[Dynamic Programming] Count Ways to Paint Houses - Medium

Dynamic programming is a method we use to solve hard problems. We do this by breaking them into smaller problems. We save the results of these smaller problems. This way, we don’t have to do the same calculation again.

When we talk about painting houses, the problem is about finding the number of ways to paint a line of houses. We have a limited number of colors and some rules to follow. The goal is to make sure that no two houses next to each other have the same color. We can solve this problem well by using dynamic programming.

In this article, we will look closely at the problem. We will talk about different dynamic programming methods to count the ways to paint houses. We will show you how to do this in Java, Python, and C++. We will also share tips on how to make our space use better in dynamic programming. Furthermore, we will check different methods and give advice on how to test edge cases. This will help us understand the problem completely.

Here is a list of topics we will cover:
- [Dynamic Programming] Ways to Count Painting Houses - Medium
- Understanding the Problem Statement
- Dynamic Programming Approach to Count Ways to Paint Houses
- Java Implementation for Count Ways to Paint Houses
- Python Solution for Count Ways to Paint Houses
- C++ Code Example for Count Ways to Paint Houses
- Optimizing Space Complexity in Dynamic Programming
- Comparative Analysis of Different Approaches
- Testing and Edge Cases for Painting Houses Problem
- Frequently Asked Questions

If you want to read more about dynamic programming, check these articles: Dynamic Programming: Fibonacci Number and Dynamic Programming: Climbing Stairs.

Understanding the Problem Statement

We have a problem about counting how many ways we can paint houses. This is a common problem in dynamic programming. We have n houses in a row and k colors. Our goal is to find out how many ways we can paint all the houses. We must make sure that no two houses next to each other have the same color.

Problem Constraints:

  • Each house can be painted with one of k colors.
  • Houses that are next to each other must not have the same color.
  • We need to calculate the total number of valid ways to paint the n houses.

Example:

For example, if we have 3 houses and 2 colors (let’s say Color 1 and Color 2), some valid ways to paint the houses could be: - House 1: Color 1, House 2: Color 2, House 3: Color 1 - House 1: Color 2, House 2: Color 1, House 3: Color 2

The answer for this example is 6. We can find this by: - For the first house, we have k choices. - For each house after the first one, we have k-1 choices. This is to make sure it is different from the house before it.

Mathematical Representation:

We can let dp[i] show how many ways to paint i houses. We can write the relationship like this:

[ dp[i] = dp[i-1] (k - 1) + dp[i-2] (k - 1) ]

Base cases: - dp[1] = k - dp[2] = k \times (k - 1)

This relationship helps us build a dynamic programming method to solve the problem in a smart way.

For more details on dynamic programming methods, check out articles like Count Ways to Climb Stairs and Paint House Problem.

Dynamic Programming Approach to Count Ways to Paint Houses

In the “Count Ways to Paint Houses” problem, we want to find out how many ways we can paint a row of houses. We have some colors to choose from. We need to make sure that no two houses next to each other have the same color. We can solve this problem well with dynamic programming.

Problem Statement

We have n houses and k colors. Our task is to find out how many ways we can paint these houses so that the houses next to each other do not have the same color.

Dynamic Programming Solution

  1. State Definition: We say dp[i] is the number of ways to paint the first i houses.

  2. Recurrence Relation:

    • For the first house, we have k choices. So, dp[1] = k.
    • For the second house, it can be painted in k - 1 ways because it cannot be the same color as the first house. Thus, dp[2] = k * (k - 1).
    • For any house after the second, we can paint the i-th house differently from the (i-1)-th house. We look at all the ways we painted the previous houses: [ dp[i] = (k - 1) (dp[i - 1] + dp[i - 2]) ]

  3. Base Cases:

    • dp[0] = 1 because there are no houses to paint.
    • dp[1] = k.

