House Robber

Problem

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

Solution

/**
 * @param {number[]} nums
 * @return {number}
 */
var rob = function(nums) {
    const dp = [0, nums[0]];
    for (let i = 1; i < nums.length; i++) {
        dp.push(Math.max(dp[i - 1] + nums[i], dp[i]));
    }
    return dp[nums.length];
};

We will implement a DP solution. Let dp[i] be the maximum amount of money that can be robbed with houses nums[0:i].

For the base case, if i = 0, then 0 amount of money can be robbed, since nums[0:0] is the empty set, so there are no houses to rob; if i = 1, then nums[0] can be robbed, since there's only a single house with money nums[0].
For the recursive case, because consecutive houses cannot be robbed, the max is either including the current house nums[i] plus the max of nums[0:i - 1] (ie. max of previous houses not consecutive to the current house); or the max of not including the current house nums[0:i].

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