Python Clean sultion( DFS, hashmaps, serialization) with explaination
Intuition\n Describe your first thoughts on how to solve this problem. \nThis code is devised to address the problem of identifying and removing duplicate folde
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis code is devised to address the problem of identifying and removing duplicate folder structures within a given set of folder paths. A folder structure is considered duplicate if there exists another folder structure with the exact same structure and folder names.\n\nThe initial intuition for solving this problem is to use a trie (prefix tree) data structure, which allows us to efficiently represent the folder paths in a hierarchical structure. By traversing this trie, we can serialize the folder structures into strings and identify duplicate structures by looking for identical serialized strings.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe approach followed in the code can be broken down into several steps:\n\n**1. Building the Trie:**\n\n- A trie data structure is built to represent the folder paths. Each node in the trie contains a dictionary representing its children, a boolean variable indicating if it\'s the end of a path, and another boolean variable to mark if it needs to be deleted (if it\'s part of a duplicated folder structure).\n- The insert function is used to insert a path into the trie. It starts from the root and iterates through the folders in the path, adding new nodes as necessary.\n\n**2. Serialization:**\n\n- The serialize function is used to create a string representation of each folder structure in the trie. This function is a recursive function that starts from the root and generates a string representation for each node by concatenating the folder names and the serialized strings of the child nodes (in lexicographical order).\n- All serialized strings are stored in a dictionary, where the key is the serialized string and the value is a list of nodes that have that serialized structure.\n- The serialization process converts the folder structures in the trie into string representations, which can be used to easily identify duplicates. Here\'s a detailed explanation of this process:\n\n    - It starts with an opening bracket ( to denote the start of a new folder structure.\n    - It then iterates over the child nodes in lexicographical order (hence the sorted function usage). Sorting ensures that two identical folder structures will have the exact same serialized string, facilitating the identification of duplicates.\n    - For each child node, it appends the folder name and the serialized string of the child node to the current serialized string.\n    - It ends with a closing bracket ) to denote the end of the current folder structure.\n    - If the serialized string is not () (which represents an empty folder), it adds the node to the list of nodes with that serialized structure in the dictionary.\n    - Finally, it returns the serialized string for the current node.\n\nSerialization is an essential step in this solution because it allows us to convert the hierarchical folder structures into string representations, which can be easily compared to identify duplicates. By representing each folder structure as a unique string, we can efficiently find duplicate structures by looking for identical serialized strings in the dictionary.\n\nSorting the keys (folder names) during the serialization process ensures that two identical folder structures will have the exact same serialized string. Without sorting, the order of child nodes might vary between two identical structures, resulting in different serialized strings and failing to identify them as duplicates. Sorting guarantees a consistent order of child nodes in the serialized string, facilitating the identification of duplicates.\n\n**3. Marking Duplicates:**\n\n- After serialization, nodes that have more than one occurrence in the dictionary (indicating duplicated structures) are marked for deletion by setting their deleted attribute to True.\n\n**4. Extracting Non-Duplicate Paths:**\n\n- Finally, the dfs (Depth-First Search) function traverses the trie to extract the non-duplicate paths. It starts from the root and recursively visits all nodes that are not marked for deletion, collecting the folder names to construct the paths. Here\'s a breakdown of its operations:\n\n    - It first checks whether the current node is marked for deletion (deleted = True). If so, it terminates the traversal for the current path since it\'s part of a duplicated folder structure.\n    - If the node is not marked for deletion and the current path is not empty, it adds the current path to the result list.\n    - It then recursively visits the child nodes that are not marked for deletion, extending the current path with the folder name of each child node.\n\n**5. Result Compilation:**\n\n- The paths collected by the dfs function are then returned as the result.\n\n# Complexity\n**Time complexity:**\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time and space complexity of the given code can be analyzed based on the operations being performed at each step. Here\'s the breakdown:\n\n**1.Inserting Paths into the Trie (Step 1):**\n\nIterating over each path and inserting it into the trie takes O(nm) time, where n is the number of paths and m is the maximum length of a path.\n\n**2.Serialization (Step 2):**\n\nThe serialization process involves a depth-first traversal of the trie, visiting each node once. Additionally, at each node, we sort the keys of the children dictionary, which takes \nO(klogk) time, where k is the number of children. Therefore, the time complexity for the serialization process can be approximated to \nO(nmlogm), where \nn is the number of nodes in the trie and \nm is the maximum branching factor (maximum number of children of a node).\n\n**3.Marking Duplicates (Step 3):**\n\nThis step involves iterating over the serialized strings and, for each string with more than one node, marking those nodes as deleted. This takes O(n) time, where n is the number of nodes in the trie.\n\n**4.Extracting Non-Duplicate Paths (Step 5):**\nThis step involves a depth-first traversal of the trie to extract non-duplicate paths, visiting each node once. This takes O(n) time, where n is the number of nodes in the trie.\n\nCombining these, the total time complexity is approximately O(nmlogm).\n\n**Space complexity:**\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n**1. Trie Structure:**\n\nThe trie structure itself takes O(nm) space to store all the nodes, where n is the number of paths and m is the maximum length of a path.\n\n**2. Serialized Dictionary:**\n\nThe serialized dictionary stores serialized strings for each node in the trie. The total length of all serialized strings can be \nO(nm), where n is the number of nodes and m is the maximum branching factor.\n\n**3. Result List:**\n\nThe result list stores the non-duplicate paths, which can take up to O(nm) space in the worst case.\n\nCombining these, the total space complexity is O(nm).\n\nReference Video: https://www.youtube.com/watch?v=5UGb889i21Y\n\n# Code\n```\nclass TrieNode:\n    def __init__(self):\n        self.children = {} # key is  or folder name, sub directories or the TrieNode is the value\n        self.is_end = False # check if it is the child node of the tree\n        self.deleted = False # True ifthe subtree should be deleted\n\nclass Solution:\n    def deleteDuplicateFolder(self, paths: List[List[str]]) -> List[List[str]]:\n        \n        # Setting Up a Basic trie Structure\n        def insert(path):\n            node = root\n            for folder in path:\n                if folder not in node.children:\n                    node.children[folder] = TrieNode()\n                node = node.children[folder]\n            node.is_end = True\n        \n        # Serialize or hash the subfolders to find the duplicate ones\n        def serialize(node):\n            ans = \'(\'\n            for folder, child in sorted(node.children.items()):\n                ans += folder + serialize(child)\n            ans += \')\'\n            if ans != "()": \n                serialized[ans].append(node)\n            return ans\n\n        # Travers the trie and append non-duplicated folders to the result\n        def dfs(node, path):\n            if node.deleted:\n                return\n            if path:\n                result.append(path)\n            for folder, child in node.children.items():\n                if not child.deleted:\n                    dfs(child, path + [folder])\n\n        # Step 1: Insert paths into the trie and build the Trie\n        root = TrieNode()\n\n        for path in paths:\n            insert(path)\n        \n        #step 2: initiliaze the serialization and serilize the nodes\n        serialized = defaultdict(list)\n        serialize(root)\n        # Step 3: Find the duplciated and mark them\n        for nodes in serialized.values():\n            if len(nodes) > 1:\n                for node in nodes:\n                    node.deleted = True\n\n        # Step 5: travers the trie and print the non duplicated sub trees\n        result = []\n        dfs(root, [])\n\n        return result\n\n```
['Python3']