Remove Invasive Species

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Unit 8 Session 2 Standard (Click for link to problem statements)

Unit 8 Session 2 Advanced (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Medium-Hard
• ⏰ Time to complete: 25-30 mins
• 🛠️ Topics: Trees, Binary Search Trees, Deletion, Inorder Successor

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Q: What is the structure of the tree?
  • A: The tree is a Binary Search Tree (BST) where each node represents a species in the ecosystem, organized alphabetically by name.
• Q: What operation needs to be performed?
  • A: The function needs to remove a specified node from the BST while maintaining BST properties.
• Q: What should be returned?
  • A: The function should return the root of the updated BST after removing the specified species.

HAPPY CASE
Input: 
      ecosystem = TreeNode("Dugong", 
                  TreeNode("BrainCoral", None, TreeNode("Clownfish")),
                  TreeNode("Lionfish", TreeNode("GiantClam"), TreeNode("Seagrass"))), 
      name = "Lionfish"
Output: Updated BST with "Lionfish" removed and replaced by its inorder successor.
Explanation: "Lionfish" has two children, so it's replaced by the inorder successor "GiantClam".

EDGE CASE
Input: 
      ecosystem = TreeNode("Dugong", 
                  TreeNode("BrainCoral", None, TreeNode("Clownfish")),
                  TreeNode("Lionfish", TreeNode("GiantClam"), TreeNode("Seagrass"))), 
      name = "Dugong"
Output: Updated BST with "Dugong" removed and replaced by its inorder successor.
Explanation: "Dugong" is the root with two children, so it's replaced by the inorder successor "GiantClam".

2: M-atch

Match what this problem looks like to known categories of problems, e.g., Linked List or Dynamic Programming, and strategies or patterns in those categories.

For Binary Search Tree (BST) problems, we want to consider the following approaches:

• BST Deletion: Removing a node from a BST while maintaining its properties involves several cases, such as no child, one child, or two children. For the two-child case, the node is replaced by its inorder successor.
• Inorder Traversal: Utilized to find the inorder successor when the node to be removed has two children.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Traverse the BST to find the node to remove. Handle the different cases based on the node's children, and update the tree accordingly.

1) If the current node (`ecosystem`) is `None`, return `None`.
2) If `name` is less than the current node's value, recurse on the left subtree.
3) If `name` is greater than the current node's value, recurse on the right subtree.
4) If `name` matches the current node's value:
    - If the node has no children, return `None`.
    - If the node has one child, return that child.
    - If the node has two children:
        - Find the inorder successor (leftmost node in the right subtree).
        - Replace the current node's value with the inorder successor's value.
        - Recursively remove the inorder successor from the right subtree.
5) Return the root of the updated tree.

⚠️ Common Mistakes

• Not correctly handling the case where the node to be removed has two children.
• Forgetting to update parent pointers when removing a node.

4: I-mplement

Implement the code to solve the algorithm.

class TreeNode:
    def __init__(self, value, left=None, right=None):
        self.val = value
        self.left = left
        self.right = right

def remove_species(ecosystem, name):
    if not ecosystem:
        return None
    
    if name < ecosystem.val:
        ecosystem.left = remove_species(ecosystem.left, name)
    elif name > ecosystem.val:
        ecosystem.right = remove_species(ecosystem.right, name)
    else:
        # Node to be deleted is found
        if not ecosystem.left:
            return ecosystem.right
        elif not ecosystem.right:
            return ecosystem.left
        else:
            # Node has two children
            successor = find_min(ecosystem.left)
            ecosystem.val = successor.val
            ecosystem.left = remove_species(ecosystem.left, successor.val)
    
    return ecosystem

def find_min(node):
    while node.left:
        node = node.left
    return node

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

- Example 1:
    - Input: 
        `ecosystem = TreeNode("Dugong", TreeNode("BrainCoral", None, TreeNode("Clownfish")), TreeNode("Lionfish", TreeNode("GiantClam"), TreeNode("Seagrass")))`, 
        `name = "Lionfish"`
    - Execution: 
        - Start at root "Dugong".
        - "Lionfish" > "Dugong", move to the right child "Lionfish".
        - Node "Lionfish" found with two children, find inorder successor "GiantClam".
        - Replace "Lionfish" with "GiantClam" and remove "GiantClam".
    - Output: 
    ```
    Dugong
    BrainCoral GiantClam
    Clownfish Seagrass
    ```

- Example 2:
    - Input: 
        `ecosystem = TreeNode("Dugong", TreeNode("BrainCoral", None, TreeNode("Clownfish")), TreeNode("Lionfish", TreeNode("GiantClam"), TreeNode("Seagrass")))`, 
        `name = "Dugong"`
    - Execution: 
        - Start at root "Dugong".
        - Node "Dugong" found with two children, find inorder successor "GiantClam".
        - Replace "Dugong" with "GiantClam" and remove "GiantClam".
    - Output: 
    ```
    GiantClam
    BrainCoral Lionfish
    Clownfish Seagrass
    ```

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Assume N represents the number of nodes in the tree.

• Time Complexity: O(H) where H is the height of the tree. In a balanced BST, this is O(log N).
• Space Complexity: O(H) due to the recursive call stack. In the worst case for a skewed tree, this could be O(N).