LeetCode #3259 — MEDIUM

Maximum Energy Boost From Two Drinks

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[1,3,1]
[3,1,1]

Example 2

[4,1,1]
[1,1,3]

Step 02

Core Insight

What unlocks the optimal approach

Can we solve it using dynamic programming?
Define <code>dpA[i]</code> as the maximum energy boost if we consider only the first <code>i + 1</code> hours such that in the last hour, we drink the energy drink A.
Similarly define <code>dpB[i]</code>.
<code>dpA[i] = max(dpA[i - 1], dpB[i - 2]) + energyDrinkA[i]</code>
Similarly, fill <code>dpB</code>.
The answer is <code>max(dpA[n - 1], dpB[n - 1])</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3259: Maximum Energy Boost From Two Drinks
class Solution {
    public long maxEnergyBoost(int[] energyDrinkA, int[] energyDrinkB) {
        int n = energyDrinkA.length;
        long[][] f = new long[n][2];
        f[0][0] = energyDrinkA[0];
        f[0][1] = energyDrinkB[0];
        for (int i = 1; i < n; ++i) {
            f[i][0] = Math.max(f[i - 1][0] + energyDrinkA[i], f[i - 1][1]);
            f[i][1] = Math.max(f[i - 1][1] + energyDrinkB[i], f[i - 1][0]);
        }
        return Math.max(f[n - 1][0], f[n - 1][1]);
    }
}

// Accepted solution for LeetCode #3259: Maximum Energy Boost From Two Drinks
func maxEnergyBoost(energyDrinkA []int, energyDrinkB []int) int64 {
	n := len(energyDrinkA)
	f := make([][2]int64, n)
	f[0][0] = int64(energyDrinkA[0])
	f[0][1] = int64(energyDrinkB[0])
	for i := 1; i < n; i++ {
		f[i][0] = max(f[i-1][0]+int64(energyDrinkA[i]), f[i-1][1])
		f[i][1] = max(f[i-1][1]+int64(energyDrinkB[i]), f[i-1][0])
	}
	return max(f[n-1][0], f[n-1][1])
}

# Accepted solution for LeetCode #3259: Maximum Energy Boost From Two Drinks
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])

// Accepted solution for LeetCode #3259: Maximum Energy Boost From Two Drinks
/**
 * [3259] Maximum Energy Boost From Two Drinks
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn max_energy_boost(energy_drink_a: Vec<i32>, energy_drink_b: Vec<i32>) -> i64 {
        let n = energy_drink_a.len();
        let mut dp_drink_a = vec![0; n];
        let mut dp_drink_b = vec![0; n];

        for i in 0..n {
            if i >= 2 {
                dp_drink_a[i] = dp_drink_a[i - 1].max(dp_drink_b[i - 2]) + energy_drink_a[i] as i64;
                dp_drink_b[i] = dp_drink_b[i - 1].max(dp_drink_a[i - 2]) + energy_drink_b[i] as i64;
            } else if i >= 1 {
                dp_drink_a[i] = dp_drink_a[i - 1] + energy_drink_a[i] as i64;
                dp_drink_b[i] = dp_drink_b[i - 1] + energy_drink_b[i] as i64;
            } else {
                dp_drink_a[i] = energy_drink_a[i] as i64;
                dp_drink_b[i] = energy_drink_b[i] as i64;
            }
        }

        dp_drink_a[n - 1].max(dp_drink_b[n - 1])
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3259() {
        assert_eq!(5, Solution::max_energy_boost(vec![1, 3, 1], vec![3, 1, 1]));
        assert_eq!(7, Solution::max_energy_boost(vec![4, 1, 1], vec![1, 1, 3]));
    }
}

// Accepted solution for LeetCode #3259: Maximum Energy Boost From Two Drinks
function maxEnergyBoost(energyDrinkA: number[], energyDrinkB: number[]): number {
    const n = energyDrinkA.length;
    const f: number[][] = Array.from({ length: n }, () => [0, 0]);
    f[0][0] = energyDrinkA[0];
    f[0][1] = energyDrinkB[0];
    for (let i = 1; i < n; i++) {
        f[i][0] = Math.max(f[i - 1][0] + energyDrinkA[i], f[i - 1][1]);
        f[i][1] = Math.max(f[i - 1][1] + energyDrinkB[i], f[i - 1][0]);
    }
    return Math.max(...f[n - 1]!);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.