LeetCode 213 House Robber II

Table of contents

Description

Question Links: LeetCode 213

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

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.

Example 3:

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

Constraints:

1 <= nums.length <= 100
0 <= nums[i] <= 1000

Solution 1: Two Linear Sub-Problems

Idea

Since the houses are arranged in a circle, house 0 and house n-1 are adjacent — we cannot rob both. This gives us a clean decomposition:

• Case A: exclude the last house → rob from [0, n-2]
• Case B: exclude the first house → rob from [1, n-1]

Each case is a standard linear House Robber problem. The answer is the maximum of the two.

houses (circle):    [2, 3, 2]
                     ^     ^--- these two are neighbors

Case A: [2, 3]    → rob house 1 → 3
Case B: [3, 2]    → rob house 1 → 3
answer: max(3, 3) = 3

houses (circle):    [1, 2, 3, 1]

Case A: [1, 2, 3]  → rob houses 0,2 → 1+3 = 4
Case B: [2, 3, 1]  → rob house 1    → 3
answer: max(4, 3) = 4

For each linear sub-problem, we use the two-variable DP from House Robber I:

• robPrev: max money if we robbed the previous house
• nRobPrev: max money if we skipped the previous house

Complexity: Time $O(n)$ — two passes over the array. Space $O(1)$ — two variables per pass.

Java

// O(n) time, O(1) space.
public int rob(int[] nums) {
    if (nums.length == 1) return nums[0];
    return Math.max(
        HouseRobber.rob(nums, 0, nums.length - 1),
        HouseRobber.rob(nums, 1, nums.length)
    );
}

// HouseRobber.rob — linear house robber on range [i, j)
public static int rob(int[] nums, int i, int j) {
    int robPrev = 0, nRobPrev = 0;
    for (int k = i; k < j; k++) { // O(n)
        int currRobbed = nRobPrev + nums[k];
        nRobPrev = Math.max(nRobPrev, robPrev);
        robPrev = currRobbed;
    }
    return Math.max(robPrev, nRobPrev);
}

Python

# O(n) time, O(1) space.
class Solution:
    def rob(self, nums: list[int]) -> int:
        def simple_rob(nums, l, r):
            did_not, robbed = 0, 0
            for i in range(l, r):  # O(n)
                t = did_not + nums[i]
                did_not = max(did_not, robbed)
                robbed = t
            return max(robbed, did_not)

        if len(nums) == 1:
            return nums[0]
        return max(simple_rob(nums, 0, len(nums) - 1),
                   simple_rob(nums, 1, len(nums)))

C++

// O(n) time, O(1) space.
class HouseRobberII {
    int robRange(vector<int> &nums, int lo, int hi) {
        int robPrev = 0, nRobPrev = 0;
        for (int i = lo; i < hi; i++) { // O(n)
            int robCur = nRobPrev + nums[i];
            nRobPrev = max(nRobPrev, robPrev);
            robPrev = robCur;
        }
        return max(robPrev, nRobPrev);
    }
public:
    int rob(vector<int> &nums) {
        int n = nums.size();
        if (n == 1) return nums[0];
        return max(robRange(nums, 0, n - 1), robRange(nums, 1, n));
    }
};

Rust

/// O(n) time, O(1) space.
pub fn rob(nums: Vec<i32>) -> i32 {
    let n = nums.len();
    if n == 0 { return 0; }
    if n == 1 { return nums[0]; }
    Self::rob_linear(&nums[..n - 1]).max(Self::rob_linear(&nums[1..]))
}

fn rob_linear(nums: &[i32]) -> i32 {
    let (mut rob_prev, mut n_rob_prev) = (0, 0);
    for &n in nums { // O(n)
        let rob_cur = n_rob_prev + n;
        n_rob_prev = n_rob_prev.max(rob_prev);
        rob_prev = rob_cur;
    }
    rob_prev.max(n_rob_prev)
}