Paint House III

Hard

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Dynamic Programming

There is a row of m houses in a small city, each house must be painted with one of the n colors (labeled from 1 to n), some houses that have been painted last summer should not be painted again.

A neighborhood is a maximal group of continuous houses that are painted with the same color.

For example: houses = [1,2,2,3,3,2,1,1] contains 5 neighborhoods [{1}, {2,2}, {3,3}, {2}, {1,1}].

Given an array houses, an m x n matrix cost and an integer target where:

houses[i]: is the color of the house i, and 0 if the house is not painted yet.
cost[i][j]: is the cost of paint the house i with the color j + 1.

Return the minimum cost of painting all the remaining houses in such a way that there are exactly target neighborhoods. If it is not possible, return -1.

Example 1:

Input: houses = [0,0,0,0,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
Output: 9
Explanation: Paint houses of this way [1,2,2,1,1]
This array contains target = 3 neighborhoods, [{1}, {2,2}, {1,1}].
Cost of paint all houses (1 + 1 + 1 + 1 + 5) = 9.

Example 2:

Input: houses = [0,2,1,2,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
Output: 11
Explanation: Some houses are already painted, Paint the houses of this way [2,2,1,2,2]
This array contains target = 3 neighborhoods, [{2,2}, {1}, {2,2}]. 
Cost of paint the first and last house (10 + 1) = 11.

Example 3:

Input: houses = [3,1,2,3], cost = [[1,1,1],[1,1,1],[1,1,1],[1,1,1]], m = 4, n = 3, target = 3
Output: -1
Explanation: Houses are already painted with a total of 4 neighborhoods [{3},{1},{2},{3}] different of target = 3.

Constraints:

m == houses.length == cost.length
n == cost[i].length
1 <= m <= 100
1 <= n <= 20
1 <= target <= m
0 <= houses[i] <= n
1 <= cost[i][j] <= 10⁴

Paint House III

Hard

PayPal

0 views

Topics:

Dynamic Programming

There is a row of m houses in a small city, each house must be painted with one of the n colors (labeled from 1 to n), some houses that have been painted last summer should not be painted again.

A neighborhood is a maximal group of continuous houses that are painted with the same color.

For example: houses = [1,2,2,3,3,2,1,1] contains 5 neighborhoods [{1}, {2,2}, {3,3}, {2}, {1,1}].

Given an array houses, an m x n matrix cost and an integer target where:

houses[i]: is the color of the house i, and 0 if the house is not painted yet.
cost[i][j]: is the cost of paint the house i with the color j + 1.

Return the minimum cost of painting all the remaining houses in such a way that there are exactly target neighborhoods. If it is not possible, return -1.

Example 1:

Input: houses = [0,0,0,0,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
Output: 9
Explanation: Paint houses of this way [1,2,2,1,1]
This array contains target = 3 neighborhoods, [{1}, {2,2}, {1,1}].
Cost of paint all houses (1 + 1 + 1 + 1 + 5) = 9.

Example 2:

Input: houses = [0,2,1,2,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3
Output: 11
Explanation: Some houses are already painted, Paint the houses of this way [2,2,1,2,2]
This array contains target = 3 neighborhoods, [{2,2}, {1}, {2,2}]. 
Cost of paint the first and last house (10 + 1) = 11.

Example 3:

Input: houses = [3,1,2,3], cost = [[1,1,1],[1,1,1],[1,1,1],[1,1,1]], m = 4, n = 3, target = 3
Output: -1
Explanation: Houses are already painted with a total of 4 neighborhoods [{3},{1},{2},{3}] different of target = 3.

Constraints:

m == houses.length == cost.length
n == cost[i].length
1 <= m <= 100
1 <= n <= 20
1 <= target <= m
0 <= houses[i] <= n
1 <= cost[i][j] <= 10⁴

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 are the constraints on n (number of houses), m (number of colors), and target (number of neighborhoods)?
If it's impossible to paint the houses with exactly `target` neighborhoods, what value should I return?
Are the costs in the `cost` matrix always non-negative integers?
Is it guaranteed that the input `houses` array will only contain values in the range [0, m]?
Could you clarify the definition of a 'neighborhood'? Specifically, if house i and house i+1 have the same color, do they always belong to the same neighborhood?