  4. Final Result: The answer we want is dp[n], which tells us how many ways we can paint n houses.

Implementation

Here is a Java code for the dynamic programming approach:

public class PaintHouses {
    public static int countWaysToPaintHouses(int n, int k) {
        if (n == 0) return 0;
        if (n == 1) return k;

        int[] dp = new int[n + 1];
        dp[1] = k;
        dp[2] = k * (k - 1);

        for (int i = 3; i <= n; i++) {
            dp[i] = (k - 1) * (dp[i - 1] + dp[i - 2]);
        }
        
        return dp[n];
    }

    public static void main(String[] args) {
        int n = 3; // number of houses
        int k = 2; // number of colors
        System.out.println("Ways to paint houses: " + countWaysToPaintHouses(n, k));
    }
}

Python Implementation

Here is a Python code for the same problem:

def count_ways_to_paint_houses(n, k):
    if n == 0:
        return 0
    if n == 1:
        return k

    dp = [0] * (n + 1)
    dp[1] = k
    dp[2] = k * (k - 1)

    for i in range(3, n + 1):
        dp[i] = (k - 1) * (dp[i - 1] + dp[i - 2])

    return dp[n]

# Example usage
n = 3  # number of houses
k = 2  # number of colors
print("Ways to paint houses:", count_ways_to_paint_houses(n, k))

C++ Implementation

Here is a C++ code for counting ways to paint houses:

#include <iostream>
using namespace std;

int countWaysToPaintHouses(int n, int k) {
    if (n == 0) return 0;
    if (n == 1) return k;

    int dp[n + 1];
    dp[1] = k;
    dp[2] = k * (k - 1);

    for (int i = 3; i <= n; i++) {
        dp[i] = (k - 1) * (dp[i - 1] + dp[i - 2]);
    }

    return dp[n];
}

int main() {
    int n = 3; // number of houses
    int k = 2; // number of colors
    cout << "Ways to paint houses: " << countWaysToPaintHouses(n, k) << endl;
    return 0;
}

This dynamic programming way helps us to find the number of ways to paint the houses while following the rules. It works good for larger values of n and k. For more insights about dynamic programming, check Dynamic Programming: Ways to Count Painting Houses.

Java Implementation for Count Ways to Paint Houses

To solve the problem of counting how many ways we can paint houses using dynamic programming in Java, we can do these steps:

  1. Define the Problem: We have n houses and k colors. We must count how many ways we can paint the houses. We cannot paint two adjacent houses with the same color.

  2. Dynamic Programming State: We use dp[i][j] to show how many ways we can paint the first i houses. The i-th house will be painted with color j.

  3. Transition Formula: We can define the transition like this: [ dp[i][j] = _{c=0}^{k-1} dp[i-1][c] c j ] This means the number of ways to paint the i-th house with color j is the total ways to paint the previous house using all colors except j.

  4. Base Case: For the first house, we can paint it with any of the k colors: [ dp[1][j] = 1 j ]

  5. Final Calculation: The total ways to paint all n houses is the sum of ways to paint the last house with any of the k colors.

Here is the Java code:

public class PaintHouses {
    public static int countWaysToPaintHouses(int n, int k) {
        if (n == 0 || k == 0) return 0;
        if (n == 1) return k;

        // dp[i] is the number of ways to paint i houses
        int[] dp = new int[n + 1];
        dp[1] = k; // First house can be painted in k ways

        // For the second house and more
        for (int i = 2; i <= n; i++) {
            dp[i] = (k - 1) * dp[i - 1] + (k - 1) * dp[i - 2];
        }

        return dp[n];
    }

    public static void main(String[] args) {
        int n = 4; // Number of houses
        int k = 3; // Number of colors
        System.out.println("The number of ways to paint " + n + " houses with " + k + " colors is: " + countWaysToPaintHouses(n, k));
    }
}

Explanation of the Code:

  • The function countWaysToPaintHouses takes the number of houses n and the number of colors k.
  • We use an array dp to store how many ways we can paint from 1 to n houses.
  • The loop calculates the ways for each house based on the last two houses. We make sure adjacent houses do not have the same color.
  • Finally, the main method tests the function with an example of 4 houses and 3 colors.

This code calculates the number of ways to paint houses using dynamic programming. It works well for larger numbers too. For more examples about dynamic programming, we can check out Dynamic Programming: Paint House.

