213. 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

Solution:

class Solution {
    public int rob(int[] nums) {
        // base case 
        if (nums == null || nums.length == 0){
            return 0;
        }

        if (nums.length == 1 ){
            return nums[0];
        }

        if (nums.length == 2){
            return Math.max(nums[0], nums[1]);
        }




        int start1 = 0;
        int end1 = nums.length - 2;
        int start2 = 1;
        int end2= nums.length -1;

        int rob1 = helper(start1, end1, nums);
        int rob2 = helper(start2, end2, nums);

        return Math.max(rob1, rob2);

    }

    private int helper(int start, int end, int[] nums){
       //  [1,2,3,     1]
       //       i
       //     f
       //   s
        int[] dp = new int[nums.length];
        dp[start] = nums[start];
        dp[start+1] = Math.max(nums[start+1], nums[start]); 
        for (int i = start + 2; i <= end; i++){

            dp[i] = Math.max(dp[i-1], dp[i-2] + nums[i]);

        }

        return dp[end];
    }
}

// TC: O(n)
// SC: O(n)

class Solution {
    public int rob(int[] nums) {
        // base case 
        if (nums == null || nums.length == 0){
            return 0;
        }

        if (nums.length == 1 ){
            return nums[0];
        }

        if (nums.length == 2){
            return Math.max(nums[0], nums[1]);
        }

        int start1 = 0;
        int end1 = nums.length - 2;
        int start2 = 1;
        int end2= nums.length -1;

        int rob1 = helper(start1, end1, nums);
        int rob2 = helper(start2, end2, nums);

        return Math.max(rob1, rob2);

    }

    private int helper(int start, int end, int[] nums){
        int fast = 0;
        int slow = 0;
        for (int i = start; i <= end; i++){
            int temp = fast; // 0 // 1// 2 
            int cur = nums[i];// 1 // 2  // 3
            // update fast
            fast = Math.max(temp, slow + cur);/// Math.(0, 1)
             /// (1, 0+2) = 2 // 1 + 3 = 4 , 2
            slow = temp;//  2 

        }

        return dp[end];
    }
}

// TC: O(n)
// SC: O(1)