89. House Robber II

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Step 1 / 33

Speed: 1.0x

Circular Setup

🏠$4

House 0

🏠$2

House 1

🏠$3

House 2

🏠$1

House 3

Why this works

Since the houses are in a circle, we cannot rob both the first and last houses.

This circular problem can be simplified into two linear subproblems (House Robber I):

Case 1: Rob houses 0 to n-2 (skipping the last house).
Case 2: Rob houses 1 to n-1 (skipping the first house).

Taking the maximum of these two linear cases yields the optimal circular solution.

Step Explanation

Function started. Since the houses are circular, robbing both the first and last houses is forbidden.

Variables

nums[4, 2, 3, 1]

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Example 1:

Input: nums = [2,3,2]

Output: 3

Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.

Example 2:

Input: nums = [1,2,3,1]

Output: 4

Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). Total amount you can rob = 1 + 3 = 4.

Constraints:

•
•

Houses are circular: can't rob both first and last. Split into two cases: (1) rob houses 0..n-2, (2) rob houses 1..n-1. Apply House Robber I logic to each case and take maximum.

Time Complexity

O(n)

Space Complexity

O(1)

89. House Robber II

Understand Before Coding

Visualize This Problem

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Example 1:

Example 2:

Constraints:

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89. House Robber II

Understand Before Coding

Why this works

Step Explanation

Variables

Visualize This Problem

Draw Your Approach

Example 1:

Example 2:

Constraints:

Learn

Topics

Companies

Hint

Use Cases

Why this works

Step Explanation

Variables