feat(algorithms, dynamic-programming): house robber 3

Author: BrianLusina

algorithms/dynamic_programming/house_robber/README.md

@@ -26,3 +26,158 @@ Total amount you can rob = 2 + 9 + 1 = 12.
 
 - Array
 - Dynamic Programming
+
+---
+# House Robber III
+
+The thief has found himself a new place for his thievery again. There is only one entrance to this area, called root.
+
+Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that all houses
+in this place form a binary tree. It will automatically contact the police if two directly-linked houses were broken
+into on the same night.
+
+Given the root of the binary tree, return the maximum amount of money the thief can rob without alerting the police.
+
+## Constraints
+
+- The number of nodes in the tree is in the range [1, 10^4].
+- 0 <= Node.val <= 10^4
+
+## Examples
+
+![Example 1](./images/examples/house_robber_3_example_1.png)
+![Example 1.1](./images/examples/house_robber_3_example_1.1.png)
+![Example 1.2](./images/examples/house_robber_3_example_1.2.png)
+![Example 1.3](./images/examples/house_robber_3_example_1.3.png)
+![Example 1.4](./images/examples/house_robber_3_example_1.4.png)
+![Example 2](./images/examples/house_robber_3_example_2.png)
+
+![Example 3](./images/examples/house_robber_3_example_3.png)
+```text
+Input: root = [3,2,3,null,3,null,1]
+Output: 7
+Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
+```
+
+![Example 4](./images/examples/house_robber_3_example_4.png)
+```text
+Input: root = [3,4,5,1,3,null,1]
+Output: 9
+Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.
+```
+
+## Topics
+
+- Dynamic Programming
+- Tree
+- Depth-First Search
+- Binary Tree
+
+## Solution(s)
+
+1. [Recursion](#recursion)
+2. [Dynamic Programming(Memoization) - Top Down Approach](#dynamic-programmingmemoization-top-down-approach)
+3. [Dynamic Programming(Optimal)](#dynamic-programming-optimal--bottom-up-approach)
+
+### Recursion
+
+This is a tree version of the classic house robber problem. At each node, we have two choices: rob it or skip it. If we
+rob the current node, we cannot rob its immediate children, so we must skip to the grandchildren. If we skip the current
+node, we can consider robbing its children. We take the maximum of these two options.
+
+#### Algorithm
+
+- If the node is null, return 0.
+- Calculate the value if we rob the current node: add the node's value plus the result from its grandchildren 
+  (left.left, left.right, right.left, right.right).
+- Calculate the value if we skip the current node: add the results from robbing the left and right children.
+- Return the maximum of these two values.
+
+#### Time Complexity
+
+O(2^n)
+
+#### Space complexity 
+
+O(n) for recursion stack.
+
+### Dynamic Programming(Memoization) Top-Down Approach
+
+The recursive solution recomputes results for the same nodes multiple times. By storing computed results in a cache
+(hash map), we avoid redundant work. Each node is processed at most once, significantly improving efficiency.
+
+#### Algorithm
+
+- Create a cache (hash map) to store computed results for each node.
+- Define a recursive function that checks the cache before computing.
+- If the node is in the cache, return the cached result.
+- Otherwise, compute the result using the same logic as the basic recursion: max of robbing current node
+  (plus grandchildren) vs skipping current node (plus children).
+- Store the result in the cache and return it.
+
+#### Complexity Analysis
+
+- Time complexity: O(n)
+- Space complexity: O(n)
+
+### Dynamic Programming (Optimal)- Bottom-Up Approach
+
+Instead of caching all nodes, we can return two values from each subtree: the maximum if we rob this node, and the
+maximum if we skip it. This eliminates the need for a hash map. For each node, "with root" equals the node value plus
+the "without" values of both children. "Without root" equals the sum of the maximum values (either with or without) from
+both children.
+
+#### Algorithm
+
+- Define a recursive function that returns a pair: [maxWithNode, maxWithoutNode].
+- For a null node, return [0, 0].
+- Recursively get the pairs for left and right children.
+- Calculate withRoot as the node's value plus leftPair[1] plus rightPair[1] (children must be skipped).
+- Calculate withoutRoot as max(leftPair) plus max(rightPair) (children can be robbed or skipped).
+- Return [include_root, exclude_root].
+- The final answer is the maximum of the two values returned for the root.
+
+![Solution 1](./images/solutions/house_robber_iii_solution_dp_bottom_up_1.png)
+![Solution 2](./images/solutions/house_robber_iii_solution_dp_bottom_up_2.png)
+![Solution 3](./images/solutions/house_robber_iii_solution_dp_bottom_up_3.png)
+![Solution 4](./images/solutions/house_robber_iii_solution_dp_bottom_up_4.png)
+![Solution 5](./images/solutions/house_robber_iii_solution_dp_bottom_up_5.png)
+![Solution 6](./images/solutions/house_robber_iii_solution_dp_bottom_up_6.png)
+![Solution 7](./images/solutions/house_robber_iii_solution_dp_bottom_up_7.png)
+![Solution 8](./images/solutions/house_robber_iii_solution_dp_bottom_up_8.png)
+![Solution 9](./images/solutions/house_robber_iii_solution_dp_bottom_up_9.png)
+![Solution 10](./images/solutions/house_robber_iii_solution_dp_bottom_up_10.png)
+![Solution 11](./images/solutions/house_robber_iii_solution_dp_bottom_up_11.png)
+![Solution 12](./images/solutions/house_robber_iii_solution_dp_bottom_up_12.png)
+
+#### Complexity Analysis O(n)
+
+The time complexity of this solution is O(n), where n is the number of nodes in the tree, since we visit all nodes once.
+
+##### Space complexity: O(n) 
+
+The space complexity of this solution is O(n), since the maximum depth of the recursive call tree is the height of the
+tree. Which is n in the worst case, and each call stores its data on the stack.
+
+### Common Pitfalls
+
+#### Confusing "Skip to Grandchildren" with "Must Rob Grandchildren"
+
+When robbing the current node, you cannot rob its immediate children, so you recursively consider the grandchildren.
+However, a common mistake is thinking you must rob the grandchildren. In reality, for each grandchild, you still have
+the choice to rob it or skip it. The recursive call on grandchildren will make this decision optimally. The constraint
+only prevents robbing adjacent nodes (parent-child), not skipping multiple levels.
+
+#### Not Handling Null Children Properly
+
+When calculating the value of robbing the current node, you need to add values from grandchildren. If a child is null,
+accessing child.left or child.right will cause a null pointer error. Always check if the left or right child exists
+before attempting to access their children. A null child contributes 0 to the total, and its non-existent children also
+contribute 0.
+
+#### Forgetting to Take Maximum in the Final Answer
+
+The optimal DFS solution returns a pair [withRoot, withoutRoot] representing the maximum money if we rob or skip the
+current node. At the root level, the final answer is the maximum of these two values, not just one of them. Forgetting
+to take this maximum and returning only withRoot or withoutRoot will give an incorrect result whenever the optimal
+strategy at the root differs from what you returned.

