How do you solve Maximum Energy Boost From Two Drinks on LeetCode?

Maximum Energy Boost From Two Drinks (LeetCode #3259) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Switching drinks incurs a penalty of one hour without energy.

What is the brute force solution for Maximum Energy Boost From Two Drinks?

The brute force approach for Maximum Energy Boost From Two Drinks is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying every possible combination of drinking the energy drinks over the hours. This means calculating the total energy for every possible sequence of drinks, which can be inefficient but helps understand the problem. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Energy Boost From Two Drinks?

Maximum Energy Boost From Two Drinks uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Array.

What are the interview tips for Maximum Energy Boost From Two Drinks?

Clarify the rules about switching drinks and the cleanse hour. Think about how to break the problem into smaller subproblems. Practice explaining your thought process as you code.

What are common mistakes when solving Maximum Energy Boost From Two Drinks?

Not accounting for the cleanse hour when switching drinks. Failing to initialize the dp arrays correctly.

#3259

Maximum Energy Boost From Two Drinks

Medium

ArrayDynamic ProgrammingDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to keep track of the maximum energy boost possible at each hour, considering whether the last drink was from A or B. This avoids redundant calculations and significantly reduces the time complexity.

⚙️

Algorithm

3 steps

1Step 1: Initialize two arrays dpA and dpB to store maximum energy boosts when the last drink is from A or B respectively.
2Step 2: Set dpA[0] and dpB[0] to the first elements of energyDrinkA and energyDrinkB.
3Step 3: Iterate through the hours, updating dpA and dpB based on the previous values, considering the cleanse hour when switching drinks.

solution.py11 lines

1def maxEnergyBoost(energyDrinkA, energyDrinkB):
2    n = len(energyDrinkA)
3    dpA = [0] * n
4    dpB = [0] * n
5    dpA[0] = energyDrinkA[0]
6    dpB[0] = energyDrinkB[0]
7    for i in range(1, n):
8        dpA[i] = max(dpA[i-1] + energyDrinkA[i], dpB[i-1])
9        dpB[i] = max(dpB[i-1] + energyDrinkB[i], dpA[i-1])
10    return max(dpA[n-1], dpB[n-1])
11

ℹ

Complexity note: The time complexity is linear because we only make a single pass through the arrays, updating our dp arrays in constant time for each hour. The space complexity is also linear due to the storage of the dp arrays.

1Switching drinks incurs a penalty of one hour without energy.
2Dynamic programming allows us to efficiently track maximum energy boosts.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.