Painting the Walls

Hard

Snowflake

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Topics:

Dynamic Programming

You are given two 0-indexed integer arrays, cost and time, of size n representing the costs and the time taken to paint n different walls respectively. There are two painters available:

A paid painter that paints the i^th wall in time[i] units of time and takes cost[i] units of money.
A free painter that paints any wall in 1 unit of time at a cost of 0. But the free painter can only be used if the paid painter is already occupied.

Return the minimum amount of money required to paint the n walls.

Example 1:

Input: cost = [1,2,3,2], time = [1,2,3,2]
Output: 3
Explanation: The walls at index 0 and 1 will be painted by the paid painter, and it will take 3 units of time; meanwhile, the free painter will paint the walls at index 2 and 3, free of cost in 2 units of time. Thus, the total cost is 1 + 2 = 3.

Example 2:

Input: cost = [2,3,4,2], time = [1,1,1,1]
Output: 4
Explanation: The walls at index 0 and 3 will be painted by the paid painter, and it will take 2 units of time; meanwhile, the free painter will paint the walls at index 1 and 2, free of cost in 2 units of time. Thus, the total cost is 2 + 2 = 4.

Constraints:

1 <= cost.length <= 500
cost.length == time.length
1 <= cost[i] <= 10⁶
1 <= time[i] <= 500

Try coding it in LeetCode

Painting the Walls

Hard

Snowflake

2 views

Topics:

Dynamic Programming

You are given two 0-indexed integer arrays, cost and time, of size n representing the costs and the time taken to paint n different walls respectively. There are two painters available:

A paid painter that paints the i^th wall in time[i] units of time and takes cost[i] units of money.
A free painter that paints any wall in 1 unit of time at a cost of 0. But the free painter can only be used if the paid painter is already occupied.

Return the minimum amount of money required to paint the n walls.

Example 1:

Input: cost = [1,2,3,2], time = [1,2,3,2]
Output: 3
Explanation: The walls at index 0 and 1 will be painted by the paid painter, and it will take 3 units of time; meanwhile, the free painter will paint the walls at index 2 and 3, free of cost in 2 units of time. Thus, the total cost is 1 + 2 = 3.

Example 2:

Input: cost = [2,3,4,2], time = [1,1,1,1]
Output: 4
Explanation: The walls at index 0 and 3 will be painted by the paid painter, and it will take 2 units of time; meanwhile, the free painter will paint the walls at index 1 and 2, free of cost in 2 units of time. Thus, the total cost is 2 + 2 = 4.

Constraints:

1 <= cost.length <= 500
cost.length == time.length
1 <= cost[i] <= 10⁶
1 <= time[i] <= 500

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 are the possible ranges for the costs and times associated with each painter?
Can the cost or time values be zero or negative?
If it is impossible to paint all the walls within the given time, what value should I return?
Are there any constraints on the number of walls or the number of painters?
If there are multiple possible solutions to minimize cost, should I return any of them or is there a specific criterion to choose one?

Brute Force Solution

Approach

The brute force strategy involves examining every possible combination to find the absolute best solution. We'll explore every possible combination of paying or not paying each painter. We will then calculate the minimum cost amongst these combinations.

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

Consider each painter one at a time.
For the first painter, we have two choices: either we pay them, or we don't.
If we pay the first painter, we need to determine how much time they will save.
Next, consider the second painter, again considering if we pay them or not.
We need to look at every combination of paying and not paying painters.
For each combination, figure out the total cost of paying the selected painters.
Also, for each combination, determine the total time saved by paying those painters.
Calculate the final cost for each combination by adding the cost of not covering the walls (which depends on the time saved).
Compare the total cost of every single combination.
Select the combination that results in the absolute minimum cost - that is our final answer.

Code Implementation

def painting_the_walls_brute_force(cost, time):	number_of_painters = len(cost)
	minimum_total_cost = float('inf')

	# Iterate through all possible combinations of painters
	for i in range(2**number_of_painters):
		painters_paid_cost = 0
		time_saved = 0

		# Determine which painters are paid in this combination
		for painter_index in range(number_of_painters):
			# Check if the painter is paid based on the bitmask
			if (i >> painter_index) & 1:
				painters_paid_cost += cost[painter_index]
				time_saved += time[painter_index]

		# Calculate the time remaining after paying selected painters
		time_remaining = number_of_painters - time_saved

		# Ensure non-negative time remaining
		if time_remaining < 0:
			time_remaining = 0

		# Calculate the cost of not covering the walls
		cost_of_not_covering = time_remaining

		# Calculate total cost
		total_cost = painters_paid_cost + cost_of_not_covering

		# Update the minimum total cost
		minimum_total_cost = min(minimum_total_cost, total_cost)

	return minimum_total_cost

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach considers every possible combination of paying or not paying each of the n painters. For each painter, there are two choices: pay or don't pay. Thus, there are 2 * 2 * ... * 2 (n times) possible combinations, which equals 2^n. For each of these 2^n combinations, we calculate the cost and time saved, which takes O(n) time. However, the dominant factor is the enumeration of all subsets, leading to a time complexity of O(2^n).