algorithms/dynamic_programming/house_robber/__init__.py

@@ -1,4 +1,6 @@
-from typing import List
+from typing import List, Optional, Dict
+
+from datastructures.trees.binary.node import BinaryTreeNode
 
 
 def rob(nums: List[int]) -> int:
@@ -10,3 +12,69 @@ def rob(nums: List[int]) -> int:
         current, previous = max(previous + house, current), current
 
     return current
+
+
+def rob_iii_recursion(root: Optional[BinaryTreeNode]) -> int:
+    if not root:
+        return 0
+    res = root.data
+
+    if root.left:
+        res += rob_iii_recursion(root.left.left) + rob_iii_recursion(root.left.right)
+    if root.right:
+        res += rob_iii_recursion(root.right.left) + rob_iii_recursion(root.right.right)
+
+    res = max(res, rob_iii_recursion(root.left) + rob_iii_recursion(root.right))
+    return res
+
+
+def rob_iii_dynamic_programming_top_down(root: Optional[BinaryTreeNode]) -> int:
+    if not root:
+        return 0
+
+    cache: Dict[BinaryTreeNode, int] = {}
+
+    def dfs(node: Optional[BinaryTreeNode]) -> int:
+        if not node:
+            return 0
+
+        if node in cache:
+            return cache[node]
+
+        cache[node] = node.data
+
+        if node.left:
+            cache[node] += dfs(node.left.left) + dfs(node.left.right)
+        if node.right:
+            cache[node] += dfs(node.right.left) + dfs(node.right.right)
+
+        cache[node] = max(cache[node], dfs(node.left) + dfs(node.right))
+        return cache[node]
+
+    return dfs(root)
+
+
+def rob_iii_dynamic_programming_bottom_up(root: Optional[BinaryTreeNode]) -> int:
+    if not root:
+        return 0
+
+    def dfs(node: Optional[BinaryTreeNode]) -> List[int]:
+        # Empty tree case
+        if not node:
+            return [0, 0]
+
+        # Recursively calculating the maximum amount that can be robbed from the left subtree of the root
+        left_subtree = dfs(node.left)
+        # Recursively calculating the maximum amount that can be robbed from the right subtree of the root
+        right_subtree = dfs(node.right)
+
+        # include_root contains the maximum amount of money that can be robbed with the parent node included
+        include_root = node.data + left_subtree[1] + right_subtree[1]
+
+        # exclude_root contains the maximum amount of money that can be robbed with the parent node excluded
+        exclude_root = max(left_subtree) + max(right_subtree)
+
+        return [include_root, exclude_root]
+
+    # Returns maximum value from the pair: [include_root, exclude_root]
+    return max(dfs(root))