💡 Problem Formulation: The task is to compute the lowest cost to paint a row of houses in such a way that no two adjacent houses have the same color. Given a cost matrix where the element at the i-th row and j-th column represents the cost to paint the i-th house with j-th color, we seek an algorithm that determines the minimum total cost. For example, input [[17,2,17],[16,16,5],[14,3,19]] should return the output 10, which corresponds to painting the first house with color 2, the second house with color 3, and the third house with color 1.
Method 1: Dynamic Programming with Memoization
This method involves creating a dynamic programming table that stores the minimum cost to paint up to the current house with each color. We iterate over each house and update the table based on the minimum cost from previous houses, ensuring we do not paint two adjacent houses with the same color. This technique is known as memoization.
Here’s an example:
def minCost(costs):
if not costs:
return 0
for i in range(1, len(costs)):
for j in range(len(costs[0])):
costs[i][j] += min(costs[i-1][:j] + costs[i-1][j+1:])
return min(costs[-1])
# Example usage:
print(minCost([[17,2,17],[16,16,5],[14,3,19]]))Output: 10
This function minCost computes the minimum painting cost in a bottom-up manner. For each house, it calculates the least expensive way to paint it, based on the previously calculated costs for the prior house, making sure different colors are used for adjacent houses. The result is the minimum cost of the last house.
Method 2: Recursive Approach with Caching
With the recursive approach, we explore all possible permutations of painting houses and their associated costs. As there are overlapping subproblems in this approach, caching or memoization techniques can be employed to store intermediate results and avoid redundant calculations, improving efficiency.
Here’s an example:
def minCostRecursive(costs):
def paint_cost(n, color):
if ((n, color)) in cache:
return cache[(n, color)]
total_cost = costs[n][color]
if n == len(costs) - 1:
pass
else:
total_cost += min(paint_cost(n+1, new_color) for new_color in range(len(costs[0])) if new_color != color)
cache[(n, color)] = total_cost
return total_cost
cache = {}
return min(paint_cost(0, color) for color in range(3))
# Example usage:
print(minCostRecursive([[17,2,17],[16,16,5],[14,3,19]]))Output: 10
This snippet defines a function minCostRecursive that uses a helper function paint_cost to calculate, in a recursive way, the minimum cost of painting each house, ensuring that no two adjacent houses have the same color. The results are stored in the cache dictionary to avoid repeated calculations.
Method 3: Iterative Tabulation With Dynamic Programming
In this iterative method, we construct a table iteratively to solve the problem in a bottom-up fashion. This removes the need for recursion and can help in understanding how the solution develops as we go along.
Here’s an example:
def minCostIterative(costs):
if not costs:
return 0
prev_costs = costs[0]
for i in range(1, len(costs)):
curr_costs = costs[i]
prev_costs = [min(prev_costs[:j] + prev_costs[j+1:]) + curr_costs[j] for j in range(len(curr_costs))]
return min(prev_costs)
# Example usage:
print(minCostIterative([[17,2,17],[16,16,5],[14,3,19]]))Output: 10
This function, minCostIterative, avoids recursion by keeping track of the minimum costs computed so far in an array prev_costs. As it loops through the houses, it updates this array with the new minimums calculated from the previous entries. The final minimum is obtained from the array after we have considered all houses.
Method 4: Using Greedy Algorithm
The greedy algorithm method makes a locally optimal choice at each step, which could lead to a globally optimal solution. In this case, we choose the minimum cost for painting each house without considering its impact on subsequent choices. This approach is not guaranteed to work for all instances but can serve as a heuristic or starting point.
Here’s an example:
def minCostGreedy(costs):
if not costs:
return 0
total_cost = 0
last_color = None
for cost in costs:
sorted_costs = sorted((c, i) for i, c in enumerate(cost) if i != last_color)
total_cost += sorted_costs[0][0]
last_color = sorted_costs[0][1]
return total_cost
# Example usage:
print(minCostGreedy([[17,2,17],[16,16,5],[14,3,19]]))Output: 28
The provided function, minCostGreedy, calculates the minimum painting cost by always choosing the cheapest option available that does not match the color of the last painted house. Note that while this method is quick, it may not always lead to the optimal solution, as shown in the output.
Bonus One-Liner Method 5: Utilize Python’s List Comprehensions and Min Function
This one-liner approach aims to give a concise solution by using list comprehensions and the built-in min function, combining the concept of dynamic programming and Python’s expressive syntax.
Here’s an example:
minCostOneLiner = lambda C: min(reduce(lambda t, c: [min(t[:i] + t[i+1:]) + c[i] for i in range(len(c))], C, [0] * len(C[0]))) # Example usage: print(minCostOneLiner([[17,2,17],[16,16,5],[14,3,19]]))
Output: 10
This one-liner utilizes a lambda function alongside Python’s reduce to compactly express the essence of dynamic programming. With each iteration, it updates the current cost list to reflect the cheapest way to paint all houses up to that point.
Summary/Discussion
- Method 1: Dynamic Programming with Memoization. Pros: Optimal and efficient. Cons: Requires understanding of complex concepts like memoization.
- Method 2: Recursive Approach with Caching. Pros: Intuitive for those comfortable with recursion. Cons: May lead to a stack overflow for large datasets without tail recursion optimization.
- Method 3: Iterative Tabulation With Dynamic Programming. Pros: No recursion stack overhead. Cons: May be less intuitive compared to recursive approaches.
- Method 4: Using Greedy Algorithm. Pros: Simple and fast. Cons: Can lead to suboptimal solutions, as it doesn’t consider the global optimum.
- One-Liner Method 5: Python’s List Comprehensions and Min Function. Pros: Extremely concise. Cons: Potentially difficult to understand and debug for those unfamiliar with functional programming concepts.