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Binary Tree Cameras

Hard

Amazon

3 views

Topics:

TreesRecursionDynamic ProgrammingGreedy Algorithms

You are given the root of a binary tree. We install cameras on the tree nodes where each camera at a node can monitor its parent, itself, and its immediate children.

Return the minimum number of cameras needed to monitor all nodes of the tree.

Example 1:

Consider the following binary tree:

    0
   / \
  0   null
 / \
0   0

Input: root = [0,0,null,0,0] Output: 1 Explanation: One camera is enough to monitor all nodes if placed at the root node.

Example 2:

Consider the following binary tree:

Input: root = [0,0,null,0,null,0,null,null,0] Output: 2 Explanation: At least two cameras are needed to monitor all nodes of the tree. One camera at the bottom most left node, and another at the node two levels above.

Constraints:

The number of nodes in the tree is in the range [1, 1000].
Node.val == 0

Binary Tree Cameras

Hard

Amazon

3 views

Topics:

TreesRecursionDynamic ProgrammingGreedy Algorithms

You are given the root of a binary tree. We install cameras on the tree nodes where each camera at a node can monitor its parent, itself, and its immediate children.

Return the minimum number of cameras needed to monitor all nodes of the tree.

Example 1:

Consider the following binary tree:

    0
   / \
  0   null
 / \
0   0

Input: root = [0,0,null,0,0] Output: 1 Explanation: One camera is enough to monitor all nodes if placed at the root node.

Example 2:

Consider the following binary tree:

Constraints:

The number of nodes in the tree is in the range [1, 1000].
Node.val == 0

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the tree be empty or contain null nodes?
What is the range of values for the nodes in the binary tree?
Do all nodes have unique values, or can there be duplicates?
If no cameras are needed (e.g., a single node tree), what should the function return?
Is the tree guaranteed to be a binary tree, or could it be a more general graph structure represented in a tree-like manner?

Brute Force Solution

Approach

The brute force approach to the binary tree cameras problem involves trying out every conceivable arrangement of cameras on the tree's nodes. We explore all possible placements and then determine which placement uses the fewest cameras while still ensuring every node is covered. This is done by exhaustively considering all combinations and directly checking the coverage for each.

Here's how the algorithm would work step-by-step:

Start by considering all possible sets of nodes where we could place cameras. This means trying placing a camera on every node individually.
For each possible set of camera placements, check if every node in the tree is either directly monitored by a camera on itself, or is monitored by a camera on one of its immediate neighbors (parent or children).
If a set of camera placements does not cover all nodes, discard that set and move on to the next one.
If a set of camera placements does cover all nodes, record the number of cameras used in that set.
After trying all possible sets of camera placements, compare the number of cameras used in each valid (covering all nodes) set.
Select the set of camera placements that uses the fewest number of cameras. This is your solution.

Code Implementation

class TreeNode:
    def __init__(self, value=0, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

def binary_tree_cameras_brute_force(root):
    all_nodes = []
    def traverse(node):
        if node:
            all_nodes.append(node)
            traverse(node.left)
            traverse(node.right)

    traverse(root)
    number_of_nodes = len(all_nodes)
    minimum_cameras = float('inf')

    for i in range(1 << number_of_nodes):
        camera_placements = []
        number_of_cameras = 0

        for j in range(number_of_nodes):
            if (i >> j) & 1:
                camera_placements.append(all_nodes[j])
                number_of_cameras += 1

        # For this placement of cameras, check for coverage
        is_covered = True
        for node in all_nodes:
            if not is_node_covered(node, camera_placements):
                is_covered = False
                break

        if is_covered:
            minimum_cameras = min(minimum_cameras, number_of_cameras)

    return minimum_cameras

def is_node_covered(node, camera_placements):
    # Check if the node is covered by any camera placement.
    if node in camera_placements:
        return True

    for camera in camera_placements:
        if camera.left == node or camera.right == node or get_parent(node, camera) : 
            return True

    return False

def get_parent(node, possible_parent):
    # Helper to determine parent relationship
    if possible_parent is None: 
        return False

    if possible_parent.left == node or possible_parent.right == node:
        return True

    return False

Big(O) Analysis

Time Complexity

O(2^n * n) – The brute force approach iterates through all possible subsets of nodes to place cameras. With n nodes, there are 2^n possible subsets. For each subset, we must check if the camera placement covers all n nodes in the tree, which takes O(n) time. Thus, the overall time complexity is O(2^n * n) because we examine each of the 2^n subsets and spend O(n) work to check coverage.

