Great explanation