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House Robber | Leetcode 198 | leetcode solution

198. House Robber

 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 : 


suppose we have a state of dp , where dp[n] represents the maximum amount of money collected 
in n days without police alert.
 we can clearly say that dp[n] = dp[n-2] + nums[n] .. i.e maximum amount of money collected the day before Tomorow plus money collected on current day(nth day). 

class Solution {
public:
    int rob(vector<int>& nums) {
        
        int n = nums.size();
        vector<int>dp(n);
        
        dp[0] = nums[0];
         if(n == 1 )
        {
           return dp[n-1];
        }
        dp[1] = max(nums[1], nums[0]);
          if(n == 2 )
        {
           return dp[n-1];
        }    
        dp[2] = nums[2] + dp[0];
        
        for(int i = 3 ; i < n ; i++)
        {
             dp[i] = max(dp[i-2],dp[i-3]) + nums[i];
        }
        
        return max({dp[n-1], dp[n-2]});
    }
};


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