Description
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 <= 1000 <= nums[i] <= 1000
Solutions
Solution 1: Dynamic Programming
The circular arrangement means that at most one of the first and last houses can be chosen for theft, so this circular arrangement problem can be reduced to two single-row house problems.
The time complexity is O(n), where n is the length of the array. The space complexity is O(1).
PythonJavaC++GoTypeScriptRust
class Solution:
def rob(self, nums: List[int]) -> int:
def _rob(nums):
f = g = 0
for x in nums:
f, g = max(f, g), f + x
return max(f, g)
if len(nums) == 1:
return nums[0]
return max(_rob(nums[1:]), _rob(nums[:-1]))(code-box)
class Solution {
public int rob(int[] nums) {
int n = nums.length;
if (n == 1) {
return nums[0];
}
return Math.max(rob(nums, 0, n - 2), rob(nums, 1, n - 1));
}
private int rob(int[] nums, int l, int r) {
int f = 0, g = 0;
for (; l <= r; ++l) {
int ff = Math.max(f, g);
g = f + nums[l];
f = ff;
}
return Math.max(f, g);
}
}(code-box)
class Solution {
public:
int rob(vector<int>& nums) {
int n = nums.size();
if (n == 1) {
return nums[0];
}
return max(robRange(nums, 0, n - 2), robRange(nums, 1, n - 1));
}
int robRange(vector<int>& nums, int l, int r) {
int f = 0, g = 0;
for (; l <= r; ++l) {
int ff = max(f, g);
g = f + nums[l];
f = ff;
}
return max(f, g);
}
};(code-box)
func rob(nums []int) int {
n := len(nums)
if n == 1 {
return nums[0]
}
return max(robRange(nums, 0, n-2), robRange(nums, 1, n-1))
}
func robRange(nums []int, l, r int) int {
f, g := 0, 0
for _, x := range nums[l : r+1] {
f, g = max(f, g), f+x
}
return max(f, g)
}(code-box)
function rob(nums: number[]): number {
const n = nums.length;
if (n === 1) {
return nums[0];
}
const robRange = (l: number, r: number): number => {
let [f, g] = [0, 0];
for (; l <= r; ++l) {
[f, g] = [Math.max(f, g), f + nums[l]];
}
return Math.max(f, g);
};
return Math.max(robRange(0, n - 2), robRange(1, n - 1));
}(code-box)
impl Solution {
pub fn rob(nums: Vec<i32>) -> i32 {
let n = nums.len();
if n == 1 {
return nums[0];
}
let rob_range = |l, r| {
let mut f = [0, 0];
for i in l..r {
f = [f[0].max(f[1]), f[0] + nums[i]];
}
f[0].max(f[1])
};
rob_range(0, n - 1).max(rob_range(1, n))
}
}(code-box)
