Problem
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
Solution
/**
* @param {number[]} nums
* @return {number}
*/
var rob = function(nums) {
var arr1 = Array(nums.length); // do not rob the first house
var arr2 = Array(nums.length); // rob the first house
for (var i = nums.length - 1; i > 0; i--) {
arr1[i] = Math.max(nums[i] + (arr1[i + 2] || 0), arr1[i + 1] || 0);
arr2[i] = i === nums.length - 1 ? 0 : Math.max(nums[i] + (arr2[i + 2] || 0), arr2[i + 1] || 0);
}
return Math.max(nums[0] + (arr2[2] || 0), arr1[1] || 0);
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).