Python Solution for Count Ways to Paint Houses

To solve the problem of counting the ways to paint houses using dynamic programming in Python, we can follow this simple approach. The main idea is to keep a DP array. In this array, dp[i][j] shows the number of ways to paint the i-th house with color j. We must make sure that no two adjacent houses are the same color.

Problem Statement

We have n houses in a line. Each house can be painted in k different colors. The rule is that no two adjacent houses can have the same color. Our task is to find the total number of ways to paint these houses.

Dynamic Programming Approach

  1. Initialization: First, we define a DP array with size n x k.
  2. Base Case: For the first house, we can paint it with any of the k colors. So we set dp[0][j] = 1 for all j from 0 to k-1.
  3. Recurrence Relation:

    • For each house i (from 1 to n-1), we calculate:

      dp[i][j] = sum(dp[i-1][m]) for all m != j

    • This means for each color j, we add all the ways to paint the previous house with colors that are not j.

Python Code Implementation

Here is how we can write this logic in Python:

def countWaysToPaintHouses(n, k):
    if n == 0 or k == 0:
        return 0
    
    # dp[i][j] will show the number of ways to paint house i with color j
    dp = [[0] * k for _ in range(n)]
    
    # Base case: For the first house, we can use any of the k colors
    for j in range(k):
        dp[0][j] = 1
    
    # Fill the dp table
    for i in range(1, n):
        for j in range(k):
            dp[i][j] = sum(dp[i-1][m] for m in range(k) if m != j)
    
    # Total ways to paint all houses
    return sum(dp[n-1][j] for j in range(k))

# Example Usage
n = 3  # Number of houses
k = 2  # Number of colors
print(countWaysToPaintHouses(n, k))  # Output: 6

Explanation of the Code

  • The function countWaysToPaintHouses takes n (number of houses) and k (number of colors) as input.
  • We create a DP table to store the number of ways to paint each house.
  • The nested loops fill the DP table based on the rules we discussed.
  • In the end, we add the values of the last row of the DP table to find the total ways to paint all houses.

This Python solution counts the ways to paint the houses while following the rules of the problem. We can also look into more related problems using dynamic programming, like the Paint Fence Problem and the Paint House Problem.

C++ Code Example for Count Ways to Paint Houses

In this section, we share a C++ code to count the ways to paint houses. We use dynamic programming to solve this problem. The problem says we have n houses in a row. Each house can be painted in k colors. But we cannot paint two adjacent houses the same color.

Problem Statement

Given n (number of houses) and k (number of colors), we want to count how many ways we can paint the houses with these rules.

C++ Implementation

#include <iostream>
using namespace std;

int countWaysToPaintHouses(int n, int k) {
    if (n == 0) return 0; // No houses
    if (n == 1) return k; // Only one way to paint one house

    // For the first house, there are k options. For the second house, there are (k-1) options.
    int same = 0, diff = k; // We set same and diff for the first two houses
    for (int i = 2; i <= n; ++i) {
        same = diff; // Current house can be painted same as the one before
        diff = (k - 1) * (same + diff); // Current house can be painted different
    }
    return same + diff; // Total ways is sum of both cases
}

int main() {
    int n = 3; // Number of houses
    int k = 2; // Number of colors
    cout << "Number of ways to paint " << n << " houses with " << k << " colors: " << countWaysToPaintHouses(n, k) << endl;
    return 0;
}

Explanation of the Code

  • The function countWaysToPaintHouses takes n and k as inputs.
  • We start two variables: same for ways to paint the house the same color as before, and diff for ways to paint it a different color.
  • We use a loop to calculate ways for each house based on the previous counts.
  • At the end, we return the total ways by adding both cases.

This way, we get the answer fast with O(n) time and O(1) space.

For more reading and similar problems in dynamic programming, check the Dynamic Programming: Paint House - Medium.

