[Dynamic Programming] House Robber II - Medium

The House Robber II problem is a step up from the House Robber I problem. It is about getting the most money a robber can take from houses that are in a circle. The tricky part is that the first and last houses are next to each other. This means the robber can’t rob both.

To solve this problem, we can use dynamic programming. This method helps us deal with the circle of houses. It also makes sure we handle the important parts of the problem well.

In this article, we will look at the dynamic programming method for the House Robber II problem. We will explain it, show how to implement it in Java, Python, and C++. We will also talk about how to make space usage better, how to deal with special cases, and compare solutions in different programming languages. We will answer common questions about the House Robber II problem too.

Dynamic Programming House Robber II Problem Explanation
Understanding the Dynamic Programming Approach for House Robber II
Java Implementation of House Robber II Solution
Python Code for House Robber II Dynamic Programming
C++ Solution for House Robber II Problem
Optimizing Space Complexity in House Robber II
Handling Edge Cases in House Robber II
Comparative Analysis of Solutions in Different Languages
Frequently Asked Questions

If you want to learn more about dynamic programming, you can check out other articles like Dynamic Programming Fibonacci Number and Dynamic Programming Climbing Stairs.

Understanding the Dynamic Programming Approach for House Robber II

The House Robber II problem is a step up from the House Robber I problem. In this case, houses are in a circle. The main problem is that if we rob the first house, we cannot rob the last house. So, we have two situations to look at:

Rob houses from the first to the second last (do not rob the last house).
Rob houses from the second to the last (do not rob the first house).

The dynamic programming approach helps us find the most money we can rob without getting caught by the police.

Dynamic Programming Formulation

We can define dp[i] as the maximum amount of money we can rob from the first i houses. We can write the rules like this:

If we rob the house at index i: [ dp[i] = nums[i] + dp[i-2] ]
If we do not rob the house at index i: [ dp[i] = dp[i-1] ]

So, we have this rule: [ dp[i] = (dp[i-1], nums[i] + dp[i-2]) ]

Base Cases

dp[0] = nums[0] (only one house)
dp[1] = \max(nums[0], nums[1]) (choose the bigger amount from the first two houses)

Implementation Steps

We need to create a helper function that finds the maximum money we can rob using the dynamic programming approach.
We will handle both situations mentioned before and return the bigger amount from the two results.

Example Code (Java)

public class HouseRobberII {
    public int rob(int[] nums) {
        if (nums.length == 1) return nums[0];
        return Math.max(robLinear(Arrays.copyOfRange(nums, 0, nums.length - 1)),
                        robLinear(Arrays.copyOfRange(nums, 1, nums.length)));
    }

    private int robLinear(int[] nums) {
        int n = nums.length;
        if (n == 0) return 0;
        if (n == 1) return nums[0];
        
        int[] dp = new int[n];
        dp[0] = nums[0];
        dp[1] = Math.max(nums[0], nums[1]);
        
        for (int i = 2; i < n; i++) {
            dp[i] = Math.max(dp[i - 1], nums[i] + dp[i - 2]);
        }
        
        return dp[n - 1];
    }
}

Example Code (Python)

class HouseRobberII:
    def rob(self, nums):
        if len(nums) == 1:
            return nums[0]
        return max(self.rob_linear(nums[:-1]), self.rob_linear(nums[1:]))

    def rob_linear(self, nums):
        n = len(nums)
        if n == 0:
            return 0
        if n == 1:
            return nums[0]

        dp = [0] * n
        dp[0] = nums[0]
        dp[1] = max(nums[0], nums[1])

        for i in range(2, n):
            dp[i] = max(dp[i - 1], nums[i] + dp[i - 2])

        return dp[n - 1]

Example Code (C++)

class HouseRobberII {
public:
    int rob(vector<int>& nums) {
        if (nums.size() == 1) return nums[0];
        return max(robLinear(vector<int>(nums.begin(), nums.end() - 1)),
                   robLinear(vector<int>(nums.begin() + 1, nums.end())));
    }

    int robLinear(vector<int>& nums) {
        int n = nums.size();
        if (n == 0) return 0;
        if (n == 1) return nums[0];

        vector<int> dp(n);
        dp[0] = nums[0];
        dp[1] = max(nums[0], nums[1]);

        for (int i = 2; i < n; ++i) {
            dp[i] = max(dp[i - 1], nums[i] + dp[i - 2]);
        }

        return dp[n - 1];
    }
};

This method helps us find the most money we can rob while following the rules. If we want to learn more about dynamic programming, we can read articles like Dynamic Programming Fibonacci Number or Dynamic Programming House Robber I.

