Maximal Score After Applying K Operations

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ArraysGreedy AlgorithmsDynamic Programming

You are given a 0-indexed integer array nums and an integer k. You have a starting score of 0.

In one operation:

choose an index i such that 0 <= i < nums.length,
increase your score by nums[i], and
replace nums[i] with ceil(nums[i] / 3).

Return the maximum possible score you can attain after applying exactly k operations.

The ceiling function ceil(val) is the least integer greater than or equal to val.

Example 1:

Input: nums = [10,10,10,10,10], k = 5
Output: 50
Explanation: Apply the operation to each array element exactly once. The final score is 10 + 10 + 10 + 10 + 10 = 50.

Example 2:

Input: nums = [1,10,3,3,3], k = 3
Output: 17
Explanation: You can do the following operations:
Operation 1: Select i = 1, so nums becomes [1,4,3,3,3]. Your score increases by 10.
Operation 2: Select i = 1, so nums becomes [1,2,3,3,3]. Your score increases by 4.
Operation 3: Select i = 2, so nums becomes [1,2,1,3,3]. Your score increases by 3.
The final score is 10 + 4 + 3 = 17.

Constraints:

1 <= nums.length, k <= 10⁵
1 <= nums[i] <= 10⁹

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:

What are the constraints on the size of the input array `nums` and the value of `k`?
Can the elements in the `nums` array be negative, zero, or non-integer?
If after applying k operations, the maximal score is zero, should I return zero or is there an expectation for a specific return when no score can be gained?
Are there any duplicate values within the `nums` array, and if so, how should they be handled during the operations?
Could you clarify the type of 'operation' being applied? Is it adding to the specific index, modifying the general array in any way, etc.?

Brute Force Solution

Approach

The brute force method for maximizing a score involves exploring absolutely every possible combination of actions. We simulate all possible sequences of applying these operations and calculate the resulting score for each. Then, we pick the highest score we observed.

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

Imagine you have a list of numbers, and you can perform an operation on any number a certain number of times.
Start by doing the operation on the first number once, and record the score after that change.
Next, still doing the operation once total, try doing it on the second number instead of the first, and record the score.
Keep going through each number, applying the operation one time each and recording all of the scores you get.
Now, move on to doing the operation twice. You could do it twice on the first number, once on the first and once on the second, and so on.
Repeat this process for every possible way to apply the operation up to the maximum allowed number of times.
After exploring all possible combinations of operations, compare all the scores you wrote down, and select the very highest one.

Code Implementation

def maximal_score_after_k_operations_brute_force(numbers, k_operations):
    maximum_score = 0

    def calculate_score(current_numbers):
        score = 0
        for number in current_numbers:
            score += number
        return score

    def apply_operations(index, remaining_operations, current_numbers):
        nonlocal maximum_score

        # If we've used all our operations,
        # calculate the score and update the maximum.
        if remaining_operations == 0:
            score = calculate_score(current_numbers)
            maximum_score = max(maximum_score, score)
            return

        # Iterate through each number in the list.
        for i in range(len(numbers)):

            # Apply an operation to the current number.
            new_numbers = current_numbers[:]
            new_numbers[i] = numbers[i] // 2

            # Recursively call the function to apply
            # the remaining operations.
            apply_operations(i, remaining_operations - 1, new_numbers)

    # Initiate the recursive process
    # starting with the original numbers.
    apply_operations(0, k_operations, numbers)

    return maximum_score

Big(O) Analysis

Time Complexity

O(n^k) – The algorithm explores all possible combinations of applying k operations to n numbers. Imagine a decision tree where each level represents an operation applied. Each node in the tree has n branches, corresponding to the n possible numbers to apply the operation to. Since we apply k operations, the depth of the tree is k. Therefore, the total number of possible combinations, and hence the number of branches we explore, grows as n raised to the power of k. Thus the time complexity is O(n^k).

