198. House Robber¶

Problem Description¶

LeetCode Problem 198: 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.

Clarification¶

No two adjacent houses can be robbed

Assumption¶

-

Solution¶

To solve the problem, we need to figure out:

Define state of the problem using a function or array. We can use rob(i) to represent the maximum amount of money at house i.
Find recurrence relation to transit between states. For house i, a robber has 2 options:
1. Rob the house. It means it can't rob the house before (i.e., house i - 1). He gets all cumulative money from house i - 2 and the money from house i.
2. Do not rob the house. He gets all cumulative money from house i - 1. So we can represent the above mentioned relation in the following equation:
\[ \text{rob}(i) = \max(\text{rob}(i - 2) + \text{nums}[i], \text{rob}(i - 1)) \]
Determine base cases to stop recurrence relation:
- \(\text{rob}(0) = \text{nums}[0]\)
- \(\text{rob}(1) = \max(\text{nums}[0], \text{nums}[1])\)

Approach 1: Recursive + Memoization¶

Based on the relationship, we can use recursive method with memoization to calculate the max amount of money for each house.

Python

class Solution:
def rob(self, nums: List[int]) -> int:
    self.memo = {}
    return self.rob_i(nums, len(nums) - 1)

def rob_i(self, nums: List[int], i: int) -> int:
    # Base case
    if i < 0:
        return 0

    # Memoization
    if i in self.memo:
        return self.memo[i]

    max_prev1 = self.rob_i(nums, i - 1)
    max_prev2 = self.rob_i(nums, i - 2)

    max_current = max(max_prev1, max_prev2 + nums[i])
    self.memo[i] = max_current  # Update memo
    return max_current

Complexity Analysis of Approach 1¶

Time complexity: \(O(n)\)
The recursive calls take at most \(n\) times due to memoization and each call takes \(O(1)\) time. So the time complexity is \(O(n)\).
Space complexity: \(O(n)\)
- The memoization table takes \(O(n)\) space to store max amount of money for each house.
- The recursion stack takes \(O(n)\) space since the recursion depth is at most \(n\).
- The total space complexity is \(O(n) + O(n) = O(n)\).

Approach 2: Iteration¶

From the relationship equation, to calculate the maximum amount of money of the current house, we only need to know the maximum amount of money of the previous two houses. So we can use iteration to calculate the maximum amount of money for each house by using two variables prev1 and prev2 (no need to use an array).

python

class Solution:
    def rob(self, nums: List[int]) -> int:
        n = len(nums)
        if n == 0:
            return 0
        prev1 = 0
        prev2 = 0
        for num in nums:
            curr = max(prev1, prev2 + num)
            prev1, prev2 = curr, prev1

        return curr

Complexity Analysis of Approach 2¶

Time complexity: \(O(n)\)
Go through \(n\) numbers using a for-loop and each iteration takes \(O(1)\) time. So the time complexity is \(O(n)\).
Space complexity: \(O(1)\)
Use 3 variables, prev1, prev2, and curr, to do the calculations. So the space complexity is \(O(1)\).

Comparison of Different Approaches¶

The table below summarize the time complexity and space complexity of different approaches:

Approach	Time Complexity	Space Complexity
Approach 1 - Recursion + Memoization	\(O(n)\)	\(O(n)\)
Approach 2 - Iteration	\(O(n)\)	\(O(1)\)

Test¶

Test empty list
Test single house
Test two houses
Test multiple houses