Easy DP

Adobe

Amazon

Amazon

Apple

Cisco

Google

Microsoft

eBay

eBay

2020-09-29

198. House Robber

Similar Question: LeetCode Question 213

Question:

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

Solution:

The DP formula is that curr_max = Math.max(num[curr - 2] + num[curr], num[curr - 1]).

class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        if (nums.length == 1) return nums[0];
        nums[1] = Math.max(nums[0], nums[1]);
        for (int i = 2; i < nums.length; i++) {
            nums[i] = Math.max(nums[i - 2] + nums[i], nums[i - 1]);                
        }   
        return nums[nums.length - 1];
    }
}

Or using two pointer to record the prev (prevRob) and the one before prev(prevNotRob)

class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        if (nums.length == 1) return nums[0];
        int prevRob = 0;
        int prevNotRob = 0;
        for (int i = 0; i < nums.length; i++) {
            int curr = prevNotRob + nums[i];
            prevNotRob = Math.max(prevRob, prevNotRob);
            prevRob = Math.max(prevRob, curr);
        }
        return Math.max(prevNotRob, prevRob);
    }
}