Optimizing Space Complexity in Dynamic Programming

In dynamic programming, we can often reduce space complexity using simple methods. This is important when our solution needs a straight line of states. The aim is to use less memory while keeping our calculations fast. Here are some easy ways to optimize space complexity in dynamic programming problems. We will look at the “Count Ways to Paint Houses” problem as an example.

1. Iterative Approach with Rolling Arrays

Instead of keeping a whole table of past states, we can often store only the last few states we need. For instance, if our problem needs n past states, we just keep the last two or three.

Example: In the “Count Ways to Paint Houses” problem, if we only need the results from the last two houses to find the current house, we can cut the space complexity from O(n) to O(1).

public int countWays(int n, int k) {
    if (n == 0) return 0;
    if (n == 1) return k;

    int prev1 = k; // ways to paint the last house
    int prev2 = k * (k - 1); // ways to paint the second last house

    for (int i = 3; i <= n; i++) {
        int current = (prev1 + prev2) * (k - 1);
        prev2 = prev1;
        prev1 = current;
    }

    return prev1;
}

2. In-Place Updates

For problems where we can fill the state table, we can change the entries in the table while we find new values. This way, we do not need to keep old values.

3. Constant Space Solutions

Sometimes, we can find a clear formula to calculate the result. This means we do not need to keep any extra states. This is good for problems with linear or polynomial properties.

4. Recursive with Memoization

Recursion can use a lot of space because of the stack depth. But we can use memoization to save results of smaller problems. This helps reduce the need to recalculate. Still, we should remember that this can keep a cache that may take up space.

5. Bit Manipulation

For some problems, we can use bit manipulation to store states in a small way. This is helpful for combinatorial problems or when we have few states.

By using these methods, we can improve space complexity in dynamic programming. This makes our algorithms work better, especially when we deal with large inputs. For more details on dynamic programming methods and tips, we can check the article on Dynamic Programming: Count Ways to Paint Houses.

Comparative Analysis of Different Approaches

When we solve the problem of counting ways to paint houses with dynamic programming, we can use different methods. The main ways are the simple recursive solution, memoization, and the bottom-up dynamic programming method. Let’s look at these methods side by side.

Naive Recursive Approach

  • Time Complexity: Exponential, O(2^n).
  • Space Complexity: O(n) because of the call stack.
  • Description: This method calculates the number of ways to paint houses again and again. It does not keep any results, so it repeats the same work.
public int countWays(int n, int k) {
    if (n == 0) return 0;
    if (n == 1) return k;
    return (k - 1) * (countWays(n - 1, k) + countWays(n - 2, k));
}

Memoization Approach

  • Time Complexity: O(n).
  • Space Complexity: O(n) to store results.
  • Description: This method makes the recursive method better by saving the results of smaller problems. This way, it does not do the same calculations again.
public int countWays(int n, int k) {
    Integer[] memo = new Integer[n + 1];
    return helper(n, k, memo);
}

private int helper(int n, int k, Integer[] memo) {
    if (n == 0) return 0;
    if (n == 1) return k;
    if (memo[n] != null) return memo[n];
    memo[n] = (k - 1) * (helper(n - 1, k, memo) + helper(n - 2, k, memo));
    return memo[n];
}

Bottom-Up Dynamic Programming Approach

  • Time Complexity: O(n).
  • Space Complexity: O(1) when we do it smartly.
  • Description: This method builds the solution step by step from the base cases. It saves space by only using the last two values we calculate.
public int countWays(int n, int k) {
    if (n == 0) return 0;
    if (n == 1) return k;
    
    int same = 0, diff = k, total = k;
    for (int i = 2; i <= n; i++) {
        same = diff;
        diff = (total) * (k - 1);
        total = same + diff;
    }
    return total;
}

Summary of Comparison

  • Efficiency: The naive recursive method is not good for big inputs because it takes a long time. But memoization and bottom-up dynamic programming run much faster in linear time.
  • Space Usage: Memoization needs extra space to keep results. The bottom-up method can use less space if we optimize it.
  • Implementation Complexity: The naive method is easy to understand. But memoization and bottom-up are a bit more complex. They do give big improvements in performance.