Space Complexity

O(1) – The provided brute force strategy iterates through all possible combinations of paying or not paying each painter. While it considers each painter and calculates costs and time saved for each combination, it does not explicitly mention storing these combinations or intermediate results in auxiliary data structures. The algorithm primarily involves calculations and comparisons. Therefore, the space complexity is constant, as no extra space that scales with the input size (N, the number of painters) is used.

Optimal Solution

Approach

The core idea is to use dynamic programming to determine the minimum cost to paint all walls. We consider each wall and decide whether to paint it directly or skip it and have another painter paint it within the time limit provided by the skipped wall's free painting time.

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

Think of this as a decision-making process for each wall: either you pay to paint it now, or you skip it and someone else paints it for free later.
Create a table to keep track of the minimum cost to paint up to each wall. Each entry in the table represents the cheapest way to paint all walls up to that point.
Start from the beginning: The cost to paint no walls is zero.
Now, for each wall, consider two options: paint it or skip it.
If you paint the current wall, add its cost to the minimum cost from the earlier point and update the table accordingly.
If you skip the current wall, that means someone else paints it for free within a certain time period. Calculate the minimum cost of painting up to the previous walls, accounting for this free painting time, and update the table.
Choose the option (paint or skip) that results in the lowest total cost and record it in the table.
Move to the next wall and repeat the process of deciding whether to paint or skip, always choosing the option with the lowest cost.
Continue this process until you've considered all the walls. The final entry in the table represents the minimum cost to paint all the walls.

Code Implementation

def min_cost_to_paint(costs, times):
    number_of_walls = len(costs)
    dp_table = [0] * (number_of_walls + 1)

    for i in range(1, number_of_walls + 1):
        # Consider painting the current wall.
        paint_cost = costs[i - 1] + dp_table[i - 1]
        
        # Consider skipping the current wall.
        skip_time = times[i - 1]
        skipped_wall_index = max(0, i - 1 - skip_time)

        # Determine the minimum cost between painting and skipping.
        skip_cost = dp_table[skipped_wall_index]
        dp_table[i] = min(paint_cost, skip_cost)

    # The last element holds the minimum cost to paint all walls.
    return dp_table[number_of_walls]

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each of the n walls once to decide whether to paint it or skip it. For each wall, it performs a constant amount of work, calculating the cost of painting it or skipping it based on previous calculations stored in a table. The size of the table is directly proportional to the number of walls, which is n. Therefore, the time complexity is O(n).

Space Complexity

O(N) – The algorithm uses a table (dynamic programming table) to store the minimum cost to paint up to each wall. This table has a size proportional to the number of walls, N. No other significant auxiliary data structures are used. Therefore, the auxiliary space complexity is O(N) due to the dynamic programming table storing intermediate results for each wall.

Edge Cases

Case	How to Handle
Null or empty costs/times arrays	Return 0 cost if both arrays are null or empty, indicating no walls to paint.
Costs or times array of length zero.	Return 0 if either input array has zero length signifying no work or no cost.
Times array contains zero values	Handle zero times by short-circuiting, effectively making the corresponding wall free if possible.
Times array contains very large values leading to potential integer overflow during calculations	Use long data types for intermediate calculations to mitigate potential integer overflow issues.
Costs array contains negative values (invalid input)	Throw an IllegalArgumentException or return an error value indicating invalid input.
Sum of all times is less than or equal to number of walls	Consider special optimization to take all walls for the minimum sum, since the paid painter can complete all walls with just one choice.
Costs array contains extremely large values approaching maximum integer value	Ensure that the DP table is large enough and utilize long data types to prevent potential integer overflow when calculating the minimum cost.
Number of walls (n) is very large, causing memory constraints in DP table allocation	Consider optimizing the space complexity, such as using a 1D DP array by iterating in reverse order to avoid overwriting previous values needed for calculations.