Design an Iterative Path Tree and Clarify File Lookup Semantics

Question

The question asks the candidate to design an iterative path tree that can manage a series of file paths, allowing for the lookup of complete file paths while handling potential duplicate directory names.

Codemia · Accepted Answer

To design an iterative path tree, we need to manage a list of file paths and allow for efficient lookups. The tree must accommodate duplicate directory names, ensuring that complete paths remain unique. The solution should avoid recursion, using an iterative approach instead.

We will create a tree structure where each node represents a directory or file. Each node will maintain a list of its children. To build the tree, we will iterate through the list of paths, splitting each path into its components and adding them to the tree. We will use a stack to manage the current path as we traverse and build the tree iteratively.

The iterative construction of the tree involves using a stack to keep track of the current node. For each path, we split it by '/' and iterate through the components. If a component does not exist in the current node's children, we create a new node and add it. We then push this node onto the stack. Once we reach the end of the path, we can store the file reference in the final node.

Using an iterative approach prevents stack overflow issues that can occur with deep recursion, especially with long paths. However, it may require more careful management of the stack. This design can scale well with the number of paths, but performance may degrade with excessive depth or breadth due to memory constraints.

Here is a sample implementation in Python:

```python
class TreeNode:
    def __init__(self, name):
        self.name = name
        self.children = {}
        self.is_file = False

class PathTree:
    def __init__(self):
        self.root = TreeNode('')

def insert(self, path):
        components = path.strip('/').split('/')
        current = self.root
        for component in components:
            if component not in current.children:
                current.children[component] = TreeNode(component)
            current = current.children[component]
        current.is_file = True

def lookup(self, path):
        components = path.strip('/').split('/')
        current = self.root
        for component in components:
            if component not in current.children:
                return None
            current = current.children[component]
        return current.is_file

# Example usage:
tree = PathTree()
tree.insert('/a/c/file1.txt')
print(tree.lookup('/a/c/file1.txt'))  # Output: True
```

Design an Iterative Path Tree and Clarify File Lookup Semantics

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