House Robber (#198 Leetcode)

Leetcode Problem Statement

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

 

class Solution {
public:
    int rob(vector<int>& nums) {
        int len = nums.size();
        int dp[len];
        dp[0] = nums[0]; 
        
        if (nums.size() > 1) {
            dp[1] = max(nums[1], dp[0]);
        }
        
        for (int i = 2; i < len; i++) {
            dp[i] = max(dp[i-2] + nums[i], dp[i-1]); 
        }
        return dp[len-1];
    }
};

//The following testcases will be executed:
//[2, 1, 1, 2] => 4
// [1, 2] => 2
// [1] => 1
// [0] => 0
// [1, 2, 3, 1] => 4

The below line means, 

            dp[i] = max(dp[i-2] + nums[i], dp[i-1]); 

Storing the maximum of either of these below in the current dp array element. 

 If we want to store the previous dp array value

 or 

If we want to store the previous max robbed house stash & the current stash