Learn and practice the Backtracking and Subsets patterns to solve coding interview problems.

Educative-99 in JavaScript: Accelerate Your Coding Interview Prep

The optimal solution to certain problems requires examining all possibilities. If any possibility is ignored, we are unable to say with mathematical certainty that our solution is optimal. In such situations, we use a technique known as backtracking to exhaustively traverse the entire solution space and collect all the information needed to compute the optimal solution. The Subsets pattern groups together those problems where we need to systematically construct all, or, a mathematically defined subset of all, the possible subsets of a given set. Again, the technique used for the construction of such a set of subsets is, in fact, backtracking.

Go With the Flow

# Statement

A thief has discovered a new neighborhood to target, where the houses can be represented as nodes in a binary tree. The money in the house is the data of the respective node. The thief can enter the neighborhood from a house represented as `root` of the binary tree. Each house has only one parent house. The thief knows that if he robs two houses that are directly connected, the police will be notified. The thief wants to know the maximum amount of money he can steal from the houses without getting caught by the police. The thief needs your help determining the maximum amount of money he can rob without alerting the police.

**Constraints:**

- The number of nodes in the tree is in the range $[1, 500]$.
- $0 \leq$ `node.data` $\leq 10^4$

# Examples

What is the maximum amount we can rob from the following array? 

$[3, 4, 5, 2, 3, null, 1]$

What is the maximum amount we can rob from the following array?
<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo stretchy="false">[</mo><mn>3</mn><mo separator="true">,</mo><mn>4</mn><mo separator="true">,</mo><mn>5</mn><mo separator="true">,</mo><mn>2</mn><mo separator="true">,</mo><mn>3</mn><mo separator="true">,</mo><mi>n</mi><mi>u</mi><mi>l</mi><mi>l</mi><mo separator="true">,</mo><mn>1</mn><mo stretchy="false">]</mo></mrow><annotation encoding="application/x-tex">[3, 4, 5, 2, 3, null, 1]</annotation></semantics></math>[3,4,5,2,3,null,1]

What is the maximum amount we can rob from the following array? 

$[3, 2, 3, null, 3, null, 1]$

What is the maximum amount we can rob from the following array?
<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mo stretchy="false">[</mo><mn>3</mn><mo separator="true">,</mo><mn>2</mn><mo separator="true">,</mo><mn>3</mn><mo separator="true">,</mo><mi>n</mi><mi>u</mi><mi>l</mi><mi>l</mi><mo separator="true">,</mo><mn>3</mn><mo separator="true">,</mo><mi>n</mi><mi>u</mi><mi>l</mi><mi>l</mi><mo separator="true">,</mo><mn>1</mn><mo stretchy="false">]</mo></mrow><annotation encoding="application/x-tex">[3, 2, 3, null, 3, null, 1]</annotation></semantics></math>[3,2,3,null,3,null,1]

House Robber III

House Robber III

from binary_tree_node import *

class BinaryTree:
 def __init__(self, *args):
 if len(args) < 1:
 self.root = None
 elif isinstance(args[0], int):
 self.root = BinaryTreeNode(args[0])
 else:
 self.root = None
 for x in args[0]:
 self.insert(x)

 # for BST insertion
 def insert(self, node_data):
 new_node = BinaryTreeNode(node_data)
 if not self.root:
 self.root = new_node
 else:
 parent = None
 temp_pointer = self.root
 while temp_pointer:
 parent = temp_pointer
 if node_data <= temp_pointer.data:
 temp_pointer = temp_pointer.left
 else:
 temp_pointer = temp_pointer.right
 if node_data <= parent.data:
 parent.left = new_node
 else:
 parent.right = new_node

 def find_in_bst_rec(self, node, node_data):
 if not node:
 return None
 if node.data == node_data:
 return node
 elif node.data > node_data:
 return self.find_in_bst_rec(node.left, node_data)
 else:
 return self.find_in_bst_rec(node.right, node_data)

 def find_in_bst(self, node_data):
 return self.find_in_bst_rec(self.root, node_data)

 def get_sub_tree_node_count(self, node):
 if not node:
 return 0
 else:
 return 1 + self.get_sub_tree_node_count(node.left) + self.get_sub_tree_node_count(node.right)

 def get_tree_deep_copy_rec(self, node):
 if node:
 new_node = BinaryTreeNode(node.data)
 new_node.left = self.get_tree_deep_copy_rec(node.left)
 new_node.right = self.get_tree_deep_copy_rec(node.right)
 return new_node
 else:
 return None

 def get_tree_deep_copy(self):
 if not self.root:
 return None
 else:
 tree_copy = BinaryTree()
 tree_copy.root = self.get_tree_deep_copy_rec(self.root)
 return tree_copy

# Template for binary tree node

class BinaryTreeNode:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None


def bfs_tree(root):
    queue = deque([root])
    while len(queue) > 0:
        node = queue.popleft()
        for child in node.children:
            queue.append(child)

Help me debug only the function House Robber III . . I don't need the corrected code.I don't want explanations or code review. Just provide me the exact and to-the-point error that helps me run the code successfully on the test cases provided.

Please help me understand why this code failed for the last failed test case. Don't give me the corrected code. . If yes, then Just provide me the exact and to-the-point error that helps me run the code successfully on the test cases provided.

Your current task is to review only the rob function and the other function(s) invoked in it/them. Do not review extra/helper files or functions not invoked in the solution function rob. Do not review the code of data structures objects instantiated like a binary tree, linked list, trie, union find, etc., in the solution functions. Just review the solution/algorithm function of the problem. Also, Answer the following questions: Is the Backtracking pattern used? Take your time and review the code.

