House Robber
The problem presents a scenario where you must choose to rob houses, but you can’t rob adjacent houses. This makes it a dynamic programming problem where you can build a solution by making optimal choices at each stage.
Here’s a Python code snippet that demonstrates a way to solve this problem using dynamic programming:


Explanation:
 Base Cases: If there is only one house, the answer is the money in that house. We also initialize the first two elements of the
dp
array with the value of the first house and the maximum of the first and second houses.  Dynamic Programming Loop: We loop from the third house onwards and calculate the maximum money for robbing up to the current house. We can either rob the current house and add its money to the solution for two houses back (
dp[i  2] + nums[i]
) or skip the current house and stick with the solution for the previous house (dp[i  1]
). We take the maximum of these two options for each house.  Final Result: The last element in the
dp
array contains the maximum money that can be robbed without alerting the police.
The time complexity of this solution is ( O(n) ), where ( n ) is the number of houses, and the space complexity is also ( O(n) ).
Identifying Problem Isomorphism
“House Robber” is a dynamic programming problem where you need to find the maximum amount of money you can rob without alerting the police, given that alerting any house directly alerts its adjacent houses.
An approximate isomorphic problem to this is “Climbing Stairs”. It’s a simpler problem where you have to find the number of distinct ways you can climb to the top of a staircase, given that you can either take one step or two steps at a time. It’s related to the “House Robber” problem because both require determining the best decision at each step considering the previous steps’ decisions.
A more complex is “House Robber II” (problem 213), where houses are arranged in a circle. This introduces additional considerations compared to the original “House Robber” problem.
Another problem of increased complexity is “House Robber III” (problem 337), which involves a binary tree layout of houses, significantly increasing the problem’s complexity.
So, in terms of complexity, we have:
 “Climbing Stairs” (problem 70)
 “House Robber”
 “House Robber II” (problem 213)
 “House Robber III” (problem 337)
These problems, while not strictly isomorphic, share the concept of making optimal decisions at each step based on previous steps. Understanding the strategies used to solve these problems can provide insight into dynamic programming and decisionmaking processes.


Language Agnostic Coding Drills
Here’s how the above code can be broken down into smaller units of learning:
Basic data types and structures: Understanding variables, lists, and list indexing in a programming language.
Control flow: Knowledge of ‘if’ conditions, and how to use them to control the flow of your program based on certain conditions.
Loops: Understanding ‘for’ loops, how to iterate over a range of numbers, and how to use the loop variable inside the loop.
Functions: Understanding how to define a function, how to pass parameters to a function, and how to return a value from a function.
List operations: Knowledge of basic list operations like finding the length of a list, accessing and modifying elements in a list, and initializing a list with a repeated value.
Dynamic programming: The concept of dynamic programming (DP), where you solve complex problems by breaking them down into simpler subproblems and storing the results of these subproblems to avoid solving the same subproblem multiple times.
Problemsolving and algorithm design: The ability to devise an algorithm to solve a given problem, in this case, the ‘house robber’ problem. This requires understanding the problem statement, formulating a strategy to solve the problem, and translating that strategy into code.
In terms of increasing level of difficulty, the topics could be ordered like this: Basic data types and structures > Control flow > Loops > Functions > List operations > Problemsolving and algorithm design > Dynamic programming. Each of these topics can be practiced and mastered separately, and then combined to understand and implement the given code.
Targeted Drills in Python
Basic data types and structures:
Declare a list and populate it with some integer values. Use indexing to access and print specific elements of the list.
1 2 3
nums = [10, 20, 30, 40, 50] print(nums[0]) # Output: 10 print(nums[1]) # Output: 50
Control flow:
Write a Python program that prints whether a number is positive or negative.
1 2 3 4 5
num = 5 if num > 0: print("Positive") else: print("Negative")
Loops:
Iterate over the list declared above, and print each element.
1 2
for num in nums: print(num)
Functions:
Define a function that takes a number as an argument and returns its square.
1 2 3 4
def square(n): return n * n print(square(5)) # Output: 25
List operations:
Initialize a list with a repeated value, and then modify an element in the list.
1 2 3
dp = [0] * 5 # Initialize a list with five zeroes dp[0] = 10 # Modify the first element print(dp) # Output: [10, 0, 0, 0, 0]
Dynamic programming:
Write a Python function to solve the Fibonacci sequence using dynamic programming.
1 2 3 4 5 6 7
def fibonacci(n): fib = [0, 1] + [0] * (n  1) for i in range(2, n + 1): fib[i] = fib[i  1] + fib[i  2] return fib[n] print(fibonacci(10)) # Output: 55
Problemsolving and algorithm design:
Solve a simpler version of the house robber problem where the houses are in a line (not a circle), so you can’t rob the first and last house at the same time.
1 2 3 4 5 6 7 8 9 10 11 12
def rob(nums): if not nums: return 0 if len(nums) <= 2: return max(nums) dp = [0] * len(nums) dp[0], dp[1] = nums[0], max(nums[0], nums[1]) for i in range(2, len(nums)): dp[i] = max(dp[i  1], dp[i  2] + nums[i]) return dp[1] print(rob([1, 2, 3, 1])) # Output: 4
These drills target the key concepts involved in understanding and writing the provided code. By working through them, you can build up to fully understanding the original code.