README_EN.md

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true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0100-0199/0198.House%20Robber/README_EN.md	Array Dynamic Programming

198. House Robber

中文文档

Description

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

Solutions

Solution 1: Dynamic Programming

We define $f[i]$ as the maximum total amount that can be robbed from the first $i$ houses, initially $f[0]=0$, $f[1]=nums[0]$.

Consider the case where $i \gt 1$, the $i$th house has two options:

• Do not rob the $i$th house, the total amount of robbery is $f[i-1]$;
• Rob the $i$th house, the total amount of robbery is $f[i-2]+nums[i-1]$;

Therefore, we can get the state transition equation:

$$ f[i]= \begin{cases} 0, & i=0 \\ nums[0], & i=1 \\ \max(f[i-1],f[i-2]+nums[i-1]), & i \gt 1 \end{cases} $$

The final answer is $f[n]$, where $n$ is the length of the array.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array.

Python3

class Solution:
    def rob(self, nums: List[int]) -> int:
        n = len(nums)
        f = [0] * (n + 1)
        f[1] = nums[0]
        for i in range(2, n + 1):
            f[i] = max(f[i - 1], f[i - 2] + nums[i - 1])
        return f[n]

Java

class Solution {
    public int rob(int[] nums) {
        int n = nums.length;
        int[] f = new int[n + 1];
        f[1] = nums[0];
        for (int i = 2; i <= n; ++i) {
            f[i] = Math.max(f[i - 1], f[i - 2] + nums[i - 1]);
        }
        return f[n];
    }
}

C++

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

Go

func rob(nums []int) int {
	n := len(nums)
	f := make([]int, n+1)
	f[1] = nums[0]
	for i := 2; i <= n; i++ {
		f[i] = max(f[i-1], f[i-2]+nums[i-1])
	}
	return f[n]
}

TypeScript

function rob(nums: number[]): number {
    const n = nums.length;
    const f: number[] = Array(n + 1).fill(0);
    f[1] = nums[0];
    for (let i = 2; i <= n; ++i) {
        f[i] = Math.max(f[i - 1], f[i - 2] + nums[i - 1]);
    }
    return f[n];
}

Rust

impl Solution {
    pub fn rob(nums: Vec<i32>) -> i32 {
        let mut f = [0, 0];
        for x in nums {
            f = [f[0].max(f[1]), f[0] + x];
        }
        f[0].max(f[1])
    }
}

Solution 2: Dynamic Programming (Space Optimization)

We notice that when $i \gt 2$, $f[i]$ is only related to $f[i-1]$ and $f[i-2]$. Therefore, we can use two variables instead of an array to reduce the space complexity to $O(1)$.

Python3

class Solution:
    def rob(self, nums: List[int]) -> int:
        f = g = 0
        for x in nums:
            f, g = max(f, g), f + x
        return max(f, g)

Java

class Solution {
    public int rob(int[] nums) {
        int f = 0, g = 0;
        for (int x : nums) {
            int ff = Math.max(f, g);
            g = f + x;
            f = ff;
        }
        return Math.max(f, g);
    }
}

C++

class Solution {
public:
    int rob(vector<int>& nums) {
        int f = 0, g = 0;
        for (int& x : nums) {
            int ff = max(f, g);
            g = f + x;
            f = ff;
        }
        return max(f, g);
    }
};

Go

func rob(nums []int) int {
	f, g := 0, 0
	for _, x := range nums {
		f, g = max(f, g), f+x
	}
	return max(f, g)
}

TypeScript

function rob(nums: number[]): number {
    let [f, g] = [0, 0];
    for (const x of nums) {
        [f, g] = [Math.max(f, g), f + x];
    }
    return Math.max(f, g);
}