Java Implementation of House Robber II Solution

We can solve the House Robber II problem using dynamic programming in Java. This problem is a bit different from the normal House Robber problem. It has houses in a circle. The main challenge is to make sure we do not rob both the first and last houses at the same time.

Java Code Implementation

public class HouseRobberII {
    public int rob(int[] nums) {
        if (nums.length == 0) return 0;
        if (nums.length == 1) return nums[0];
        
        // Calculate the maximum money for two cases:
        // 1. Skip the first house
        // 2. Skip the last house
        return Math.max(robLinear(Arrays.copyOfRange(nums, 0, nums.length - 1)),
                        robLinear(Arrays.copyOfRange(nums, 1, nums.length)));
    }

    private int robLinear(int[] nums) {
        int prev1 = 0;
        int prev2 = 0;
        for (int num : nums) {
            int temp = prev1;
            prev1 = Math.max(prev2 + num, prev1);
            prev2 = temp;
        }
        return prev1;
    }

    public static void main(String[] args) {
        HouseRobberII robber = new HouseRobberII();
        int[] houses = {2, 3, 2};
        System.out.println("Maximum amount robbed: " + robber.rob(houses)); // Output: 3
    }
}

Explanation

rob Method: This method checks some special cases like no houses or just one house. Then it finds the maximum money we can rob by looking at two cases:
- Skipping the first house.
- Skipping the last house.
robLinear Method: This helper method uses dynamic programming for the normal House Robber problem:
- It has two variables, prev1 and prev2, to remember the maximum amounts robbed until the current and the last house.
- It goes through the list and updates the values based on if we rob the current house or not.

This code works well for the circle of houses and keeps a linear time of O(n) and a space of O(1).

Python Code for House Robber II Dynamic Programming

To solve the House Robber II problem using dynamic programming in Python, we need to change the way we solve House Robber I. This is because the houses are in a circle. We will look at two cases:

Rob houses from the first house to the second last house.
Rob houses from the second house to the last house.

Then we find which case gives us more money. Here is the Python code:

def rob(nums):
    if not nums:
        return 0
    if len(nums) == 1:
        return nums[0]
    
    def rob_linear(houses):
        prev, curr = 0, 0
        for amount in houses:
            prev, curr = curr, max(prev + amount, curr)
        return curr

    # Case 1: Exclude the last house
    case1 = rob_linear(nums[:-1])
    # Case 2: Exclude the first house
    case2 = rob_linear(nums[1:])
    
    return max(case1, case2)

# Example Usage
houses = [2, 3, 2]
print(rob(houses))  # Output: 3

Explanation:

The rob function checks if the list is empty or has only one house and takes care of these cases.
The rob_linear function uses the logic from House Robber I for a straight line of houses.
We call rob_linear two times. First, we leave out the last house. Second, we leave out the first house.
In the end, we return the most money we can rob from the two cases.

This way, we handle the circular setup of the problem well. We use dynamic programming ideas to make it work better.

C++ Solution for House Robber II Problem

The House Robber II problem is a version of the House Robber problem. Here, the houses are in a circle. Our goal is to find the most money we can rob without taking from two houses that are next to each other. We use dynamic programming to find the best amount of money we can get.

Here is a C++ code for the solution:

#include <vector>
#include <algorithm>

class Solution {
public:
    int rob(std::vector<int>& nums) {
        int n = nums.size();
        if (n == 0) return 0;
        if (n == 1) return nums[0];
        
        // This function helps to calculate max money for a line of houses
        auto robLinear = [&](std::vector<int>& houses) {
            int prev1 = 0, prev2 = 0;
            for (int amount : houses) {
                int temp = prev1;
                prev1 = std::max(prev1, prev2 + amount);
                prev2 = temp;
            }
            return prev1;
        };

        // Since houses are in a circle, we can choose to rob:
        // 1. From the first house to the second last house
        // 2. From the second house to the last house
        return std::max(robLinear(std::vector<int>(nums.begin(), nums.end() - 1)), 
                        robLinear(std::vector<int>(nums.begin() + 1, nums.end())));
    }
};

Explanation of the Code:

The rob function takes a vector nums. This vector shows how much money is in each house.
If there are no houses, it returns 0. If there is only one house, it returns the money in that house.
The robLinear lambda function calculates the most money we can rob from a line of houses. It uses a dynamic programming method.
The main function looks at two cases:
1. Robbing from the first house to the second last house.
2. Robbing from the second house to the last house.
It returns the bigger amount from the two cases.

