[Dynamic Programming] House Robber I - Easy

Dynamic Programming is a strong method we use to solve optimization problems. We break these problems into smaller, easier problems. The House Robber I problem is a well-known example. In this problem, we need to find out the most money we can rob from a row of houses. The rule is that we cannot rob two houses that are next to each other. We use the ideas of dynamic programming to create a good plan that helps us get the most money while following this rule.

In this article, we will look closely at the House Robber I problem. We will start by explaining the problem. Then, we will talk about the main ideas of dynamic programming. We will discuss the optimal substructure and overlapping subproblems that are part of this problem. After that, we will show how to implement dynamic programming with tabulation in Java, Python, and C++. We will also talk about ways to use less space in these languages. Finally, we will answer some common questions to help you understand better.

Diving into Dynamic Programming for House Robber I Easy
Understanding the Problem Statement for House Robber I
Optimal Substructure and Overlapping Subproblems in House Robber I
Dynamic Programming Approach with Tabulation in Java
Dynamic Programming Approach with Tabulation in Python
Dynamic Programming Approach with Tabulation in C++
Space Optimized Approach for House Robber I in Java
Space Optimized Approach for House Robber I in Python
Space Optimized Approach for House Robber I in C++
Frequently Asked Questions

If we want to learn more about dynamic programming, we can look at topics like the Dynamic Programming Fibonacci Number, Dynamic Programming Climbing Stairs, or the Dynamic Programming Maximum Subarray (Kadane’s Algorithm). These links will help us understand more about how we can use dynamic programming in different ways.

Understanding the Problem Statement for House Robber I

The House Robber I problem is a well-known challenge in dynamic programming. We can explain the problem like this:

We are a robber planning to steal money from houses on a street. Each house has some money inside. But we must be careful. If we rob two houses next to each other, the security will call the police. Our goal is to get the most money we can without setting off any alarms.

Problem Specification:

We get an array nums. In this array, nums[i] shows how much money is in the i-th house.
We cannot rob two houses that are next to each other.
We need to give back the most money we can rob at night without calling the police.

Example:

Input: nums = [1, 2, 3, 1]
Output: 4
Explanation: If we rob houses 1 and 3, we get the most money which is 4.

Constraints:

The length of nums is from 1 to 100.
Each number in nums is a non-negative whole number.

This problem has a good structure and it shows overlapping problems. So, it is a good fit for dynamic programming. If we want to learn more about dynamic programming, we can check articles like Dynamic Programming - Fibonacci Number or Dynamic Programming - Climbing Stairs.

Optimal Substructure and Overlapping Subproblems in House Robber I

In the House Robber I problem, we need to find the most money a robber can take from a line of houses. We cannot rob two adjacent houses because of alarms. This problem shows both optimal substructure and overlapping subproblems. These features make it a good fit for dynamic programming.

Optimal Substructure

The optimal substructure property means we can build the best solution from the best solutions of smaller parts. For the House Robber I problem, we can think about each house like this:

If the robber robs house i, he can collect money from house i plus the most money from houses 0 to i-2.
If the robber skips house i, he can only take the most money from houses 0 to i-1.

We can write this as:

maxAmount(i) = max(maxAmount(i-1), nums[i] + maxAmount(i-2))

Here, nums[i] is the money in house i. The base cases are:

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

Overlapping Subproblems

The overlapping subproblems property tells us that we can break the problem into smaller parts that we use many times. In the House Robber I problem, when we calculate maxAmount(i) for different i, we will need to find maxAmount(i-1) and maxAmount(i-2) again and again.

Dynamic programming helps us save these results to avoid doing the same work over and over. We can do this using a top-down approach with memoization or a bottom-up approach with tabulation.

Example

Let’s look at the house values: nums = [2, 7, 9, 3, 1]. We can show the optimal substructure like this:

For house 0: maxAmount(0) = 2
For house 1: maxAmount(1) = max(2, 7) = 7
For house 2: maxAmount(2) = max(7, 9 + 2) = 11
For house 3: maxAmount(3) = max(11, 3 + 7) = 11
For house 4: maxAmount(4) = max(11, 1 + 11) = 12

So, the maximum money the robber can take is 12.

By using the ideas of optimal substructure and overlapping subproblems, we can solve the House Robber I problem quickly with dynamic programming. For more about dynamic programming problems, check out Dynamic Programming - Fibonacci Number.

Dynamic Programming Approach with Tabulation in Java

The tabulation method in dynamic programming is a bottom-up way to solve problems. We build a table step by step. For the House Robber I problem, we will make an array to keep the maximum money we can rob up to each house.

