The challenge
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 a list of non-negative integers 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 = [0] Output: 0
Constraints:
1 <= nums.length <= 100
0 <= nums[i] <= 1000
The solution in Java code
class Solution { public int rob(int[] nums) { 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 start, int end) { int a = 0, b = 0; for (int i = start; i <= end; i++) { int temp = b; if (nums[i] + a > b) b = nums[i] + a; a = temp; } return b; } }
An Alternate View at the Acceptance Criteria
If we take the first example and say that:
Input: nums = [2,3,2] Output: 4 Explanation: You can rob house 1 (money = 2) and rob house 3 (money = 2), because they are NOT adjacent houses. As they have house 2 (money = 3) inbetween them.
If we take this new viewpoint, then we can write the following code to resolve the problem:
class Solution { public int rob(int[] nums) { // start from 0, then skip until end, build a total int total1 = 0; // start from 1, then skip until end, build a total int total2 = 0; for (int i=0; i<=nums.length; i++) { int val = (i==nums.length) ? 0 : nums[i]; if (i%2==0) total1 += val; else total2 += val; } // compare the totals and give the higher one return Math.max(total1, total2); } }