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;
}
};