Paint House IV

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Dynamic Programming

You are given an even integer n representing the number of houses arranged in a straight line, and a 2D array cost of size n x 3, where cost[i][j] represents the cost of painting house i with color j + 1.

The houses will look beautiful if they satisfy the following conditions:

No two adjacent houses are painted the same color.
Houses equidistant from the ends of the row are not painted the same color. For example, if n = 6, houses at positions (0, 5), (1, 4), and (2, 3) are considered equidistant.

Return the minimum cost to paint the houses such that they look beautiful.

Example 1:

Input: n = 4, cost = [[3,5,7],[6,2,9],[4,8,1],[7,3,5]]

Output: 9

Explanation:

The optimal painting sequence is [1, 2, 3, 2] with corresponding costs [3, 2, 1, 3]. This satisfies the following conditions:

No adjacent houses have the same color.
Houses at positions 0 and 3 (equidistant from the ends) are not painted the same color (1 != 2).
Houses at positions 1 and 2 (equidistant from the ends) are not painted the same color (2 != 3).

The minimum cost to paint the houses so that they look beautiful is 3 + 2 + 1 + 3 = 9.

Example 2:

Input: n = 6, cost = [[2,4,6],[5,3,8],[7,1,9],[4,6,2],[3,5,7],[8,2,4]]

Output: 18

Explanation:

The optimal painting sequence is [1, 3, 2, 3, 1, 2] with corresponding costs [2, 8, 1, 2, 3, 2]. This satisfies the following conditions:

No adjacent houses have the same color.
Houses at positions 0 and 5 (equidistant from the ends) are not painted the same color (1 != 2).
Houses at positions 1 and 4 (equidistant from the ends) are not painted the same color (3 != 1).
Houses at positions 2 and 3 (equidistant from the ends) are not painted the same color (2 != 3).

The minimum cost to paint the houses so that they look beautiful is 2 + 8 + 1 + 2 + 3 + 2 = 18.

Constraints:

2 <= n <= 10⁵
n is even.
cost.length == n
cost[i].length == 3
0 <= cost[i][j] <= 10⁵

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for `costs[i][j]` and `target`? Can they be negative or zero?
Are there any constraints on the values of `m` (number of houses) and `n` (number of colors)?
If it's impossible to paint the houses with exactly `target` neighborhoods, what value should I return?
If `houses[i]` is already painted (not 0), do I still incur the cost associated with painting it a different color?
Is there a specific meaning to the value 0 in the `houses` array, and is it guaranteed that 0 only appears to denote unpainted houses?

Brute Force Solution

Approach

The core idea is to try every single way to paint the houses. For each house, we consider all available color choices and calculate the total cost of that painting combination. We then choose the painting option with the lowest overall cost.

Here's how the algorithm would work step-by-step:

For the first house, consider painting it each of the available colors, one color at a time.
For each color we tried for the first house, consider all possible colors for the second house.
Continue this process for all the houses, creating every possible combination of house colors.
For each complete combination (each house painted a certain color), calculate the total cost to paint all houses in that specific way.
Compare the total cost of all possible color combinations.
Choose the combination with the lowest total cost. This is the best way to paint the houses.

Code Implementation

def paint_house_iv_brute_force(costs):
    number_of_houses = len(costs)
    number_of_colors = len(costs[0]) if number_of_houses > 0 else 0
    minimum_cost = float('inf')

    def calculate_total_cost(colors):
        total_cost = 0
        for house_index, color_index in enumerate(colors):
            total_cost += costs[house_index][color_index]
        return total_cost

    def find_minimum_cost(house_index, current_colors):
        nonlocal minimum_cost

        # Base case: all houses have been painted
        if house_index == number_of_houses:
            total_cost = calculate_total_cost(current_colors)
            minimum_cost = min(minimum_cost, total_cost)
            return

        # Try each color for the current house
        for color_index in range(number_of_colors):
            # Explore the possibility of painting the current house this color
            find_minimum_cost(house_index + 1, current_colors + [color_index])

    # Initiate the recursive calls to try every painting combination
    find_minimum_cost(0, [])

    #If minimum cost is unchanged, no houses or colors
    if minimum_cost == float('inf'):
        return 0

    return minimum_cost

Big(O) Analysis

Time Complexity

O(k^n) – The algorithm explores every possible coloring combination. If there are n houses and each house can be painted with one of k colors, then there are k options for the first house, k options for the second house, and so on. This creates k * k * ... * k (n times) possible combinations. Therefore, the total number of combinations is k^n, leading to O(k^n) time complexity as we evaluate the cost of each combination.