Space Complexity

O(K^N) – The brute force method described explores all possible combinations of applying K operations to N numbers. This involves implicitly creating a call stack for each possible combination of operations. In the worst case, this could lead to a recursion tree where each node represents a choice of which number to apply an operation to, resulting in potentially K^N branches where N is the input size and K is the maximum number of operations. Thus, the maximum depth of the call stack can grow up to K^N, leading to auxiliary space usage that is O(K^N).

Optimal Solution

Approach

To maximize the score, we need to repeatedly pick the largest number, increase the score by that number, and then update that number. A smart way to do this efficiently is to always know what the largest number is without re-searching the entire collection every time.

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

First, put all the numbers into a special data structure that lets you quickly find the biggest one.
Then, repeat the following steps a set number of times, based on how many operations you're allowed to do.
Find and grab the current biggest number from the special data structure.
Add this biggest number to your total score.
Update the biggest number by using the provided formula and place the updated number back into the special data structure, so it stays organized to quickly locate the next biggest.
After doing all the steps, the total score will be the maximum possible score.

Code Implementation

import heapq

def find_maximal_score(number_list, operation_count):
    maximal_score = 0

    # Use a max heap to efficiently retrieve the largest element.
    max_heap = [-number for number in number_list]
    heapq.heapify(max_heap)

    for _ in range(operation_count):
        # Extract the current maximum value.
        current_maximum = -heapq.heappop(max_heap)

        maximal_score += current_maximum

        # Update the maximum value as per the formula.
        updated_maximum = current_maximum // 3

        # Add the updated value back to the heap. 
        heapq.heappush(max_heap, -updated_maximum)

    return maximal_score

Big(O) Analysis

Time Complexity

O(k log n) – We begin by inserting n elements into a priority queue (or heap), which takes O(n log n) time. However, since we are given that we perform k operations, the initialization time becomes irrelevant if k is larger than n. The dominant cost comes from performing k iterations. In each iteration, we extract the maximum element from the priority queue (O(log n)), add it to the score, update the element, and insert it back into the priority queue (O(log n)). Therefore, each operation takes O(log n) time. Consequently, k operations will take O(k log n) time, which dominates the initialization cost.

Space Complexity

O(N) – The solution uses a data structure, such as a heap or priority queue, to efficiently find the largest number. This data structure stores all N numbers from the input array. The space used by the priority queue grows linearly with the input size N, as it must hold all the elements to maintain the ability to quickly retrieve the maximum. Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input array	If the input array is empty, return 0 as no operations can be performed.
K equals 0	If K is 0, return 0 as no operations can be performed.
All elements in the array are zero	The algorithm should handle the case where all elements are zero correctly, resulting in a score of 0 if K > 0.
Very large K compared to the array size	The algorithm should perform the operation at most the number of times the array contains elements greater than zero, and only allow this many operations.
Array contains negative numbers	The algorithm should correctly process negative numbers; consider using absolute value if the problem description requires that.
Large array size and large K leading to potential integer overflow when calculating the score	Use a data type that can accommodate larger numbers (e.g., long) to prevent integer overflow during score calculation.
Array contains very large numbers	Ensure the chosen data type can handle the magnitude of array values without overflow.
K is negative	If K is negative, throw an exception or return an error code as the number of operations cannot be negative.

Empty input array

How to Handle:

If the input array is empty, return 0 as no operations can be performed.

K equals 0

How to Handle:

If K is 0, return 0 as no operations can be performed.

All elements in the array are zero

How to Handle:

The algorithm should handle the case where all elements are zero correctly, resulting in a score of 0 if K > 0.

Very large K compared to the array size

How to Handle:

The algorithm should perform the operation at most the number of times the array contains elements greater than zero, and only allow this many operations.

Array contains negative numbers

How to Handle:

The algorithm should correctly process negative numbers; consider using absolute value if the problem description requires that.

Large array size and large K leading to potential integer overflow when calculating the score

How to Handle:

Use a data type that can accommodate larger numbers (e.g., long) to prevent integer overflow during score calculation.

Array contains very large numbers

How to Handle:

Ensure the chosen data type can handle the magnitude of array values without overflow.

K is negative

How to Handle:

If K is negative, throw an exception or return an error code as the number of operations cannot be negative.