#3259

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Maximum Energy Boost From Two Drinks

Maximize energy boost from two drinks with a state transition dynamic programming approach.

Problem Statement

You are given two arrays, energyDrinkA and energyDrinkB, representing energy boosts per hour from two different energy drinks. Both arrays are of equal length, n. In each hour, you can choose one of the drinks, but switching between drinks requires an hour of cleansing, meaning you will not get an energy boost during that hour.

Your task is to maximize the total energy boost you can receive over n hours by making the best drink choices while accounting for the cleansing time when switching between drinks.

Examples

Example 1

Input: energyDrinkA = [1,3,1], energyDrinkB = [3,1,1]

Output: 5

To gain an energy boost of 5, drink only the energy drink A (or only B).

Example 2

Input: energyDrinkA = [4,1,1], energyDrinkB = [1,1,3]

Output: 7

To gain an energy boost of 7:

Constraints

n == energyDrinkA.length == energyDrinkB.length
3 <= n <= 105
1 <= energyDrinkA[i], energyDrinkB[i] <= 105

Solution Approach

State Transition Dynamic Programming

Use dynamic programming to track the maximum energy boost at each hour, considering whether to stay with the same drink or switch. Define two states: one for choosing drink A and one for choosing drink B, and optimize transitions between these states.

State Transition with Cleansing Period

When switching between drinks, introduce a cleansing period (no energy boost). Ensure the dynamic programming approach handles this by tracking the last hour you switched and adjusting the energy accordingly.

Iterative Solution with Time Complexity Optimization

Iterate through the hours while keeping track of the best possible energy boost for each drink choice. Use space optimization by only storing the current and previous states, reducing the space complexity.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the dynamic programming approach, usually O(n) for each array, and the space complexity can be optimized to O(1) if only the last two states are stored.

What Interviewers Usually Probe

Can the candidate demonstrate an understanding of dynamic programming with state transitions?
Does the candidate optimize space complexity by using only relevant states?
Does the candidate effectively incorporate the cleansing time when switching between drinks?

Common Pitfalls or Variants

Common pitfalls

Failing to handle the cleansing period correctly when switching drinks.
Overcomplicating the dynamic programming approach and not optimizing space.
Not considering the impact of switching drinks at each step, leading to suboptimal solutions.

Follow-up variants

Solving the problem with more than two energy drinks.
Optimizing the solution further using sliding window techniques.
Adapting the problem to handle a different number of hours or different energy drink properties.

FAQ

What is the primary approach for solving the Maximum Energy Boost From Two Drinks problem?

The primary approach is using state transition dynamic programming, where you track the maximum energy boost while considering possible switches between drinks and the cleansing period.

How do I handle the cleansing time when switching between energy drinks?

In the dynamic programming approach, you account for a cleansing hour by adjusting the energy boost during transitions, ensuring no boost is gained during the switching hour.

What is the time complexity of the dynamic programming approach in this problem?

The time complexity of this approach is O(n), where n is the number of hours or length of the energy drink arrays.

Can the problem be solved using a greedy algorithm?

A greedy algorithm is not suitable for this problem because it doesn't account for the required cleansing time when switching drinks, which can lead to suboptimal solutions.

What other dynamic programming problems have a similar state transition pattern?

Problems like the 'House Robber' and 'Word Break' share a similar state transition dynamic programming approach, where you optimize decisions based on previous states.

terminal

Solution

Solution 1: Dynamic Programming

We define $f[i][0]$ to represent the maximum boost energy obtained by choosing energy drink A at the $i$-th hour, and $f[i][1]$ to represent the maximum boost energy obtained by choosing energy drink B at the $i$-th hour. Initially, $f[0][0] = \textit{energyDrinkA}[0]$, $f[0][1] = \textit{energyDrinkB}[0]$. The answer is $\max(f[n - 1][0], f[n - 1][1])$.

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

Solution 2: Dynamic Programming (Space Optimization)

We notice that the state $f[i]$ is only related to $f[i - 1]$ and not to $f[i - 2]$. Therefore, we can use only two variables $f$ and $g$ to maintain the state, thus optimizing the space complexity to $O(1)$.

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

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