This C++ solution works well for the circle of houses and makes sure we do not rob two houses next to each other. If you want to learn more about dynamic programming, you can check the Dynamic Programming - House Robber I article.

Optimizing Space Complexity in House Robber II

In the House Robber II problem, we can make space usage better while still using a dynamic programming method. The main point is that we only need to remember the last two states. We do not need to keep a whole array for all houses.

Space Optimization Strategy

Instead of using a big array dp[] with size n, we can use two simple variables prev1 and prev2. These will hold the maximum money we can rob from the last two houses.
This change makes the space usage go down from O(n) to O(1).

Implementation

Here is a Java method that shows the better way to do it:

public class HouseRobberII {
    public int rob(int[] nums) {
        int n = nums.length;
        if (n == 1) return nums[0];
        return Math.max(robLinear(Arrays.copyOfRange(nums, 0, n - 1)), 
                         robLinear(Arrays.copyOfRange(nums, 1, n)));
    }

    private int robLinear(int[] nums) {
        int prev1 = 0, prev2 = 0;
        for (int num : nums) {
            int temp = prev1;
            prev1 = Math.max(prev1, prev2 + num);
            prev2 = temp;
        }
        return prev1;
    }
}

Python Version

This is how we can do the same space saving in Python:

def rob(nums):
    def rob_linear(nums):
        prev1, prev2 = 0, 0
        for num in nums:
            prev1, prev2 = max(prev1, prev2 + num), prev1
        return prev1

    n = len(nums)
    if n == 1: return nums[0]
    return max(rob_linear(nums[:-1]), rob_linear(nums[1:]))

C++ Implementation

The C++ version also shows the same space saving:

class Solution {
public:
    int rob(vector<int>& nums) {
        int n = nums.size();
        if (n == 1) return nums[0];
        return max(robLinear(vector<int>(nums.begin(), nums.end() - 1)),
                   robLinear(vector<int>(nums.begin() + 1, nums.end())));
    }

    int robLinear(vector<int>& nums) {
        int prev1 = 0, prev2 = 0;
        for (int num : nums) {
            int temp = prev1;
            prev1 = max(prev1, prev2 + num);
            prev2 = temp;
        }
        return prev1;
    }
};

By using this O(1) space method, we cut down the memory use a lot. We still solve the House Robber II problem well. This change helps a lot when we have big inputs. It keeps our solution quick and able to handle more data.

Handling Edge Cases in House Robber II

When we work on the House Robber II problem with dynamic programming, we must think about different edge cases. This way, our solution will be strong. Here are some cases we should look at:

Empty Input: If the list of houses is empty, we cannot rob anything. So, the maximum amount we can rob is 0.
```
if (nums.length == 0) return 0;
```
Single House: If there is only one house, we can just rob that house.
```
if (nums.length == 1) return nums[0];
```
Two Houses: If there are two houses, we choose the one that has more value.
```
if (nums.length == 2) return Math.max(nums[0], nums[1]);
```

More than Two Houses: Since the houses are in a circle, we need to think in two ways:

Rob houses from index 0 to n-2.
Rob houses from index 1 to n-1.

def rob(nums):
    if not nums:
        return 0
    if len(nums) == 1:
        return nums[0]
    if len(nums) == 2:
        return max(nums[0], nums[1])

    def simple_rob(houses):
        prev, curr = 0, 0
        for amount in houses:
            prev, curr = curr, max(prev + amount, curr)
        return curr

    return max(simple_rob(nums[:-1]), simple_rob(nums[1:]))

Negative Values: If any house has a negative value, we should not count it. The algorithm treats it as 0 when we calculate the maximum amount.
All Houses with Zero Value: If all houses have a value of 0, we still get 0 as the result.
Large Input Size: We must make sure the algorithm works well with large input sizes. It should run in O(n) time and O(1) space to be the best.

By thinking about these edge cases, we make our House Robber II solution better and more correct. The dynamic programming method will help us get the most money while considering the circle of houses and stopping us from robbing two houses next to each other.

Comparative Analysis of Solutions in Different Languages

In this part, we look at solutions for the House Robber II problem. We will see how it is done in three programming languages: Java, Python, and C++. Each solution uses dynamic programming to solve the problem well. We also see how each language works.

