🟨 House Robber (#198)
🔗 Problem Link
📋 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.
💡 Examples
Example 1
Input: nums = [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). Total = 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 = 2 + 9 + 1 = 12.
🔑 Key Insights & Approach
Core Observation: At each house, choose max of (rob current + skip previous) vs (skip current, keep previous max).
Why DP?
- dp[i] = max(dp[i-1], dp[i-2] + nums[i])
- O(n) time, O(1) space optimized
- Decision: rob or skip
Pattern: "Linear DP with Skip" pattern.
🐍 Solution: Python
Approach: Space-Optimized DP
Time Complexity: O(n) | Space Complexity: O(1)
from typing import List
class Solution:
def rob(self, nums: List[int]) -> int:
if not nums:
return 0
if len(nums) == 1:
return nums[0]
prev2, prev1 = 0, 0
for num in nums:
current = max(prev1, prev2 + num)
prev2 = prev1
prev1 = current
return prev1
🔵 Solution: Golang
Approach: O(1) Space
Time Complexity: O(n) | Space Complexity: O(1)
func rob(nums []int) int {
if len(nums) == 0 {
return 0
}
if len(nums) == 1 {
return nums[0]
}
prev2, prev1 := 0, 0
for _, num := range nums {
current := max(prev1, prev2+num)
prev2 = prev1
prev1 = current
}
return prev1
}
func max(a, b int) int {
if a > b {
return a
}
return b
}