Maximum Score from Performing Multiplication Operations

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ArraysDynamic Programming

You are given two 0-indexed integer arrays nums and multipliers of size n and m respectively, where n >= m.

You begin with a score of 0. You want to perform exactly m operations. On the i^th operation (0-indexed) you will:

Choose one integer x from either the start or the end of the array nums.
Add multipliers[i] * x to your score.
- Note that multipliers[0] corresponds to the first operation, multipliers[1] to the second operation, and so on.
Remove x from nums.

Return the maximum score after performing m operations.

Example 1:

Input: nums = [1,2,3], multipliers = [3,2,1]
Output: 14
Explanation: An optimal solution is as follows:
- Choose from the end, [1,2,3], adding 3 * 3 = 9 to the score.
- Choose from the end, [1,2], adding 2 * 2 = 4 to the score.
- Choose from the end, [1], adding 1 * 1 = 1 to the score.
The total score is 9 + 4 + 1 = 14.

Example 2:

Input: nums = [-5,-3,-3,-2,7,1], multipliers = [-10,-5,3,4,6]
Output: 102
Explanation: An optimal solution is as follows:
- Choose from the start, [-5,-3,-3,-2,7,1], adding -5 * -10 = 50 to the score.
- Choose from the start, [-3,-3,-2,7,1], adding -3 * -5 = 15 to the score.
- Choose from the start, [-3,-2,7,1], adding -3 * 3 = -9 to the score.
- Choose from the end, [-2,7,1], adding 1 * 4 = 4 to the score.
- Choose from the end, [-2,7], adding 7 * 6 = 42 to the score. 
The total score is 50 + 15 - 9 + 4 + 42 = 102.

Constraints:

n == nums.length
m == multipliers.length
1 <= m <= 300
m <= n <= 10⁵
-1000 <= nums[i], multipliers[i] <= 1000

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 possible ranges for the values in `nums` and `multipliers`, and what are the maximum lengths of the `nums` and `multipliers` arrays?
Can the numbers in `nums` and `multipliers` be negative, zero, or only positive?
If there are multiple ways to achieve the maximum score, is any valid maximum score acceptable, or is there a specific criteria for choosing among them?
If the `multipliers` array is longer than the `nums` array, such that not all multiplication operations can be performed, what should the result be?
Is the order of operations fixed, meaning I must always start with `multipliers[0]` and proceed sequentially?

Brute Force Solution

Approach

The brute force method for this problem involves trying every single possible combination of multiplication operations. We explore each choice we can make at each step, calculating the score for each path and eventually picking the best one. It's like trying every route on a map to find the shortest one.

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

Start by considering your first multiplication choice. You can either multiply the first number in your list or the last number by the first multiplier.
Calculate the score you get from that choice.
Now, move on to the next multiplier. For *each* of the previous choices, consider *both* multiplying the current multiplier with the new first number *and* multiplying it with the new last number.
Keep going, creating two new possibilities for every existing one, and calculating the score along each 'path'.
Do this until you've used all the multipliers.
At the end, you'll have a collection of scores, one for each possible sequence of multiplication choices.
Compare all of these final scores, and select the highest one. That's your answer.

Code Implementation

def maximum_score_brute_force(
    numbers_list,
    multipliers_list
):

    def calculate_maximum_score(
        current_index,
        left_pointer,
        current_score
    ):
        # If we have used all the multipliers, return the current score.
        if current_index == len(multipliers_list):
            return current_score

        right_pointer = len(numbers_list) - 1 - (current_index - left_pointer)

        # Explore the option of multiplying with the left element
        left_score =
            calculate_maximum_score(
                current_index + 1,
                left_pointer + 1,
                current_score
                + numbers_list[left_pointer]
                * multipliers_list[current_index]
            )

        # Explore the option of multiplying with the right element
        right_score =
            calculate_maximum_score(
                current_index + 1,
                left_pointer,
                current_score
                + numbers_list[right_pointer]
                * multipliers_list[current_index]
            )

        # Return the maximum score between the two options
        return max(left_score, right_score)

    # Initiate the recursive calls to start the brute force calculation
    maximum_score = calculate_maximum_score(0, 0, 0)

    return maximum_score

Big(O) Analysis

Time Complexity

O(2^m) – The brute force approach explores every possible combination of multiplying numbers with multipliers. At each step, for each multiplier, we have two choices: multiply it with the leftmost or rightmost remaining number. Given m multipliers, this generates a binary decision tree with a depth of m, resulting in 2^m possible paths (sequences of multiplication operations). Each path represents a unique combination, and we need to evaluate the score for each path. Therefore, the time complexity is O(2^m), where m is the number of multipliers.

Space Complexity

O(2^M) – The brute force solution explores every possible combination of multiplication operations using recursion. At each step, it considers two choices (multiply with the first or last number), effectively doubling the number of possibilities. Since there are M multipliers (where M is the number of multiplication operations to perform), this leads to a binary tree-like structure of recursive calls. Therefore, the number of possibilities, and thus the number of function calls stored on the call stack at any given point, can grow up to 2^M in the worst case, leading to an auxiliary space complexity of O(2^M).