In cases where speed is very important, we usually choose the bottom-up dynamic programming method. It is fast and needs less space. For more learning, we can check out related topics like Dynamic Programming: Fibonacci Numbers and Dynamic Programming: Climbing Stairs.

Testing and Edge Cases for Painting Houses Problem

When we test the “Count Ways to Paint Houses” problem, we need to think about different edge cases and situations. These can affect if the solution works well or not. Here are some important test cases and edge cases we should check:

  1. Minimum Input Values:
    • We can test with n = 0 (no houses). The output should be 0. This is because no houses can be painted.
    • We can test with n = 1 (one house). The output should be k, where k is the number of colors.
  2. Single Color:
    • We test with k = 1 (only one color) and n > 1. The output should be 0. This is because we cannot paint adjacent houses the same color.
  3. Multiple Houses with Colors:
    • We can test with n = 3 and k = 2. The output should be 2:
      • Possible combinations are: RGY, RGB (R: Red, G: Green, B: Blue).
  4. Large Input Values:
    • We should test with large values for n (like n = 1000) and reasonable values for k (like k = 10). This checks if the algorithm runs well without crashing or taking too much time.
  5. All Houses Same Color:
    • We test with n = 5 and k = 3. The output should be 30 because all houses can be painted in different color combinations.
  6. Performance Under Constraints:
    • We need to check how the algorithm deals with maximum values for both n and k. This is to make sure it runs in time and does not use too much memory.
  7. Randomized Input:
    • We can create random values for n and k. Then we verify if the output matches the expected number of ways to paint according to the rules.

Example Code for Testing

Here is a simple example in Python to test the function that counts the ways to paint houses:

def countWaysToPaintHouses(n, k):
    if n == 0:
        return 0
    if n == 1:
        return k
    return k * ((k - 1) ** (n - 1))

# Testing the function
test_cases = [
    (0, 3),  # Expected: 0
    (1, 3),  # Expected: 3
    (3, 2),  # Expected: 2
    (5, 1),  # Expected: 0
    (5, 3),  # Expected: 30
    (1000, 10)  # Performance test
]

for n, k in test_cases:
    print(f"n = {n}, k = {k} => Count Ways: {countWaysToPaintHouses(n, k)}")

In conclusion, we need to test edge cases, minimum and maximum inputs, and performance situations. This is very important to make sure the solution to the painting houses problem is strong. This testing will help us find any problems in the algorithm and keep it within the rules.

Frequently Asked Questions

1. What is the dynamic programming approach to count ways to paint houses?

The dynamic programming approach to count ways to paint houses is about breaking the problem into smaller parts. Each house can be painted in many ways. But we have rules that say adjacent houses can’t be the same color. We use a table to keep track of how many ways we can paint up to each house while following these rules. This way, we can quickly find the total ways to paint all the houses.

2. How can I implement the count ways to paint houses problem in Java?

To implement the count ways to paint houses problem in Java, we can make a dynamic programming array. This array will record how many ways we can paint each house. We will use a recurrence relation to find the valid combinations based on previous houses. You can look at the Java implementation section in the article for a complete code example that shows this method well.

3. What are the edge cases to consider when counting ways to paint houses?

When we count ways to paint houses, we should think about edge cases. For example, if there are no houses to paint, we should say there are zero ways. If there is only one house, the number of ways to paint it is the same as the number of colors we have. These cases help us to make sure our dynamic programming solution is strong.

4. How does the space complexity affect the count ways to paint houses solution?

Space complexity in the count ways to paint houses solution is very important, especially with big inputs. The usual dynamic programming approach uses O(n) space, where n is the number of houses. But we can make it better to O(1) space. We only need to keep track of the last two states because each house’s state only depends on the previous two. This makes our solution more efficient while still being correct.

The count ways to paint houses problem is similar to other dynamic programming problems like climbing stairs and the house robber problem. They all involve making choices under certain rules. The main difference is in the specific rules for the choices next to each other. Looking at these different problems helps us understand dynamic programming better. You can find more about similar problems in our article on the dynamic programming house robber problem.