198. House Robber

1. Description

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.

2. Example

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.

3. Constraints

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

4. Solutions

Dynamic Programming

n = nums.size()
Time complexity: O(n)
Space complexity: O(1)

// dp[i] represents the maximum amount of money that can be robbed
// from houses [0 .. i] (inclusive)
//
// Base cases:
//   dp[0] = nums[0]                         // only one house
//   dp[1] = max(nums[0], nums[1])           // can only rob one of the two
//
// Transition:
//   dp[i] = max(dp[i - 2] + nums[i], dp[i - 1])
//   - dp[i - 2] + nums[i]: rob the current house, so we must skip i - 1
//   - dp[i - 1]:           do not rob the current house, keep the best so far
//
// Space optimization:
//   Instead of a full dp array, we use rolling variables:
//   a = dp[i - 2], b = dp[i - 1]
class Solution {
public:
    int rob(const vector<int> &nums) {
        if (nums.size() == 1) {
            return nums[0];
        }

        int a = nums[0], b = max(nums[0], nums[1]);
        for (int i = 2; i < nums.size(); ++i) {
            const int backup = b;
            b = max(a + nums[i], b);
            a = backup;
        }
        return b;
    }
};