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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0101_0200.s0198_house_robber;

// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Array #Dynamic_Programming
// #Algorithm_I_Day_12_Dynamic_Programming #Dynamic_Programming_I_Day_3
// #Level_2_Day_12_Dynamic_Programming #Udemy_Dynamic_Programming #Big_O_Time_O(n)_Space_O(n)
// #2022_06_28_Time_0_ms_(100.00%)_Space_39.9_MB_(85.30%)

/**
 * 198 - House Robber\.
 *
 * Medium
 *
 * 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`
**/
public class Solution {
    public int rob(int[] nums) {
        if (nums.length == 0) {
            return 0;
        }
        if (nums.length == 1) {
            return nums[0];
        }
        if (nums.length == 2) {
            return Math.max(nums[0], nums[1]);
        }
        int[] profit = new int[nums.length];
        profit[0] = nums[0];
        profit[1] = Math.max(nums[1], nums[0]);
        for (int i = 2; i < nums.length; i++) {
            profit[i] = Math.max(profit[i - 1], nums[i] + profit[i - 2]);
        }
        return profit[nums.length - 1];
    }
}