LeetCode 3259: Maximum Energy Boost From Two Drinks Solution
Master LeetCode problem 3259 (Maximum Energy Boost From Two Drinks), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
3259. Maximum Energy Boost From Two Drinks
Problem Explanation
Explanation:
To maximize the total energy boost, we need to consider the best strategy for each hour. We can switch between energy drinks every hour, but we have to wait for one hour to switch drinks. We can track the maximum energy boost we can get at each hour by considering the options of continuing with the same drink or switching to the other drink.
- Initialize two variables
drinkAanddrinkBto store the maximum energy boost we can get if we drink A or B at the current hour. - Iterate through each hour and update
drinkAanddrinkBbased on the maximum energy boost we can get considering the current hour. - At each hour, we consider two options:
- If we drink A at the current hour, the maximum energy boost is
drinkA + energyDrinkA[i]ordrinkB + energyDrinkA[i]if we switch from B to A. - If we drink B at the current hour, the maximum energy boost is
drinkB + energyDrinkB[i]ordrinkA + energyDrinkB[i]if we switch from A to B.
- If we drink A at the current hour, the maximum energy boost is
- The final result is the maximum value between
drinkAanddrinkB.
Time Complexity: O(n) where n is the length of the input arrays. Space Complexity: O(1)
:
Solution Code
class Solution {
public int maxEnergy(int[] energyDrinkA, int[] energyDrinkB) {
int drinkA = 0, drinkB = 0;
for (int i = 0; i < energyDrinkA.length; i++) {
int nextDrinkA = Math.max(drinkA, drinkB + energyDrinkA[i]);
int nextDrinkB = Math.max(drinkB, drinkA + energyDrinkB[i]);
drinkA = nextDrinkA;
drinkB = nextDrinkB;
}
return Math.max(drinkA, drinkB);
}
}Try It Yourself
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