How do you solve House Robber V on LeetCode?

House Robber V (LeetCode #3840) 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: Non-adjacent houses must have different colors.

What is the brute force solution for House Robber V?

The brute force approach for House Robber V is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all possible combinations of houses to rob, checking if they are valid based on color and adjacency. This will give us the maximum amount of money. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve House Robber V?

House Robber V uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for House Robber V?

Explain your thought process clearly. Discuss edge cases and constraints upfront. Practice similar problems to build confidence.

What are common mistakes when solving House Robber V?

Forgetting to check color constraints. Not handling edge cases like single house.

#3840

House Robber V

Medium

ArrayDynamic ProgrammingHash MapArray

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)

We can use dynamic programming to keep track of the maximum money that can be robbed up to each house while respecting the color and adjacency constraints.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP array where dp[i] represents the maximum money robbed up to house i.
2Step 2: For each house, decide whether to rob it or skip it based on the previous house's color and amount.
3Step 3: Return the last element of the DP array as the result.

solution.py11 lines

1def rob(nums, colors):
2    n = len(nums)
3    dp = [0] * n
4    dp[0] = nums[0]
5    for i in range(1, n):
6        dp[i] = dp[i - 1]
7        for j in range(i - 1, -1, -1):
8            if colors[j] != colors[i]:
9                dp[i] = max(dp[i], dp[j] + nums[i])
10                break
11    return max(dp)

ℹ

Complexity note: The time complexity is quadratic due to the nested loop checking previous houses, while space is linear for the DP array.

1Non-adjacent houses must have different colors.
2Dynamic programming can efficiently track maximum values.

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