Code Implementation

Here is how we can use the tabulation approach for the House Robber I problem in Java:

public class HouseRobber {
    public int rob(int[] nums) {
        if (nums.length == 0) return 0;
        if (nums.length == 1) return nums[0];

        int n = nums.length;
        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], dp[i - 2] + nums[i]);
        }

        return dp[n - 1];
    }

    public static void main(String[] args) {
        HouseRobber robber = new HouseRobber();
        int[] houses = {2, 7, 9, 3, 1};
        System.out.println("Maximum amount robbed: " + robber.rob(houses)); // Output: 12
    }
}

Explanation of the Code

Initialization: We first take care of special cases. This is when there are no houses or just one house.
Dynamic Programming Array: We create an array dp. In this array, dp[i] shows the maximum money we can rob from the first i+1 houses.
Recurrence Relation: For each house starting from the third one, we choose to rob the current house or not. If we rob it, we add its value to the money robbed from houses up to i-2. If we skip it, we keep the value from i-1.
Result: The last part of the dp array has the most money we can rob.

This method works in O(n) time and uses O(n) space. The tabulation method is good for problems with overlapping parts and good structure. So, it fits well for the House Robber I problem.

Dynamic Programming Approach with Tabulation in Python

We can use the dynamic programming approach with tabulation for the House Robber I problem. This method builds a table to keep track of the most money we can rob from each house. It helps us avoid repeating calculations for the same problems by storing results step by step.

Problem Statement

We have a list of non-negative numbers. Each number shows how much money is in each house. We cannot rob two houses that are next to each other. Our goal is to find out the most money we can rob without getting caught by the police.

Tabulation Approach

Initialization: We create an array dp. In this array, dp[i] shows the most money we can rob from the first i houses.
State Transition: For each house i, we have two choices:
- Rob house i: We add the money from house i to the maximum money we can rob from houses up to i-2 (which is dp[i-2] + nums[i]).
- Skip house i: We take the maximum amount we can rob from houses up to i-1 (which is dp[i-1]).
- So, we can say:
```
dp[i] = max(dp[i-1], dp[i-2] + nums[i])
```
Base Cases:
- If there are no houses, the most money is 0.
- If there is one house, the most money is the amount in that house.

Python Implementation

def rob(nums):
    if not nums:
        return 0
    n = len(nums)
    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], dp[i-2] + nums[i])
    
    return dp[-1]

Example

If we have the list nums = [2, 7, 9, 3, 1], the rob function will give us 12. This is the most money we can rob by taking money from house 1, 2, and 4 (which is 2 + 9 + 1).

This dynamic programming method is good because it has a time complexity of O(n) and a space complexity of O(n) because we store the dp array. To use less space, we can change the space complexity to O(1) by only keeping track of the last two values we computed.

Dynamic Programming Approach with Tabulation in C++

We can use the tabulation method for the House Robber I problem. This method creates a table to store the most money we can rob from each house. The main idea is to choose at each house if we will rob it or skip it. We must make sure we do not rob two houses next to each other.

C++ Implementation

Here is how we can write the tabulation approach in C++:

#include <vector>
#include <iostream>

using namespace std;

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

    // Create a DP array
    vector<int> dp(n);
    dp[0] = nums[0];
    dp[1] = max(nums[0], nums[1]);

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

    return dp[n - 1]; // Maximum money that can be robbed
}

int main() {
    vector<int> houses = {2, 7, 9, 3, 1};
    cout << "Maximum amount robbed: " << rob(houses) << endl;
    return 0;
}

Explanation of the Code

Input: We have a vector nums which shows how much money is in each house.
Base Cases:
- If there are no houses, we return 0.
- If there is only one house, we return its money.
DP Array Initialization:
- dp[0] means the maximum money from the first house.
- dp[1] is the maximum money we can get from the first and second house.
DP Array Filling:
- For each house from the third one, we calculate the maximum money. We either skip this house or rob it and add its money to what we got from two houses before.
Return Value: The last value in the DP array gives the most money we can rob.

This method has time complexity O(n) and space complexity O(n) because of the DP array. For better ways to optimize space, we can see the space-optimized methods below.

For more on dynamic programming techniques, we can check the Dynamic Programming Fibonacci Number article.

Space Optimized Approach for House Robber I in Java

We can make the Space Optimized Approach for the House Robber I problem better. This method cuts down the space we need from O(n) to O(1). We only keep the last two biggest amounts we robbed. We do not need to save all amounts from each house. This method is good for solving the problem and uses less memory.

Code Implementation in Java

public class HouseRobber {
    public int rob(int[] nums) {
        if (nums.length == 0) return 0;
        if (nums.length == 1) return nums[0];

        int prev1 = 0; // Max amount robbed from previous house
        int prev2 = 0; // Max amount robbed from house before previous

        for (int num : nums) {
            int temp = prev1; // Store the value of prev1
            prev1 = Math.max(prev1, prev2 + num); // Max amount if robbing current house
            prev2 = temp; // Update prev2 to be the old prev1
        }
        return prev1; // Maximum amount robbed
    }

    public static void main(String[] args) {
        HouseRobber robber = new HouseRobber();
        int[] houses = {2, 7, 9, 3, 1};
        System.out.println("Maximum amount that can be robbed: " + robber.rob(houses));
    }
}

Explanation of the Code

Initialization: We start with two variables, prev1 and prev2. They save the biggest amounts robbed from the last two houses.
Iteration: For each house, we check which value is bigger:
- Not robbing the current house (we keep prev1).
- Robbing the current house (we add its value to prev2).
Updating Values: After we check, we update prev1 with the new biggest amount. Then, we set prev2 to the old value of prev1.
Result: At the end, prev1 has the biggest amount we can rob from the houses.