Brute Force Solution

Approach

The brute force method for painting houses considers every single possible color combination for all the houses. It's like trying every paint color on every house and checking if it works within the rules. The most affordable combination is then chosen.

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

Start by picking a color for the first house.
Then, for each possible color for the first house, try every possible color for the second house.
Continue doing this for each house, exploring every single color option.
Whenever you reach the last house, check if the total arrangement meets the neighborhood requirement: specifically the required number of neighborhoods.
If the coloring arrangement does meet the required number of neighborhoods, calculate the total cost of using that particular coloring.
Keep track of all the arrangements that met the neighborhood requirement and their respective cost.
Finally, compare the costs of all the valid arrangements and choose the one with the smallest cost; that's your answer.

Code Implementation

def paint_house_brute_force(
    houses, cost, number_of_neighborhoods_required
):
    number_of_houses = len(houses)
    number_of_colors = len(cost[0])
    minimum_cost = float('inf')

    def calculate_neighborhoods(colors_assigned):
        neighborhood_count = 0
        for house_index in range(1, number_of_houses):
            if colors_assigned[house_index] !=
               colors_assigned[house_index - 1]:
                neighborhood_count += 1
        return neighborhood_count + 1

    def calculate_total_cost(colors_assigned):
        total_cost = 0
        for house_index in range(number_of_houses):
            total_cost += cost[house_index][colors_assigned[house_index] - 1]
        return total_cost

    def solve(house_index, colors_assigned):
        nonlocal minimum_cost

        if house_index == number_of_houses:
            # Base case: all houses are painted.
            number_of_neighborhoods =
                calculate_neighborhoods(colors_assigned)

            if number_of_neighborhoods ==
               number_of_neighborhoods_required:
                current_cost = calculate_total_cost(colors_assigned)
                minimum_cost = min(minimum_cost, current_cost)

            return

        if houses[house_index] == 0:
            # If the house is not painted, try each color
            for color_index in range(1, number_of_colors + 1):
                colors_assigned[house_index] = color_index
                solve(house_index + 1, colors_assigned)

        else:
            # If the house is pre-painted,
            # only try the pre-assigned color
            colors_assigned[house_index] = houses[house_index]
            solve(house_index + 1, colors_assigned)

    colors_assigned = [0] * number_of_houses
    solve(0, colors_assigned)

    if minimum_cost == float('inf'):
        return -1
    else:
        return minimum_cost

Big(O) Analysis

Time Complexity

O(m^n) – The brute force approach explores every possible coloring combination for n houses, where each house can be painted with one of m colors. This leads to m options for the first house, m options for the second house, and so on, resulting in m * m * ... * m (n times) total combinations. The number of neighborhood calculations and cost calculations grows proportionally with the number of color combinations. Therefore, the time complexity is O(m^n), where n is the number of houses and m is the number of available colors.

Space Complexity

O(1) – The brute force approach, as described, primarily iterates through possible color combinations and calculates costs. It doesn't explicitly mention storing all combinations simultaneously. The space used mainly consists of variables to hold the current color selection, the number of neighborhoods, the current cost, and the minimum cost seen so far. These variables occupy a constant amount of space, irrespective of the number of houses or colors. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

This problem asks us to paint houses while minimizing cost and achieving a specific number of neighborhood blocks. The key is to break down the problem into smaller, overlapping subproblems and remember the best solutions to those subproblems to avoid redundant calculations.

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

Think of building up the solution house by house. At each house, we want to consider the cost of painting it each possible color.
As we consider each house and color, we also keep track of how many 'neighborhoods' we've created so far. A neighborhood is a continuous sequence of houses with the same color.
To avoid recalculating the same things repeatedly, we save the lowest cost to paint up to a certain house, using a specific color for that house, and ending up with a specific number of neighborhoods.
When we get to a new house, we look back at our saved costs. For each color we could paint the new house, we check the costs of painting the previous house each possible color.
If we choose the same color as the previous house, the number of neighborhoods stays the same. If we choose a different color, the number of neighborhoods increases by one.
We pick the color for the new house that gives us the lowest total cost, considering the costs from the previous house.
We repeat this process for each house, color, and number of neighborhoods. By the end, we'll know the lowest cost to paint all the houses with the desired number of neighborhoods.