Optimal Solution

Approach

The best way to solve this is to think backwards using something like a decision tree. Instead of calculating everything, we break the problem into smaller subproblems and reuse already computed results to avoid repetition, like remembering the answers to simple math problems.

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

Consider each step where you have to make a choice: either use a number from the beginning or a number from the end of the list.
For each choice, figure out what the score would be if you take that number and move forward.
Notice that if you have already calculated the score for some part of the list, remember it. You can just look up the previously calculated score instead of recomputing it.
When you get to the end of the list (or the multiplication list), compare the scores from the beginning and end, and pick the better score.
Go back up to the beginning, and remember the path of choices that gave you the best score. This is your final answer.

Code Implementation

def maximum_score_from_multiplication_operations(numbers, multipliers):
    number_length = len(numbers)
    multiplier_length = len(multipliers)

    memo = {}

    def calculate_maximum_score(left_index, multiplier_index):
        if multiplier_index == multiplier_length:
            return 0

        if (left_index, multiplier_index) in memo:
            return memo[(left_index, multiplier_index)]

        right_index = number_length - 1 - (multiplier_index - left_index)

        # Explore choosing the left element
        score_from_left =
calculate_maximum_score(left_index + 1, multiplier_index + 1) +
            numbers[left_index] * multipliers[multiplier_index]

        # Explore choosing the right element
        score_from_right = calculate_maximum_score(left_index, multiplier_index + 1) +
            numbers[right_index] * multipliers[multiplier_index]

        # Choose the maximum score
        maximum_score = max(score_from_left, score_from_right)
        memo[(left_index, multiplier_index)] = maximum_score

        return maximum_score

    # Kick off the calculation.
    result = calculate_maximum_score(0, 0)
    return result

Big(O) Analysis

Time Complexity

O(m*n) – The solution utilizes dynamic programming with memoization to avoid redundant calculations. The state is defined by the index of the multiplier used and the number of elements taken from the left side of the nums array. Given that there are m multipliers and we can take at most n elements from the left (which implicitly defines the number of elements taken from the right), the size of the memoization table is O(m*n). Each cell in the table is computed only once, and the computation involves constant time operations. Therefore, the overall time complexity is O(m*n), where m is the length of the multipliers array and n is the length of the nums array.

Space Complexity

O(m * n) – The described solution uses dynamic programming with memoization. The memoization table stores the results of subproblems to avoid recomputation. The dimensions of this table depend on the number of multiplication operations (m) and the length of the input nums array (n), since each state is defined by the index of the multiplier used and the left pointer in the nums array. More precisely the number of states are defined by m and the difference between the left and right index that can be at most n. The memoization table therefore takes O(m * n) space where m is the length of the multipliers array and n is the length of the nums array.

Edge Cases

Case	How to Handle
Null or empty nums array	Return 0 if nums or multipliers is null or empty, as there are no operations to perform.
Null or empty multipliers array	Return 0 if nums or multipliers is null or empty, as there are no operations to perform.
nums array size is smaller than multipliers array size	Only use as many elements from nums as needed according to the length of multipliers; loop condition must account for this.
Large input sizes for nums and multipliers (close to limits)	Use dynamic programming with memoization to avoid redundant calculations and ensure acceptable runtime complexity.
nums and multipliers arrays contain only negative numbers	The solution should correctly handle negative numbers, potentially leading to the highest score by strategically choosing left/right elements.
nums array contains all zeros, and multipliers contains positive and negative values	The result should always be zero since any multiplier times zero is zero, handle the edge condition by returning 0 in the algorithm.
Integer overflow during multiplication	Use a larger data type (e.g., long) to store intermediate multiplication results to prevent integer overflow.
Multipliers array contains duplicate values	The duplicate values should not affect the algorithm correctness, as the dynamic programming approach considers all possible combinations.

Null or empty nums array

How to Handle:

Return 0 if nums or multipliers is null or empty, as there are no operations to perform.

Null or empty multipliers array

How to Handle:

Return 0 if nums or multipliers is null or empty, as there are no operations to perform.

nums array size is smaller than multipliers array size

How to Handle:

Only use as many elements from nums as needed according to the length of multipliers; loop condition must account for this.

Large input sizes for nums and multipliers (close to limits)

How to Handle:

Use dynamic programming with memoization to avoid redundant calculations and ensure acceptable runtime complexity.

nums and multipliers arrays contain only negative numbers

How to Handle:

The solution should correctly handle negative numbers, potentially leading to the highest score by strategically choosing left/right elements.

nums array contains all zeros, and multipliers contains positive and negative values

How to Handle:

The result should always be zero since any multiplier times zero is zero, handle the edge condition by returning 0 in the algorithm.

Integer overflow during multiplication

How to Handle:

Use a larger data type (e.g., long) to store intermediate multiplication results to prevent integer overflow.

Multipliers array contains duplicate values

How to Handle:

The duplicate values should not affect the algorithm correctness, as the dynamic programming approach considers all possible combinations.