198. House Robber

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

Python Code
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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 Code
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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++ Code
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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 Code
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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 Code
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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 Code
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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

Python Code
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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 Code
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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++ Code
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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 Code
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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 Code
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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);
}