# Logic 1: 100 pass - 81% faster - Straighforward without considering a number of complex cases makes this pass. Just take care of .. or . alone. Considering /... kind of confuses the situation

class Solution:

def simplifyPath(self, path: str) -> str:

stack = [] # Container to hold the answer, should start with /

no_entries = set(["", "..", "."]) # No entries in result with respect to these values

elements = path.split("/") # Split by hashes to remove single or multiple hash

for ele in elements:

if stack and ele == "..": # indicates we should go back on dir

stack.pop()

elif ele not in no_entries:

stack.append(ele)

stack = "/" + "/".join(stack)

return stack