Paint House IV

The Paint House IV problem requires calculating the minimum cost to paint a set of houses based on adjacency and color constraints. Use dynamic programming to track the minimum cost at each house while ensuring no two consecutive houses are painted the same color. With an optimal solution, this problem highlights state transition dynamic programming to manage costs effectively.

Problem Statement

You are given an even number of houses, n, and a 2D array cost, where cost[i][j] represents the cost of painting house i with color j+1. Your goal is to determine the minimum cost to paint all houses such that the houses look beautiful, meaning no two adjacent houses share the same color. The sequence of painting must also ensure that adjacent houses do not have the same color, while minimizing the total cost.

Given the constraints, you need to devise an optimal strategy to compute the minimum cost while taking advantage of dynamic programming to account for previous choices at each house. The solution must adhere to the adjacency condition while minimizing the cost of painting all the houses.

Examples

Example 1

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

Output: 9

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

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: 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

Solution Approach

Dynamic Programming Approach

Use dynamic programming to keep track of the minimum cost of painting the houses while respecting the adjacency constraint. Define a DP table where each state represents the minimum cost of painting the houses up to the current house, ensuring no adjacent houses have the same color.

State Transitions

Each house can be painted with one of three colors, but consecutive houses must be painted with different colors. For each house, choose the color that minimizes the total cost while considering the previous house's color to avoid conflicts.

Optimization and Time Complexity

Optimize the dynamic programming solution by reducing the state space and ensuring that the solution is computed in O(n) time. This avoids brute-force search, making the solution feasible for large values of n up to 100,000.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the dynamic programming approach is O(n) since each house is processed once, and for each house, we make constant time decisions regarding the color choice. The space complexity is O(n) as we need to store the DP states for each house. Optimizations can reduce the space complexity further by storing only the last computed state.

What Interviewers Usually Probe

• The candidate should demonstrate an understanding of dynamic programming and how it can be applied to minimize costs while adhering to constraints.
• Expect the candidate to efficiently implement state transitions and explain how previous choices influence subsequent decisions.
• The candidate should be able to optimize the space and time complexity of their solution, making it feasible for larger inputs.

Common Pitfalls or Variants

Common pitfalls

• Overcomplicating the solution with unnecessary state space, which increases both time and space complexity.
• Incorrectly handling the adjacency constraints, which could lead to suboptimal solutions or incorrect outputs.
• Failing to optimize the solution for large inputs, which could result in time or memory limits being exceeded.

Follow-up variants

• Paint House IV can be extended to different numbers of colors, which adds complexity to the dynamic programming approach.
• If the number of houses is odd, adjustments are required to handle the constraints properly.
• The problem can also be solved with a greedy approach, but it may not always guarantee an optimal solution.

FAQ

How do I minimize the painting costs for Paint House IV?

You can minimize the painting costs by using dynamic programming to track the optimal cost at each house while ensuring adjacent houses are not painted the same color.

What is the time complexity of solving Paint House IV?

The time complexity of solving the Paint House IV problem is O(n), as you process each house once and make constant time decisions for each.

Can I use a greedy algorithm for Paint House IV?

A greedy algorithm may not guarantee the optimal solution, as it could overlook the best long-term painting sequence. Dynamic programming ensures the minimum cost is achieved.

What does state transition dynamic programming mean in Paint House IV?

State transition dynamic programming means tracking the optimal painting cost for each house while considering the previous house's color to ensure adjacent houses have different colors.

How does GhostInterview help with Paint House IV?

GhostInterview guides you in applying dynamic programming to minimize painting costs, offering feedback on common pitfalls and helping optimize your solution for large inputs.