Introduction

In this blog post, we will explore how to solve the LeetCode problem "3259" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given two integer arrays energyDrinkA and energyDrinkB of the same length n by a futuristic sports scientist. These arrays represent the energy boosts per hour provided by two different energy drinks, A and B, respectively. You want to maximize your total energy boost by drinking one energy drink per hour. However, if you want to switch from consuming one energy drink to the other, you need to wait for one hour to cleanse your system (meaning you won't get any energy boost in that hour). Return the maximum total energy boost you can gain in the next n hours. Note that you can start consuming either of the two energy drinks. Example 1: Input: energyDrinkA = [1,3,1], energyDrinkB = [3,1,1] Output: 5 Explanation: 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 Explanation: To gain an energy boost of 7: Drink the energy drink A for the first hour. Switch to the energy drink B and we lose the energy boost of the second hour. Gain the energy boost of the drink B in the third hour. Constraints: n == energyDrinkA.length == energyDrinkB.length 3 <= n <= 105 1 <= energyDrinkA[i], energyDrinkB[i] <= 105

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: Use dynamic programming to store and reuse optimal subproblem solutions. We track the maximum energy achievable at each time step, considering the last drink consumed and the possibility of a switch.
• State Definition: The DP state dp[i][j] stores the maximum energy achievable up to hour i, where j indicates the last action: 0 = drank A, 1 = drank B, 2 = switched from A to B, 3 = switched from B to A.

• Transitions: Define the transitions between states carefully, considering the energy gained from each drink and the penalty of switching.

• Runtime Complexity: O(n), Storage Complexity: O(n)

Code

    def max_energy_boost(energyDrinkA, energyDrinkB):
    n = len(energyDrinkA)
    dp = [[0] * 4 for _ in range(n)]

    # Initialize the first hour
    dp[0][0] = energyDrinkA[0]
    dp[0][1] = energyDrinkB[0]
    dp[0][2] = 0  # No switch possible at the first hour
    dp[0][3] = 0  # No switch possible at the first hour

    for i in range(1, n):
        # Drink A
        dp[i][0] = max(dp[i-1][0] + energyDrinkA[i], 
                       dp[i-1][3] + energyDrinkA[i])

        # Drink B
        dp[i][1] = max(dp[i-1][1] + energyDrinkB[i],
                       dp[i-1][2] + energyDrinkB[i])

        # Switch from A to B
        dp[i][2] = dp[i-1][0]

        # Switch from B to A
        dp[i][3] = dp[i-1][1]

    return max(dp[n-1][0], dp[n-1][1])