Java Implementation

In Java, we use an array. This array keeps track of the most money we can rob up to each house. We need to think about the circular layout of the houses. We can rob from the first house to the second last house or from the second house to the last house.

public class HouseRobberII {
    public int rob(int[] nums) {
        if (nums.length == 1) return nums[0];
        return Math.max(robLinear(Arrays.copyOfRange(nums, 0, nums.length - 1)),
                        robLinear(Arrays.copyOfRange(nums, 1, nums.length)));
    }

    private int robLinear(int[] nums) {
        int rob1 = 0, rob2 = 0;
        for (int n : nums) {
            int temp = Math.max(rob1 + n, rob2);
            rob1 = rob2;
            rob2 = temp;
        }
        return rob2;
    }
}

Python Implementation

The Python solution is almost the same as Java. It uses list slicing for the two cases. It also has a helper function for the robbery logic.

class Solution:
    def rob(self, nums):
        if len(nums) == 1:
            return nums[0]
        return max(self.rob_linear(nums[:-1]), self.rob_linear(nums[1:]))

    def rob_linear(self, nums):
        rob1, rob2 = 0, 0
        for n in nums:
            temp = max(rob1 + n, rob2)
            rob1 = rob2
            rob2 = temp
        return rob2

C++ Implementation

In C++, we do it in a similar way. We use vectors for storage and create a function for the linear robbery logic.

class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.size() == 1) return nums[0];
        return max(robLinear(vector<int>(nums.begin(), nums.end() - 1)),
                   robLinear(vector<int>(nums.begin() + 1, nums.end())));
    }

    int robLinear(vector<int>& nums) {
        int rob1 = 0, rob2 = 0;
        for (int n : nums) {
            int temp = max(rob1 + n, rob2);
            rob1 = rob2;
            rob2 = temp;
        }
        return rob2;
    }
};

Performance Comparison

Time Complexity: All three solutions have a time complexity of O(n). Here n is the number of houses.
Space Complexity: Each solution reduces space complexity to O(1). It only keeps two variables for the last two maximum values. But slicing in Python and Java may need extra space for a short time.

Conclusion

We see that the syntax and specific features are different in Java, Python, and C++. But the main dynamic programming method stays the same. Each language can solve the House Robber II problem well. This shows how flexible dynamic programming is in different programming languages. For more about similar dynamic programming problems, you can check Dynamic Programming - House Robber I.

Frequently Asked Questions

1. What is the House Robber II problem in Dynamic Programming?

The House Robber II problem is a twist on the classic House Robber problem. In this problem, the houses are in a circle. This means the first house is next to the last house. This setup creates special rules. We need to find new ways to use dynamic programming to get the most money without getting caught by the police. Understanding this problem helps us learn more about dynamic programming.

2. How does the dynamic programming approach work for House Robber II?

The dynamic programming method for House Robber II breaks the problem into smaller parts. We define states that show the most money we can take from certain houses. We build our solutions step by step. The main idea is to look at two situations. One is to rob from the first house and not the last one. The other is to rob from the last house and not the first one. This way, we make the best choices at each step.

3. Can you provide a Java implementation of the House Robber II problem?

Sure! Here is a simple Java code for the House Robber II problem:

public int rob(int[] nums) {
    if(nums.length == 1) return nums[0];
    return Math.max(robLinear(Arrays.copyOfRange(nums, 0, nums.length - 1)), 
                      robLinear(Arrays.copyOfRange(nums, 1, nums.length)));
}

private int robLinear(int[] nums) {
    int prev1 = 0, prev2 = 0;
    for (int num : nums) {
        int temp = prev1;
        prev1 = Math.max(prev2 + num, prev1);
        prev2 = temp;
    }
    return prev1;
}

This code works well with the circle rule. It uses the normal House Robber logic on two different parts.

4. Is there a Python code example for House Robber II?

Yes! Here is a Python code example for the House Robber II problem:

def rob(nums):
    if len(nums) == 1:
        return nums[0]

    def rob_linear(houses):
        prev1, prev2 = 0, 0
        for num in houses:
            temp = prev1
            prev1 = max(prev2 + num, prev1)
            prev2 = temp
        return prev1

    return max(rob_linear(nums[:-1]), rob_linear(nums[1:]))

This Python code uses a helper function. It calculates the most money we can rob while taking care of the circle of houses.

5. How do you optimize space complexity in the House Robber II solution?

To make space use better in the House Robber II problem, we can use less space for our dynamic programming states. Instead of keeping an array for past results, we only need two variables. These will keep track of the most money robbed from the last two houses. This way, we cut down space use from O(n) to O(1). This makes our solution faster.

For more information about dynamic programming, we can read more articles. Some good ones are Dynamic Programming: House Robber I and Dynamic Programming: Fibonacci Number.