House Robber II 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
House Robber II Solutions
✅Time: O(n)
✅Space: O(n)→O(1)
C++
class Solution {
public:
int rob(vector<int>& nums) {
if (nums.empty())
return 0;
if (nums.size() == 1)
return nums[0];
auto rob = [&](int l, int r) {
int prev1 = 0; // dp[i - 1]
int prev2 = 0; // dp[i - 2]
for (int i = l; i <= r; ++i) {
const int dp = max(prev1, prev2 + nums[i]);
prev2 = prev1;
prev1 = dp;
}
return prev1;
};
return max(rob(0, nums.size() - 2), rob(1, nums.size() - 1));
}
};
Java
class Solution {
public int rob(int[] nums) {
if (nums.length == 0)
return 0;
if (nums.length == 1)
return nums[0];
return Math.max(rob(nums, 0, nums.length - 2), rob(nums, 1, nums.length - 1));
}
private int rob(int[] nums, int l, int r) {
int prev1 = 0; // dp[i - 1]
int prev2 = 0; // dp[i - 2]
for (int i = l; i <= r; ++i) {
final int dp = Math.max(prev1, prev2 + nums[i]);
prev2 = prev1;
prev1 = dp;
}
return prev1;
}
}
Python
class Solution:
def rob(self, nums: List[int]) -> int:
def rob(l: int, r: int) -> int:
dp1 = 0
dp2 = 0
for i in range(l, r + 1):
temp = dp1
dp1 = max(dp1, dp2 + nums[i])
dp2 = temp
return dp1
if not nums:
return 0
if len(nums) < 2:
return nums[0]
return max(rob(0, len(nums) - 2), rob(1, len(nums) - 1))
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