Question

Formatted question description: https://leetcode.ca/all/198.html

 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 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: [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: [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.

 @tag-dp

Algorithm

Dynamic Programming to solve, maintain a 1D array dp, where dp[i] represents the maximum value that can be grabbed in the [0, i] interval.

For the current i, there are two mutually exclusive options for robbing and not robbing. Not robbing is dp[i-1] (equivalent to removing nums[i], while robbing is the maximum value of [0, i-1]), robbing is dp[i-2] + nums[i] (equivalent to removing nums[i-1]).

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

Code

Java

  • 
    public class House_Robber {
    
    
        class Solution {
            public int rob(int[] nums) {
    
                if (nums == null || nums.length == 0) {
                    return 0;
                }
    
                // dp[i] means until i, max possible amount
                int[] dp = new int[nums.length + 1];
    
                dp[0] = 0;
                dp[1] = nums[0];
    
                for (int i = 2; i <= nums.length; i++) {
                    // 2 cases: rob current house, not rob current
                    dp[i] = Math.max(nums[i - 1] + dp[i - 2], dp[i - 1]);
                }
    
                return dp[nums.length];
            }
        }
    }
    
  • rob[i + 1] = nums[i] + skip[i] // If we rob at house[i], we must skip house[i-1]
    skip[i + 1] = max(rob[i - 1], skip[i - 1]) // If we skip house[i], we can pick the maximum from robbing or skipping house[i-1]
    
  • class Solution(object):
      def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums) == 0:
          return 0
        if len(nums) <= 2:
          return max(nums)
        dp = [0 for i in range(0, 2)]
        dp[0] = nums[0]
        dp[1] = max(nums[1], nums[0])
        for i in range(2, len(nums)):
          dp[i % 2] = max(dp[(i - 1) % 2], dp[(i - 2) % 2] + nums[i])
        return dp[(len(nums) - 1) % 2]
    
    

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