Space Complexity

O(2^N) – The brute force approach described considers all possible subsets of nodes for camera placement. This implicitly requires generating and checking each subset, potentially storing them temporarily. Since there are 2^N possible subsets for a tree with N nodes, where N is the number of nodes in the binary tree, the auxiliary space used to represent and iterate through these subsets can grow exponentially with the input size. Therefore, the space complexity is O(2^N).

Optimal Solution

Approach

The most efficient way to solve this problem is by working from the bottom up and figuring out when a camera is absolutely necessary. We want to cover all nodes in the tree using the fewest number of cameras possible by strategically placing them.

Here's how the algorithm would work step-by-step:

Start at the very bottom of the tree, at the leaf nodes.
For each node, determine its status: either it needs a camera, it is covered by a camera, or it has a camera itself.
If a node's children need a camera, then that node absolutely needs to have a camera placed on it.
If a node has a camera, then its parent is automatically covered.
If a node's children are covered, then the node itself is considered covered.
Work your way up the tree, repeating these status checks and camera placements.
The key is to delay placing a camera until it's the only way to cover a node that needs it. This avoids unnecessary cameras.
Finally, check the root node. If it's not covered, you must place a camera there to ensure the whole tree is monitored.

Code Implementation

class TreeNode:
    def __init__(self, value=0, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

def binary_tree_cameras(root):
    number_of_cameras = 0

    def dfs(node):
        nonlocal number_of_cameras

        if not node:
            return 1

        left_state = dfs(node.left)
        right_state = dfs(node.right)

        #If either child needs coverage, install a camera.
        if left_state == 0 or right_state == 0:
            number_of_cameras += 1
            return 2

        #If either child has a camera, this node is covered.
        if left_state == 2 or right_state == 2:
            return 1

        #If both children are covered, this node needs coverage.
        return 0

    if dfs(root) == 0:
        number_of_cameras += 1

    return number_of_cameras

Big(O) Analysis

Time Complexity

O(n) – The algorithm traverses the binary tree in a bottom-up manner, processing each node exactly once during the recursion. The work done at each node involves a constant number of status checks and potential camera placement. Therefore, the time complexity is directly proportional to the number of nodes in the tree, which is represented by n. Thus, the overall time complexity is O(n).

Space Complexity

O(N) – The primary space complexity comes from the recursion stack during the depth-first traversal of the binary tree. In the worst-case scenario, such as a skewed tree, the recursive calls can go as deep as the number of nodes, N, in the tree. Each recursive call adds a frame to the stack, consuming memory for local variables and return addresses. Therefore, the space complexity is proportional to the height of the tree, which can be N in the worst case, resulting in O(N) auxiliary space.

Edge Cases

Case	How to Handle
Null or empty tree	Return 0, as an empty tree requires no cameras.
Single node tree	Return 1, as a single node needs a camera.
Two node tree (parent-child)	Return 1, placing a camera on the parent node covers both.
Tree where all nodes form a single chain/linked list	The algorithm should efficiently handle this skewed tree structure with minimal extra space overhead beyond recursion stack.
Deeply unbalanced tree leading to stack overflow in recursive implementations	Ensure the recursion depth is limited, or consider converting the solution to an iterative approach.
Tree with a large number of nodes (scalability)	The solution's time complexity must be linear to avoid timeouts (e.g., O(n) solution).
Tree with all nodes requiring a camera	The algorithm should correctly identify and cover all nodes even with maximum camera usage.
Root node with two leaf children	Placing camera on root covers all, returning 1 as minimum cameras.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the tree be empty or contain null nodes?
What is the range of values for the nodes in the binary tree?
Do all nodes have unique values, or can there be duplicates?
If no cameras are needed (e.g., a single node tree), what should the function return?
Is the tree guaranteed to be a binary tree, or could it be a more general graph structure represented in a tree-like manner?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start by considering all possible sets of nodes where we could place cameras. This means trying placing a camera on every node individually.
For each possible set of camera placements, check if every node in the tree is either directly monitored by a camera on itself, or is monitored by a camera on one of its immediate neighbors (parent or children).
If a set of camera placements does not cover all nodes, discard that set and move on to the next one.
If a set of camera placements does cover all nodes, record the number of cameras used in that set.
After trying all possible sets of camera placements, compare the number of cameras used in each valid (covering all nodes) set.
Select the set of camera placements that uses the fewest number of cameras. This is your solution.