Code Implementation

def paint_house_iii(houses, cost, number_of_houses, number_of_colors, target_neighborhoods):
    maximum_integer = float('inf')
    # Initialize the DP table with maximum values, indicating initially unreachable states.
    dp = [[[maximum_integer] * (target_neighborhoods + 1) for _ in range(number_of_colors + 1)] for _ in range(number_of_houses + 1)]

    # Base case: No houses painted, 0 neighborhoods, cost 0.
    dp[0][0][0] = 0

    for house_index in range(1, number_of_houses + 1):
        for last_color in range(number_of_colors + 1):
            for neighborhood_count in range(target_neighborhoods + 1):
                if dp[house_index - 1][last_color][neighborhood_count] != maximum_integer:
                    if houses[house_index - 1] == 0:
                        for current_color in range(1, number_of_colors + 1):
                            new_neighborhood_count = neighborhood_count + (current_color != last_color)

                            # Only update if neighborhood count is feasible
                            if new_neighborhood_count <= target_neighborhoods:
                                dp[house_index][current_color][new_neighborhood_count] = min(
                                    dp[house_index][current_color][new_neighborhood_count],
                                    dp[house_index - 1][last_color][neighborhood_count] + cost[house_index - 1][current_color - 1]
                                )
                    else:
                        current_color = houses[house_index - 1]
                        new_neighborhood_count = neighborhood_count + (current_color != last_color)

                        # Only update if neighborhood count is feasible
                        if new_neighborhood_count <= target_neighborhoods:
                            dp[house_index][current_color][new_neighborhood_count] = min(
                                dp[house_index][current_color][new_neighborhood_count],
                                dp[house_index - 1][last_color][neighborhood_count]
                            )

    minimum_cost = maximum_integer
    # Find the minimum cost among all possible last colors for the target neighborhoods.
    for last_color in range(1, number_of_colors + 1):
        minimum_cost = min(minimum_cost, dp[number_of_houses][last_color][target_neighborhoods])

    # If no valid solution is found, return -1.
    if minimum_cost == maximum_integer:
        return -1
    else:
        return minimum_cost

Big(O) Analysis

Time Complexity

O(m * n * N) – The algorithm iterates through each of the n houses. For each house, it considers painting it with each of the m possible colors. For each house and color combination, it iterates through all possible neighborhood counts from 1 to N (the target number of neighborhoods). Therefore, the time complexity is proportional to the product of the number of colors, the number of houses, and the number of neighborhoods, resulting in O(m * n * N).

Space Complexity

O(N * target * n) – The algorithm uses dynamic programming to store intermediate results, specifically the lowest cost to paint up to a certain house, using a specific color for that house, and ending up with a specific number of neighborhoods. This is achieved with a DP table that stores the costs for each house, color, and number of neighborhoods. The dimensions of this table are N (number of houses), target (the desired number of neighborhoods), and n (number of colors). Therefore, the auxiliary space used is proportional to the product of these dimensions, which can be represented as O(N * target * n).

Edge Cases

Case	How to Handle
houses array is null or empty	Return -1 immediately as there are no houses to paint.
cost array is null or empty	Return -1 immediately as we cannot determine the cost to paint the houses.
n (number of houses) is not consistent between houses and cost arrays	Return -1 immediately as the input is invalid.
m (number of colors) is 0	Return -1 since we cannot paint with 0 colors.
target is 0	If target is 0, return 0 if all houses are already painted the same color, else return -1.
target is greater than n (number of houses)	Return -1 immediately as it is impossible to have more neighborhoods than houses.
No valid solution exists (impossible to reach the target neighborhoods)	Return -1 after the DP is complete if the min cost is still infinity.
Integer overflow in cost calculation	Use long data type to store cost values during calculation to avoid overflow.