198. 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: [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: [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.

dp state transition: dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])

// Dynamic Programming
int rob(vector<int>& nums) { // time: O(n); space: O(n)
    if (nums.empty()) return 0;
    int n = nums.size();
    vector<int> money(n, 0); // money[i] means the total amount a rob can rob from nums[0...i]
    for (int i = 0; i < n; ++i) {
        money[i] = max((i > 0 ? money[i - 1] : 0), (i > 1 ? money[i - 2] : 0) + nums[i]);
    }
    return money.back();
}

// Space Optimized Dynamic Programming
int rob(vector<int>& nums) { // time: O(n); space: O(1)
    if (nums.empty()) return 0;
    int n = nums.size();
    int two_step = 0, one_step = 0;
    for (int i = 0; i < n; ++i) {
        int cur = max(one_step, two_step + nums[i]);
        two_step = one_step;
        one_step = cur;
    }
    return one_step;
}