# 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;
dp = nums;

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;
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;
for (int i = 1; i < n; ++i) {
int c = max(nums[i] + a, b);
a = b;
b = c;
}
return b;
}
};

• class Solution:
def rob(self, nums: List[int]) -> int:
not_rob, rob = 0, nums
for num in nums[1:]:
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 = nums
dp = max(nums, nums)
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, 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, dp] = [dp, Math.max(dp, dp + num)];
}
return dp;
}


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