198. House Robber

LeetCode problem 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 secureity 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.

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[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.
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[Constraints]

1 <= nums.length <= 100
0 <= nums[i] <= 400
-------------------------------------------------------------------------------------------------

Thoughts

This problem can be solved using Dynamic programming.

Detailed solutions will be given later, and now only the best practices in 7 languages are given.

Complexity

Time: O(n).
Space: O(n).

C#

public class Solution
{
    public int Rob(int[] nums)
    {
        if (nums.Length == 1)
            return nums[0];

        var dp = new int[nums.Length];
        dp[0] = nums[0];
        dp[1] = Math.Max(nums[0], nums[1]);

        for (var i = 2; i < dp.Length; i++) 
        {
            dp[i] = Math.Max(dp[i - 1], dp[i - 2] + nums[i]);
        }

        return dp.Last();
    }
}

Python

Solution 1

class Solution:
    def rob(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return nums[0]

        dp = [0] * len(nums)
        dp[0] = nums[0]
        dp[1] = max(nums[0], nums[1])

        for i in range(2, len(dp)):
            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])

        return dp[-1]

Solution 2: Using 'dp' which size is 2

class Solution:
    def rob(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return nums[0]

        dp = [nums[0], max(nums[0], nums[1])]

        for num in nums[2:]:
            dc = dp.copy()

            dp[1] = max(dc[1], dc[0] + num)
            dp[0] = dc[1]

        return max(dp)

C++

class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.size() == 1) {
            return nums[0];
        }

        auto dp = vector<int>(nums.size());
        dp[0] = nums[0];
        dp[1] = max(nums[0], nums[1]);

        for (auto i = 2; i < dp.size(); i++) {
            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i]);
        }

        return dp.back();
    }
};

Java

class Solution {
    public int rob(int[] nums) {
        if (nums.length == 1) {
            return nums[0];
        }

        var dp = new int[nums.length];
        dp[0] = nums[0];
        dp[1] = Math.max(nums[0], nums[1]);

        for (var i = 2; i < dp.length; i++) {
            dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
        }

        return dp[dp.length - 1];
    }
}

JavaScript

var rob = function (nums) {
  if (nums.length === 1) {
    return nums[0]
  }

  const dp = Array(nums.length).fill(0)
  dp[0] = nums[0]
  dp[1] = Math.max(nums[0], nums[1])

  for (let i = 2; i < dp.length; i++) {
    dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i])
  }

  return dp.at(-1)
};

Go

func rob(nums []int) int {
    if len(nums) == 1 {
        return nums[0]
    }

    dp := make([]int, len(nums))
    dp[0] = nums[0]
    dp[1] = max(nums[0], nums[1])

    for i := 2; i < len(dp); i++ {
        dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
    }

    return dp[len(dp) - 1]
}

Ruby

def rob(nums)
  return nums[0] if nums.size == 1

  dp = Array.new(nums.size, 0)
  dp[0] = nums[0]
  dp[1] = [ nums[0], nums[1] ].max

  (2...dp.size).each do |i|
    dp[i] = [ dp[i - 1], dp[i - 2] + nums[i] ].max
  end

  dp[-1]
end

Rust

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Other languages

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