House Robber III (Revisited) 🎯

Welcome back, fellow learners! Today, we're going to dive into an exciting problem called House Robber III. This problem is a variation of the classic House Robber problem, and it's a great way to practice your understanding of dynamic programming and recursion.

Introduction 📝

The House Robber III problem is a variation of the House Robber problem, where we aim to find the maximum amount of money that a thief can steal from a list of houses. However, this time, we're not limited to just one street. Let's see how it works!

Problem Statement 📝

Given an array arr representing the money in each house, and an integer k representing the number of streets. For each street, there is a single house at the end. The thief can only rob houses that are on the same street. The problem is to determine the maximum amount of money the thief can steal.

Understanding the Problem 💡

To understand the problem better, let's break it down:

Each street consists of one or more houses.
The thief can only rob houses on the same street.
The goal is to find the maximum amount of money the thief can steal.

Solution Approach 💡

We'll solve this problem using dynamic programming and recursion. The idea is to calculate the maximum amount of money that can be stolen from each street, and then choose the street with the highest total.

Algorithm 💡

Base Case: If k is 1 (i.e., only one street), we solve the traditional House Robber problem.
Recursive Step: For each street i (from 1 to k), we calculate max(robbed[i-1], robbed[i] + robbed[i+1]), where robbed[i] is the maximum amount of money that can be stolen from the houses on street i.
Return the maximum value calculated from all streets.

Code Example 📝

Here's a Python example of how you can implement the House Robber III algorithm:

python

def house_robber_iii(arr, k):
    n = len(arr)

    # Base Case: If k is 1, we solve the traditional House Robber problem
    if k == 1:
        return max(arr)

    # Define a table to store the maximum amount of money that can be stolen from each street
    robbed = [0] * (n + 2)

    # Calculate the maximum amount of money that can be stolen from each street
    for i in range(1, k+1):
        robbed[i] = max(robbed[i-1], arr[i-1] + (i < n and robbed[i+1]) or 0)

    # Return the maximum value calculated from all streets
    return max(robbed)

Quiz 🎯

That's all for today! Practice this algorithm, and you'll be well on your way to mastering dynamic programming and recursion. Stay tuned for more exciting lessons! 🚀