Code Implementation

class TreeNode:
    def __init__(self, value=0, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

def binary_tree_cameras_brute_force(root):
    all_nodes = []
    def traverse(node):
        if node:
            all_nodes.append(node)
            traverse(node.left)
            traverse(node.right)

    traverse(root)
    number_of_nodes = len(all_nodes)
    minimum_cameras = float('inf')

    for i in range(1 << number_of_nodes):
        camera_placements = []
        number_of_cameras = 0

        for j in range(number_of_nodes):
            if (i >> j) & 1:
                camera_placements.append(all_nodes[j])
                number_of_cameras += 1

        # For this placement of cameras, check for coverage
        is_covered = True
        for node in all_nodes:
            if not is_node_covered(node, camera_placements):
                is_covered = False
                break

        if is_covered:
            minimum_cameras = min(minimum_cameras, number_of_cameras)

    return minimum_cameras

def is_node_covered(node, camera_placements):
    # Check if the node is covered by any camera placement.
    if node in camera_placements:
        return True

    for camera in camera_placements:
        if camera.left == node or camera.right == node or get_parent(node, camera) : 
            return True

    return False

def get_parent(node, possible_parent):
    # Helper to determine parent relationship
    if possible_parent is None: 
        return False

    if possible_parent.left == node or possible_parent.right == node:
        return True

    return False

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start at the very bottom of the tree, at the leaf nodes.
For each node, determine its status: either it needs a camera, it is covered by a camera, or it has a camera itself.
If a node's children need a camera, then that node absolutely needs to have a camera placed on it.
If a node has a camera, then its parent is automatically covered.
If a node's children are covered, then the node itself is considered covered.
Work your way up the tree, repeating these status checks and camera placements.
The key is to delay placing a camera until it's the only way to cover a node that needs it. This avoids unnecessary cameras.
Finally, check the root node. If it's not covered, you must place a camera there to ensure the whole tree is monitored.

Code Implementation

class TreeNode:
    def __init__(self, value=0, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

def binary_tree_cameras(root):
    number_of_cameras = 0

    def dfs(node):
        nonlocal number_of_cameras

        if not node:
            return 1

        left_state = dfs(node.left)
        right_state = dfs(node.right)

        #If either child needs coverage, install a camera.
        if left_state == 0 or right_state == 0:
            number_of_cameras += 1
            return 2

        #If either child has a camera, this node is covered.
        if left_state == 2 or right_state == 2:
            return 1

        #If both children are covered, this node needs coverage.
        return 0

    if dfs(root) == 0:
        number_of_cameras += 1

    return number_of_cameras

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty tree	Return 0, as an empty tree requires no cameras.
Single node tree	Return 1, as a single node needs a camera.
Two node tree (parent-child)	Return 1, placing a camera on the parent node covers both.
Tree where all nodes form a single chain/linked list	The algorithm should efficiently handle this skewed tree structure with minimal extra space overhead beyond recursion stack.
Deeply unbalanced tree leading to stack overflow in recursive implementations	Ensure the recursion depth is limited, or consider converting the solution to an iterative approach.
Tree with a large number of nodes (scalability)	The solution's time complexity must be linear to avoid timeouts (e.g., O(n) solution).
Tree with all nodes requiring a camera	The algorithm should correctly identify and cover all nodes even with maximum camera usage.
Root node with two leaf children	Placing camera on root covers all, returning 1 as minimum cameras.