Please provide a brief time complexity analysis of the function rob. While calculating time complexity, include the time complexity of built-in Javascript functions invoked by the code. The Answer should be to the point and in a few lines.

Please provide a brief Space complexity analysis of the function rob. While calculating space complexity, include the space complexity of built-in Javascript functions invoked by the code. Do not include the space used for the input parameters and the space used for just returning an output(not used as a scratch space). The Answer should be to the point and in a few lines

def evaluate_results(self) -> object:
 """
 self.__inputs is the list of inputs that are passed as input to function
 self.__actual_output is a list of outputs that is produced by the learner's code
 return : self.__edu__result object
 """
 from collections import deque

 class TreeNode:
 def __init__(self, data):
 self.data = data
 self.left = None
 self.right = None

 def level_order_traversal(root):
 if not root:
 return []
 
 result = []
 q = deque([root])
 while q:
 curr = q.popleft()
 
 if curr:
 result.append(curr.data)
 q.append(curr.left)
 q.append(curr.right)
 else:
 result.append(None)

 if all(val == None for val in q):
 break 
 
 return result
 

 def create_binary_tree(level_order):
 if not level_order:
 return None
 
 root = TreeNode(level_order[0])
 q = deque([root])
 i = 1
 while i < len(level_order) and q:
 curr = q.popleft()
 
 if level_order[i] != None:
 curr.left = TreeNode(level_order[i])
 q.append(curr.left)
 i += 1
 
 if i < len(level_order) and level_order[i] != None:
 curr.right = TreeNode(level_order[i])
 q.append(curr.right)
 i += 1
 
 return root
 
 def rob_eval(root):
 return max(heist(root))
 
 def heist(root):
 if root == None: 
 return [0, 0]

 left_subtree = heist(root.left) 
 right_subtree = heist(root.right)
 include_root = root.data + left_subtree[1] + right_subtree[1] 
 exclude_root = max(left_subtree) + max(right_subtree)

 return [include_root, exclude_root] 

 # multiple outputs will be added at each index respectively for a single test_case
 expected_outputs = [] 
 inp = self.__inputs[0]
 inp_root = create_binary_tree(inp)
 expected = rob_eval(inp_root)
 expected_outputs.append(expected)
 actual1 = self.__actual_outputs[0]
 
 if json.dumps(expected) == json.dumps(actual1):
 self.__edu_result.store_result(expected_outputs, TestCaseResult.PASS)
 else:
 self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL)

 return self.__edu_result

// Definition of a binary tree node
// class TreeNode {
//   constructor(data) {
//     this.data = data;
//     this.left = null;
//     this.right = null;
//   }
// }

import {TreeNode} from './ds_v1/BinaryTree.js';

function rob(root) {
    
    // Replace this placeholder return statement with your code
    return -1;
}

export {
    rob
}

house rober iii (1).yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1 : [9, 7, 11, 1, 8, 10, 12]
    output:
    - 1 : 40

  - name: "testcase 2"
    inputs:
    - 1 : [3]
    output:
    - 1 : 3

  - name: "testcase 3"
    inputs:
    - 1 : [3,2,3,null,3,null,1]
    output:
    - 1 : 7

  - name: "testcase 4"
    inputs:
    - 1 : [8, 7, 10]
    output:
    - 1 : 17

  - name: "testcase 5"
    inputs:
    - 1 : [15, 10, 25, 5, 12, 20, 30, 3, 7, 11, 13]
    output:
    - 1 : 99   


def sum(input_1, input_2):
    return input_1+input_2

def validate_inputs(self) -> None:
 """
 self.__inputs is the list of inputs that are passed as input to function
 store the result using the self.store_result(error_msg, status)
 return : void
 """
 from collections import deque

 class TreeNode:
 def __init__(self, data):
 self.data = data
 self.left = None
 self.right = None 

 def level_order_traversal(root):
 if not root:
 return []
 
 result = []
 q = deque([root])
 while q:
 curr = q.popleft()
 
 if curr:
 result.append(curr.data)
 q.append(curr.left)
 q.append(curr.right)
 else:
 result.append(None)

 if all(val == None for val in q):
 break 
 
 return result
 
 def is_valid_level_order(level_order):
 if not level_order:
 return True # An empty list is a valid level-order traversal

 n = len(level_order)
 i = 0
 root = TreeNode(level_order[i])
 q = deque([root])
 i = 1

 while i < n and q:
 curr = q.popleft()

 if level_order[i] is not None:
 curr.left = TreeNode(level_order[i])
 q.append(curr.left)
 i += 1

 if i < n and level_order[i] is not None:
 curr.right = TreeNode(level_order[i])
 q.append(curr.right)
 i += 1
 
 # Remove None values after the last integer
 while level_order and level_order[-1] is None:
 level_order.pop()
 
 return level_order_traversal(root) == level_order

 # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1
 inp = self.__inputs[0]
 # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1
 if not all(0 <= num <= 10**4 for num in inp if num is not None):
 self.store_result ("'Node' must consist of values from 0 to 10^4 only", ValidationFunction.FAIL)
 elif (len(inp) !=0 and inp[0] is None):
 self.store_result ("Not a valid value of type TreeNode", ValidationFunction.FAIL)
 elif not (1 <= sum(1 for value in inp if value is not None) <= 500):
 self.store_result ("Binary tree is expected to have 1 <= nodes <= 10^4", ValidationFunction.FAIL)
 elif not is_valid_level_order(inp):
 self.store_result ("Not a valid value of type TreeNode", ValidationFunction.FAIL)
 else:
 self.store_result("", ValidationFunction.PASS)

Try to solve the House Robber III problem.

Backtracking

Subsets

Conclusion

House Robber III

Statement

Examples

Understand the problem

Figure it out!

Try it yourself