How do you solve Painting the Walls on LeetCode?

Painting the Walls (LeetCode #2742) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: The free painter can only be used when the paid painter is busy, which creates a dependency on the order of painting.

What is the time complexity of Painting the Walls?

The optimal solution for Painting the Walls runs in O(n²) time and uses O(n) space. The complexity is O(n²) because we are using a nested loop to fill the DP array, where n is the number of walls.

What algorithm or data structure is used to solve Painting the Walls?

Painting the Walls uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Painting the Walls?

Always clarify the constraints and rules of the problem before diving into coding. Think about edge cases, such as when all walls have the same cost or time. Explain your thought process while coding to demonstrate your understanding.

What are common mistakes when solving Painting the Walls?

Not considering the time constraints when using the free painter. Failing to initialize the DP array correctly, leading to incorrect results.

Painting the Walls — LeetCode #2742 (Hard)

Tags: Array, Dynamic Programming

Related patterns: Dynamic Programming, Greedy Algorithms

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach involves trying all possible combinations of which walls to paint with the paid painter and which to paint with the free painter. This will help us find the minimum cost, but it's inefficient due to the large number of combinations.

The complexity is O(n²) because we are iterating through all subsets of walls, which can be up to 2^n combinations, leading to a quadratic time complexity when calculating costs.

Step 1: Iterate through all possible subsets of walls to be painted by the paid painter.
Step 2: For each subset, calculate the total cost of the paid painter and the total time taken.
Step 3: Determine how many walls can be painted by the free painter within the remaining time.
Step 4: Calculate the total cost for each combination and keep track of the minimum cost found.

1. For cost = [1,2,3,2], time = [1,2,3,2], we start with r=0 (no paid walls). Total cost = 0. 2. For r=1, we try paid wall at index 0 (cost 1), total time 1. Free walls can be painted in 3 units of time. Cost = 1. 3. For r=1, paid wall at index 1 (cost 2), total time 2. Free walls can be painted in 2 units of time. Cost = 2. 4. For r=2, paid walls at indices 0 and 1 (cost 3), total time 3. Free walls can be painted in 1 unit of time. Cost = 3. 5. Continue until all combinations are checked. 6. Minimum cost found is 3.

Optimal Solution approach

Time complexity: O(n²). Space complexity: O(n).

The optimal solution uses dynamic programming to keep track of the minimum cost required to paint the walls while considering the constraints of the two painters. This approach is efficient and avoids redundant calculations.

The complexity is O(n²) because we are using a nested loop to fill the DP array, where n is the number of walls.

Step 1: Initialize a DP array where dp[i] represents the minimum cost to paint the first i walls.
Step 2: Iterate through each wall and for each wall, decide whether to use the paid painter or the free painter based on the time taken.
Step 3: Update the DP array by considering the cost of using the paid painter for the current wall and the time taken for the previous walls.
Step 4: Return the value in dp[n] which represents the minimum cost to paint all walls.

1. For cost = [1,2,3,2], time = [1,2,3,2], initialize dp = [0, ∞, ∞, ∞, ∞]. 2. For i=1, check j=0: dp[1] = min(∞, 0 + 1 + 0) = 1. 3. For i=2, check j=0: dp[2] = min(∞, 0 + 1 + 1) = 2; check j=1: dp[2] = min(2, 1 + 2 + 0) = 2. 4. For i=3, check j=0: dp[3] = min(∞, 0 + 1 + 2) = 3; check j=1: dp[3] = min(3, 1 + 2 + 1) = 3; check j=2: dp[3] = min(3, 2 + 3 + 0) = 3. 5. For i=4, check j=0: dp[4] = min(∞, 0 + 1 + 3) = 4; check j=1: dp[4] = min(4, 1 + 2 + 2) = 4; check j=2: dp[4] = min(4, 2 + 3 + 1) = 4; check j=3: dp[4] = min(4, 3 + 2 + 0) = 4. 6. Final dp = [0, 1, 2, 3, 4]. Minimum cost is 4.

Key Insights

The free painter can only be used when the paid painter is busy, which creates a dependency on the order of painting.
Dynamic programming helps in breaking down the problem into smaller subproblems, allowing us to build the solution incrementally.

Common Mistakes

Not considering the time constraints when using the free painter.
Failing to initialize the DP array correctly, leading to incorrect results.

Interview Tips

Always clarify the constraints and rules of the problem before diving into coding.
Think about edge cases, such as when all walls have the same cost or time.
Explain your thought process while coding to demonstrate your understanding.

#2742

Painting the Walls

Hard

Array↗Dynamic Programming↗Dynamic Programming↗Greedy Algorithms↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

⚙️

Algorithm

4 steps

1Step 1: Initialize a DP array where dp[i] represents the minimum cost to paint the first i walls.
2Step 2: Iterate through each wall and for each wall, decide whether to use the paid painter or the free painter based on the time taken.
3Step 3: Update the DP array by considering the cost of using the paid painter for the current wall and the time taken for the previous walls.
4Step 4: Return the value in dp[n] which represents the minimum cost to paint all walls.

solution.py11 lines

1# Full working Python code
2
3def min_cost(cost, time):
4    n = len(cost)
5    dp = [float('inf')] * (n + 1)
6    dp[0] = 0
7    for i in range(1, n + 1):
8        for j in range(i):
9            dp[i] = min(dp[i], dp[j] + cost[j] + (i - j - 1))
10    return dp[n]
11

ℹ

Complexity note: The complexity is O(n²) because we are using a nested loop to fill the DP array, where n is the number of walls.

1The free painter can only be used when the paid painter is busy, which creates a dependency on the order of painting.
2Dynamic programming helps in breaking down the problem into smaller subproblems, allowing us to build the solution incrementally.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.