Maximum Energy Boost From Two Drinks

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ArraysDynamic Programming

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the `energy1` and `energy2` arrays contain negative numbers, zero, or only positive integers?
What are the constraints on the sizes of `energy1` and `energy2`, and the value of `k`? Are they small enough to allow for a solution with O(n*k) time complexity?
If `k` is larger than the combined length of `energy1` and `energy2`, should I return the sum of all elements in both arrays, or is this considered an invalid input?
If no combination of drinks allows us to select at most `k` drinks (e.g., both arrays are empty and k > 0), what value should I return?
Are we looking for the absolute maximum energy boost, or is there a specific preference if multiple combinations yield the same maximum boost?

Brute Force Solution

Approach

We need to find two drinks that give us the maximum energy boost when combined. The brute force way is to simply check every single possible pair of drinks and see which pair yields the highest energy.

Here's how the algorithm would work step-by-step:

Take the first drink and consider it as one of the drinks in our pair.
Now, go through every other drink and pair it with the first drink.
Calculate the combined energy boost for each of these pairs.
Keep track of the highest combined energy boost you've found so far.
Next, take the second drink and repeat the process of pairing it with every other drink.
Again, calculate the combined energy boost for each of these pairs and update the highest boost if necessary.
Continue doing this for every drink in the list, pairing it with all the other drinks to explore all possible combinations.
After checking all possible pairs, the highest combined energy boost you've kept track of is the answer.

Code Implementation

def max_energy_boost(energy_levels):
    maximum_boost = 0

    # Iterate through each drink as the first drink in the pair.
    for first_drink_index in range(len(energy_levels)): 
        
        # Iterate through the remaining drinks to form pairs.
        for second_drink_index in range(first_drink_index + 1, len(energy_levels)): 
            
            current_boost = energy_levels[first_drink_index] + energy_levels[second_drink_index]

            #Update maximum_boost if current_boost exceeds it.
            if current_boost > maximum_boost:
                maximum_boost = current_boost

    return maximum_boost

Big(O) Analysis

Time Complexity

O(n²) – The described brute force approach involves iterating through each of the n drinks in the input array. For each drink, the algorithm compares it with every other drink to find the combined energy boost. This comparison process requires another loop that iterates through the remaining drinks, resulting in nested loops where the outer loop runs n times, and the inner loop runs approximately n times in the worst case. Therefore, the number of operations grows proportionally to n multiplied by n, leading to a time complexity of O(n²).

Space Complexity

O(1) – The described brute force approach only uses a few constant space variables. It keeps track of the highest combined energy boost found so far and implicitly uses loop counters which are constant in size, irrespective of the number of drinks (N). No additional data structures like arrays or hash maps are created to store intermediate results. Therefore, the space complexity is constant.

Optimal Solution

Approach

To maximize the energy boost, we want to select the two drinks that give us the highest combined energy. We can achieve this efficiently by first sorting the drinks by their energy levels and then picking the two with the highest energy.

Here's how the algorithm would work step-by-step:

First, arrange the drinks in order from the lowest energy to the highest energy.
Then, simply pick the two drinks that are now at the end of the list because they have the most energy.
Add up the energy from these two drinks. This will be the maximum energy boost you can get.

Code Implementation

def maximum_energy_boost_from_two_drinks(energy_levels):
    energy_levels.sort()

    max_energy_boost = 0
    left_index = 0
    right_index = len(energy_levels) - 1

    # Iterate until the left and right indices meet
    while left_index < right_index:
        # Calculate the current energy boost
        current_energy_boost = energy_levels[left_index] + energy_levels[right_index]

        # Update the maximum energy boost if necessary
        if current_energy_boost > max_energy_boost:
            max_energy_boost = current_energy_boost

        left_index += 1
        right_index -= 1

    return max_energy_boost

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this algorithm is sorting the array of drinks. Sorting algorithms like merge sort or quicksort have a time complexity of O(n log n), where n is the number of drinks. After sorting, selecting the last two elements takes constant time, O(1). Therefore, the overall time complexity is determined by the sorting step, resulting in O(n log n).

Space Complexity

O(1) – The provided solution sorts the input array of drinks in place. Therefore, no auxiliary data structures of significant size are used. The only extra memory required is for a few temporary variables used during the sorting process, such as those used for swapping elements. Since the number of these variables does not depend on the input size N (the number of drinks), the space complexity is constant.

Edge Cases

Case	How to Handle
energy1 and energy2 are both null or empty	Return 0 since no drinks can be selected.
k is 0	Return 0 as no drinks can be consumed.
k is larger than the sum of lengths of energy1 and energy2	Return the sum of all elements in both arrays as we can take all drinks.
Arrays contain only positive values	The standard algorithm should work correctly without special handling.
Arrays contain negative values or zeros	The algorithm must account for negative energy boosts when determining the maximum boost, and zero boosts should be treated as a valid option.
Large input arrays where k is relatively small	An efficient solution should avoid unnecessary computations by focusing on the top k elements of each array and merging them.
Integer overflow potential when summing large energy values	Use a 64-bit integer type (long in Java, long long in C++) to store intermediate and final sums to prevent overflow.
One array is significantly larger than the other and k is close to the length of the smaller array	The solution needs to correctly handle the case where all drinks from one array can be selected, plus some from the other.

energy1 and energy2 are both null or empty

How to Handle:

Return 0 since no drinks can be selected.

k is 0

How to Handle:

Return 0 as no drinks can be consumed.

k is larger than the sum of lengths of energy1 and energy2

How to Handle:

Return the sum of all elements in both arrays as we can take all drinks.

Arrays contain only positive values

How to Handle:

The standard algorithm should work correctly without special handling.

Arrays contain negative values or zeros

How to Handle:

The algorithm must account for negative energy boosts when determining the maximum boost, and zero boosts should be treated as a valid option.

Large input arrays where k is relatively small

How to Handle:

An efficient solution should avoid unnecessary computations by focusing on the top k elements of each array and merging them.

Integer overflow potential when summing large energy values

How to Handle:

Use a 64-bit integer type (long in Java, long long in C++) to store intermediate and final sums to prevent overflow.

One array is significantly larger than the other and k is close to the length of the smaller array

How to Handle:

The solution needs to correctly handle the case where all drinks from one array can be selected, plus some from the other.