198, House Robber

I Problem

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

Related Topics

• Array
• Dynamic Programming

II Solution

Approach 1: Dynamic Programming

pub fn rob(nums: Vec<i32>) -> i32 {
    //Self::dp(nums)

    Self::optimize_dp(nums)
}

fn dp(nums: Vec<i32>) -> i32 {
    let len = nums.len();
    match len {
        0 => 0,
        1 => nums[0],
        _ => {
            let mut dp = vec![0; len];
            (dp[0], dp[1]) = (nums[0], max(nums[0], nums[1]));

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

fn optimize_dp(nums: Vec<i32>) -> i32 {
    let len = nums.len();
    match len {
        0 => 0,
        1 => nums[0],
        _ => {
            let (mut first, mut second) = (nums[0], max(nums[0], nums[1]));

            for i in 2..len {
                (first, second) = (second, max(first + nums[i], second));
            }

            second
        }
    }
}

public int rob(int[] nums) {
    //return this.dp(nums);

    return this.optimizeDP(nums);
}

int dp(int[] nums) {
     switch (nums.length) {
        case 0 -> {
            return 0;
        }
         case 1 -> {
             return nums[0];
         }
         default -> {
            int[] dp = new int[nums.length];
            dp[0] = nums[0];
            dp[1] = Math.max(nums[0], nums[1]);

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

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

int optimizeDP(int[] nums) {
    switch (nums.length) {
        case 0 -> {
            return 0;
        }
        case 1 -> {
            return nums[0];
        }
        default -> {
            int first = nums[0], second = Math.max(nums[0], nums[1]);

            for (int i = 2; i < nums.length; i++) {
                int temp = second;
                second = Math.max(first + nums[i], second);
                first = temp;
            }

            return second;
        }
    }
}

func rob(nums []int) int {
    //return dp(nums)

    return optimizeDP(nums)
}

func dp(nums []int) int {
    size := len(nums)
    switch size {
    case 0:
        return 0
    case 1:
        return nums[0]
    default:
        dp := make([]int, size)
        dp[0], dp[1] = nums[0], max(nums[0], nums[1])
        
        for i := 2; i < size; i++ {
            dp[i] = max(dp[i-2]+nums[i], dp[i-1])
        }
        
        return dp[size-1]
    }
}

func optimizeDP(nums []int) int {
    size := len(nums)
    switch size {
    case 0:
        return 0
    case 1:
        return nums[0]
    default:
        first, second := nums[0], max(nums[0], nums[1])
    
        for i := 2; i < size; i++ {
            first, second = second, max(first+nums[i], second)
        }
    
        return second
    }
}