Question

Formatted question description: https://leetcode.ca/all/198.html

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems 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 = [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 2:

Input: nums = [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.

Constraints:

1 <= nums.length <= 100
0 <= nums[i] <= 400

Algorithm

Dynamic Programming to solve, maintain a 1D array dp, where dp[i] represents the maximum value that can be grabbed in the [0, i] interval.

For the current i, there are two mutually exclusive options for robbing and not robbing. Not robbing is dp[i-1] (equivalent to removing nums[i], while robbing is the maximum value of [0, i-1]), robbing is dp[i-2] + nums[i] (equivalent to removing nums[i-1]).

dp[i] = max(num[i] + dp[i - 2], dp[i - 1])

Code

public class House_Robber {


    class Solution {
        public int rob(int[] nums) {

            if (nums == null || nums.length == 0) {
                return 0;
            }

            // dp[i] means until i, max possible amount
            int[] dp = new int[nums.length + 1];

            dp[0] = 0;
            dp[1] = nums[0];

            for (int i = 2; i <= nums.length; i++) {
                // 2 cases: rob current house, not rob current
                dp[i] = Math.max(nums[i - 1] + dp[i - 2], dp[i - 1]);
            }

            return dp[nums.length];
        }
    }
}

/////////

class Solution {
    public int rob(int[] nums) {
        int a = 0, b = nums[0];
        for (int i = 1; i < nums.length; ++i) {
            int c = Math.max(nums[i] + a, b);
            a = b;
            b = c;
        }
        return b;
    }
}

// rob[i + 1] = nums[i] + skip[i] // If we rob at house[i], we must skip house[i-1]
// skip[i + 1] = max(rob[i - 1], skip[i - 1]) // If we skip house[i], we can pick the maximum from robbing or skipping house[i-1]

class Solution {
public:
    int rob(vector<int>& nums) {
        int n = nums.size();
        int a = 0, b = nums[0];
        for (int i = 1; i < n; ++i) {
            int c = max(nums[i] + a, b);
            a = b;
            b = c;
        }
        return b;
    }
};

# greedy
class Solution:
    def rob(self, nums: List[int]) -> int:
        not_rob, rob = 0, nums[0]
        for num in nums[1:]:
            # must max check
            # eg. first robbed 99999, then following ones are just 3,3,3,3,3
            not_rob, rob = rob, max(num + not_rob, rob)
        return rob

############

class Solution(object):
  def rob(self, nums):
    """
    :type nums: List[int]
    :rtype: int
    """
    if len(nums) == 0:
      return 0
    if len(nums) <= 2:
      return max(nums)
    dp = [0 for i in range(0, 2)]
    dp[0] = nums[0]
    dp[1] = max(nums[1], nums[0])
    for i in range(2, len(nums)):
      dp[i % 2] = max(dp[(i - 1) % 2], dp[(i - 2) % 2] + nums[i])
    return dp[(len(nums) - 1) % 2]

func rob(nums []int) int {
    a, b, n := 0, nums[0], len(nums)
    for i := 1; i < n; i++ {
        a, b = b, max(nums[i] + a, b)
    }
    return b
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

function rob(nums: number[]): number {
    const dp = [0, 0];
    for (const num of nums) {
        [dp[0], dp[1]] = [dp[1], Math.max(dp[1], dp[0] + num)];
    }
    return dp[1];
}

impl Solution {
    pub fn rob(nums: Vec<i32>) -> i32 {
        let mut dp = [0, 0];
        for num in nums {
            dp = [dp[1], dp[1].max(dp[0] + num)]
        }
        dp[1]
    }
}

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