How do you solve Paint House IV on LeetCode?

Paint House IV (LeetCode #3429) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The adjacency constraint requires careful consideration of the previous house's color.

What is the brute force solution for Paint House IV?

The brute force approach for Paint House IV is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible combination of painting the houses while checking the constraints. This is straightforward but inefficient as it explores all possibilities. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Paint House IV?

Paint House IV uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Array.

What are the interview tips for Paint House IV?

Clearly explain your thought process and constraints before jumping into coding. Consider edge cases, such as the minimum number of houses. Use comments to clarify your logic in the code, especially in dynamic programming solutions.

What are common mistakes when solving Paint House IV?

Failing to check both adjacency and equidistant constraints simultaneously. Overlooking the initialization of the DP table leading to incorrect results.

#3429

Paint House IV

Medium

ArrayDynamic ProgrammingDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to build up the minimum cost while ensuring the constraints are met. This approach is efficient as it avoids redundant calculations.

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Algorithm

3 steps

1Step 1: Initialize a DP table to store the minimum costs for each house and color.
2Step 2: For each house, calculate the minimum cost for each color considering the previous house's colors and the equidistant constraints.
3Step 3: Return the minimum value from the last house's costs.

solution.py9 lines

1# Full working Python code
2def minCost(n, cost):
3    dp = [[0]*3 for _ in range(n)]
4    for j in range(3):
5        dp[0][j] = cost[0][j]
6    for i in range(1, n):
7        for j in range(3):
8            dp[i][j] = cost[i][j] + min(dp[i-1][(j+1)%3], dp[i-1][(j+2)%3])
9    return min(dp[n-1])

ℹ

Complexity note: The time complexity is O(n) because we iterate through each house once, and for each house, we perform a constant amount of work (calculating the minimum cost for 3 colors).

1The adjacency constraint requires careful consideration of the previous house's color.
2The equidistant constraint can be managed by ensuring that colors at mirrored positions are different.

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