House Robber

House Robber

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 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 representing the amount of money of each house, determine 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:

• 0 <= nums.length <= 100

• 0 <= nums[i] <= 400

Solutions

🧠 Cpp

class Solution
{    
    //DP storage for function results
    map<vector<int>, int> cache;

public:    
    int rob(vector<int> money)
    {
        if(money.empty())
            return 0;

        //DP check cache
        auto found = cache.find(money);
        if(found != end(cache))
        {
            return found->second;
        }

        //if we can skip
        bool has_next_house = next(begin(money)) != end(money);

        //robbing the first house and skipping the next one (next(begin(money),2)
        int robbing_first_house_scenario =  
            money.front() + (has_next_house ? rob(vector<int>(next(begin(money),2), end(money))) : 0 );
        //second scenario is we skipping the current house and go right to the next one
        int skipping_first_house_scenario = rob(vector<int>(next(begin(money)), end(money)));

        int res = max(robbing_first_house_scenario, skipping_first_house_scenario);

        //DP add to cache
        cache.insert(make_pair(money, res));
        return res;
    }
};