Space Complexity

O(N) – The plain English explanation outlines a process that explores every possible combination of colors for the houses using recursion. This recursive approach, though not explicitly stated as code, creates a call stack to keep track of the current painting combination being explored. In the worst-case scenario, the depth of the recursion will be equal to the number of houses, denoted as N. Thus, the maximum space used by the call stack is proportional to N, and the space complexity is O(N).

Optimal Solution

Approach

We want to find the cheapest way to paint a row of houses, where each house can be painted one of several colors, and there's a cost for each color. The trick is to build up the best solution for each house one at a time, remembering the lowest costs from the previous houses.

Here's how the algorithm would work step-by-step:

Think about painting the first house. For each color, record how much it costs to paint that first house that color.
Now, for the second house, look at each color option. To figure out the cheapest way to paint the second house a specific color, add that color's cost to the smallest cost from the previous house, but you can't use the same color as the second house because of the constraint.
Keep going through each house. For each house, figure out the cheapest way to paint it each color by adding that color's cost to the best possible cost from the previous house, making sure the colors are different.
After you've considered all the houses, find the smallest cost among all the colors for the very last house. That's the overall minimum cost to paint all the houses.

Code Implementation

def min_cost(house_costs, number_of_houses, number_of_colors): 
    if not house_costs or number_of_houses == 0 or number_of_colors == 0:
        return 0

    # Initialize cost for the first house.
    previous_house_costs = house_costs[0]

    for house_index in range(1, number_of_houses): 
        current_house_costs = house_costs[house_index]
        new_costs = [0] * number_of_colors

        # Calculate the minimum cost for each color.
        for color_index in range(number_of_colors): 
            min_cost_excluding_current_color = float('inf')

            # Find the minimum cost from the previous house, excluding the current color.
            for previous_color_index in range(number_of_colors):
                if previous_color_index != color_index:
                    min_cost_excluding_current_color = min(min_cost_excluding_current_color, previous_house_costs[previous_color_index])

            # This is where we're building up the solution.
            new_costs[color_index] = current_house_costs[color_index] + min_cost_excluding_current_color

        previous_house_costs = new_costs

    # After all houses, find the minimum of the costs of the last house.
    return min(previous_house_costs)

Big(O) Analysis

Time Complexity

O(n * k) – The algorithm iterates through each of the n houses. For each house, it iterates through all k possible colors to determine the minimum cost to paint that house a particular color, considering the costs of painting the previous house a different color. Therefore, the algorithm performs approximately n * k operations, leading to a time complexity of O(n * k).

Space Complexity

O(N * K) – The algorithm builds up the best solution for each house one at a time, storing the minimum cost for each color at each house. This requires storing a cost for each color (K) for each house (N). Therefore, we maintain a data structure to hold these intermediate costs which grows linearly with both the number of houses (N) and the number of colors (K). This results in O(N * K) auxiliary space.

Edge Cases

Case	How to Handle
Empty costs array	Return 0 as no houses need painting, representing zero cost.
Empty costs for a specific house	Handle this as an invalid input, potentially throwing an exception or returning a default large value to indicate impossible cost.
Only one house to paint (costs array length of 1)	Return the minimum cost within the single row, as we only need to paint this one house.
Only one color available	Sum all the costs for the color, as we have no choice but to paint every house using the same color.
Large number of houses and colors, leading to potential integer overflow when summing costs	Use a larger data type like long to store intermediate and final cost values to prevent overflow.
All costs are extremely large values (close to the maximum integer)	Be cautious about integer overflows when calculating minimum costs and consider using a larger data type if needed.
Forbidden adjacent colors cause no possible solution	Return a sentinel value (e.g., -1 or Infinity) when no valid combination of colors satisfies the constraints and allows to paint all houses
Costs array contains negative values	Handle this as an invalid input as costs cannot be negative; throw an exception or return an error value.

Empty costs array

How to Handle:

Return 0 as no houses need painting, representing zero cost.

Empty costs for a specific house

How to Handle:

Handle this as an invalid input, potentially throwing an exception or returning a default large value to indicate impossible cost.

Only one house to paint (costs array length of 1)

How to Handle:

Return the minimum cost within the single row, as we only need to paint this one house.

Only one color available

How to Handle:

Sum all the costs for the color, as we have no choice but to paint every house using the same color.

Large number of houses and colors, leading to potential integer overflow when summing costs

How to Handle:

Use a larger data type like long to store intermediate and final cost values to prevent overflow.

All costs are extremely large values (close to the maximum integer)

How to Handle:

Be cautious about integer overflows when calculating minimum costs and consider using a larger data type if needed.

Forbidden adjacent colors cause no possible solution

How to Handle:

Return a sentinel value (e.g., -1 or Infinity) when no valid combination of colors satisfies the constraints and allows to paint all houses

Costs array contains negative values

How to Handle:

Handle this as an invalid input as costs cannot be negative; throw an exception or return an error value.