House Robber II


Problem Statement :


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 :



title-img 


                            Solution in C :

int rob(int* n, int ns)
{
	int pre,cur, t;
	int i, j, max = 0;
	for (j = 0; j < 3; ++j)
	{
		pre = cur = n[j];
		for (i = j + 2; i < (ns - (!j)); ++i)
		{
			t = cur>(pre + n[i]) ? cur : (pre + n[i]);
			pre = cur;
			cur = t;
		}
		max = max>cur ? max : cur;
	}
	return max;
}
                        


                        Solution in C++ :

class Solution {
public:
    int rob(vector<int>& nums) {
        int n = nums.size(); 
        if (n < 2) return n ? nums[0] : 0;
        return max(robber(nums, 0, n - 2), robber(nums, 1, n - 1));
    }
private:
    int robber(vector<int>& nums, int l, int r) {
        int pre = 0, cur = 0;
        for (int i = l; i <= r; i++) {
            int temp = max(pre + nums[i], cur);
            pre = cur;
            cur = temp;
        }
        return cur;
    }
};
                    


                        Solution in 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;
    int prev2 = 0;

    for (int i = l; i <= r; ++i) {
      final int dp = Math.max(prev1, prev2 + nums[i]);
      prev2 = prev1;
      prev1 = dp;
    }
    return prev1;
  }
}
                    


                        Solution in Python : 
                            
class Solution:
    def rob(self, nums: List[int]) -> int:
        if len(nums) <= 2:
            return max(nums)
        def recur(arr):
            if len(arr) <= 2:
                return max(arr)
            n_1 = max(arr[0], arr[1])
            n_2 = arr[0]
            n = max(n_1, n_2+arr[2])

            for i in range(3, len(arr)):
                n_2 = n_1
                n_1 = n
                n = max(n_1, n_2+arr[i])
            return n
        return max(recur(nums[:-1]), recur(nums[1:]))


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