This method uses constant space and makes sure the algorithm runs in O(n) time. It is good for large inputs. If you want to learn more about dynamic programming, you can check articles on Dynamic Programming: Fibonacci Number and Dynamic Programming: Climbing Stairs.

Space Optimized Approach for House Robber I in Python

The Space Optimized Approach for the House Robber I problem helps us save space. It cuts down space usage from O(n) to O(1) while keeping the time needed at O(n). We do this by only tracking the last two biggest amounts robbed instead of keeping a whole array of results.

Implementation

In this method, we use two variables. They hold the most money that we can rob up to the current house and the house before it. The way we decide to rob or skip each house stays the same.

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

    prev1, prev2 = 0, 0

    for num in nums:
        temp = prev1
        prev1 = max(prev2 + num, prev1)
        prev2 = temp

    return prev1

Explanation

Initialization: We start with prev1 and prev2 at 0. This shows the most money we can rob up to the last house and the house before that.
Iteration: We go through each house’s money:
- We keep the value of prev1 in a temporary variable.
- We update prev1 to be the bigger amount. This is either robbing the current house (adding num to prev2) or not robbing it (keeping prev1).
- We set prev2 to the old value of prev1 (that we saved in temp).
Result: After we finish going through the list, prev1 has the most money that we can rob.

Example

For the input nums = [2, 7, 9, 3, 1], the function will give back 12. The best choice is to rob houses with amounts 2, 9, and 1.

This space-optimized way works well for the House Robber I problem while keeping the good performance of dynamic programming.

For more about dynamic programming, you can look at related topics like Dynamic Programming Fibonacci Number or Dynamic Programming Climbing Stairs.

Space Optimized Approach for House Robber I in C++

To solve the House Robber I problem with a space-optimized way in C++, we can use just two variables. We do not need a whole array for dynamic programming. This lowers the space use from O(n) to O(1).

Problem Recap

In the House Robber problem, there is a robber. He wants to rob houses along a street. Each house has a certain amount of money. The robber cannot rob two houses that are next to each other on the same night. The goal is to take the most money possible.

Space Optimized Algorithm

The main idea is to track the most money we can rob up to the current house using two variables: - prev1: This is the most money robbed up to the house before the last one. - prev2: This is the most money robbed up to the last house.

C++ Implementation

Here is the C++ code for the space-optimized way:

#include <vector>
#include <iostream>
using namespace std;

class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.empty()) return 0;
        if (nums.size() == 1) return nums[0];

        int prev1 = 0, prev2 = 0;
        for (int num : nums) {
            int temp = prev1;
            prev1 = max(prev1, prev2 + num);
            prev2 = temp;
        }
        return prev1;
    }
};

int main() {
    Solution solution;
    vector<int> houses = {2, 7, 9, 3, 1};
    cout << "Maximum amount robbed: " << solution.rob(houses) << endl;
    return 0;
}

Explanation of the Code

We start with prev1 and prev2 set to zero.
We go through each house’s money value in the nums array.
For each house, we find the most money we can rob. We can either rob the current house (adding its value to prev2) or skip it (keeping prev1).
At the end, we return prev1, which has the most money that can be robbed.

This method keeps space use low while keeping the time speed at O(n). It makes it a good solution for the House Robber I problem in C++. For more about dynamic programming, see Dynamic Programming Fibonacci Number.

Frequently Asked Questions

1. What is the House Robber I problem in dynamic programming?

The House Robber I problem is a famous challenge in dynamic programming. It asks us to find the most money we can take from a line of houses. We cannot rob two houses that are next to each other. This problem teaches us about optimal substructure and overlapping subproblems. It is a good start to learn dynamic programming methods.

2. How do I implement the House Robber I solution in Java?

We can implement the House Robber I algorithm in Java using dynamic programming. We can use tabulation or a space-saving method. The main idea is to have an array. This array will hold the maximum money we can rob at each house. We must make sure not to rob two houses that are next to each other. You can see a full example in our section on Dynamic Programming Approach with Tabulation in Java.

3. What are the common dynamic programming techniques used in House Robber I?

The common methods to solve the House Robber I problem are tabulation and memoization. Tabulation is a bottom-up method. Memoization is a top-down method. Both methods use dynamic programming ideas. They break down the problem into smaller parts. This helps us to find the most money we can rob without getting caught by the police.

4. Can I solve House Robber I using Python?

Yes, we can solve the House Robber I problem easily in Python. We can use either tabulation or a space-saving method. Python’s list is very helpful for the dynamic programming solution. You can check our section on Dynamic Programming Approach with Tabulation in Python for a complete example.

5. What is the space-optimized approach for House Robber I in C++?

The space-optimized approach for the House Robber I problem in C++ helps us to use less space. It changes the space from O(n) to O(1). We only need to remember the last two values we calculated. This method lets us find the maximum money we can rob without needing a big array. Look at our detailed code example in the section on Space Optimized Approach for House Robber I in C++.