Maximum Score from Performing Multiplication Operations

class Solution {
public:
    int getMaxScore(vector<int>& numbers, vector<int>& weights) {
        int numsSize = numbers.size();
        int weightsSize = weights.size();
        vector<int> score(weightsSize + 1, 0);
            
        for (int step = weightsSize - 1; step >= 0; step--) {
            for (int l = 0; l <= step; l++) {
                score[l] = max(weights[step] * numbers[l] + score[l + 1],
                               weights[step] * numbers[numsSize - 1 - (step - l)] + score[l]);
            }
        }
            
        return score[0];
    }
};

This C++ solution is designed to find the maximum score by performing multiplication operations on two arrays, one containing numbers and the other containing weights for each operation. The program utilizes a dynamic programming approach to compute the score efficiently.

Solution Explanation:

• Initializes a score vector of size weightsSize + 1 with all elements set to zero. This vector is used to store intermediate results to facilitate dynamic programming.
• Loops backward from the last element of weights to the first. This reverse iteration helps in building up the solution from the base case.
• For each weight at position step, iterates over all positions l where l ranges from 0 to step. This loop calculates the maximum score for the current subproblem by considering two choices:
  • Multiplying the current weight with the element at position l in the numbers array and adding the resulting product to the score at position l+1.
  • Multiplying the current weight with the element from the end of the numbers array (adjusted for the current step and l) and adding this product to the score at position l.
• The maximum obtained value from the above two choices is stored back in score[l].
• Finally, it returns the first element of the score vector which holds the maximum score obtainable with all operations performed.

This approach ensures that every possible combination of multiplication operations using each weight exactly once is considered, thus guaranteeing the computation of the maximum possible score efficiently.

class Solution {
    public int maxScore(int[] elements, int[] factors) {
        int elementCount = elements.length;
        int factorCount = factors.length;
        int[] scores = new int[factorCount + 1];
            
        for (int operation = factorCount - 1; operation >= 0; operation--) {
            for (int leftIndex = 0; leftIndex <= operation; leftIndex++) {
                scores[leftIndex] = Math.max(factors[operation] * elements[leftIndex] + scores[leftIndex + 1],
                                             factors[operation] * elements[elementCount - 1 - (operation - leftIndex)] + scores[leftIndex]);
            }
        }
            
        return scores[0];
    }
}

The solution presented aims to solve the problem of calculating the maximum score by performing multiplication operations on two arrays, elements and factors. The Java program uses dynamic programming to enhance efficiency and manage complexities associated with larger inputs. The following steps outline the approach:

1. Initialize elementCount and factorCount to store the lengths of elements and factors respectively.
2. Create an array scores of size factorCount + 1 to store intermediate results and ultimately the maximum score.
3. Iterate through the factors array from the last index to the first (reverse order), using variable operation.
4. For each operation, iterate through all possible positions leftIndex that elements can be matched with factors[operation].
5. Calculate the score by taking the maximum between:
   • Multiplying factors[operation] with the elements[leftIndex] and adding the score from the next operation.
   • Multiplying factors[operation] with the element from the right end of the elements corresponding to the current operation (elementCount - 1 - (operation - leftIndex)) and adding the existing score at leftIndex.
6. Update the scores array with the computed maximum value for each leftIndex.
7. Return the first element of scores array, which contains the maximum score after evaluating all possibilities.

The program effectively computes the maximum possible score by selecting the optimal pairings of elements from both arrays based on given multiplication operations. The use of reverse looping through the operations and a two-dimensional consideration for score calculation (left and right potential operations) ensures thorough exploration of combinations, making dynamic programming ideal for this context. The implementation efficiently updates the scores while considering multiple scenarios at each operation level, leading to the finalized maximum score at scores[0].

int greater(int x, int y) { return x > y ? x : y; }
    
int calculateMaxScore(int *elements, int elementsCount, int *coefficients, int coefficientsCount) {
    int total_elements = elementsCount;
    int total_operations = coefficientsCount;
    int scores[total_operations + 1];
    for (int i = 0; i <= total_operations; i++) scores[i] = 0;
        
    for (int operation = total_operations - 1; operation >= 0; operation--) {
        for (int leftIndex = 0; leftIndex <= operation; leftIndex++) {
            scores[leftIndex] = greater(coefficients[operation] * elements[leftIndex] + scores[leftIndex + 1],
                                        coefficients[operation] * elements[total_elements - 1 - (operation - leftIndex)] + scores[leftIndex]);
        }
    }
    return scores[0];
}

The C program provided calculates the maximum score that can be obtained by performing multiplication operations on an array of integers. This task essentially leverages a dynamic programming approach to solve the problem efficiently. Here is a breakdown of how the implementation works:

• Function Definitions:

  • int greater(int x, int y) - This function returns the greater of the two integers x and y.
  • int calculateMaxScore(int *elements, int elementsCount, int *coefficients, int coefficientsCount) - This is the main function that calculates the maximum score. It takes as parameters pointers to the arrays elements and coefficients, and their respective sizes elementsCount and coefficientsCount.

• Initialization:

  • An integer total_elements is set to elementsCount, representing the total number of elements.
  • An integer total_operations is set to coefficientsCount, indicating the total number of operations that can be performed.
  • An array scores[] is created with a size of total_operations + 1 and is initialized to zero. This array is used to store the maximum scores computable at each step of the dynamic programming algorithm.

• Dynamic Programming Algorithm:

  1. The outer loop iterates from total_operations - 1 down to 0, representing each operation that can be performed.
  2. The inner loop iterates through each possible leftIndex from 0 to the current operation value. This index helps in selecting the elements from the start (leftIndex) and from the end (total_elements - 1 - (operation - leftIndex)) of the array.
  3. Inside the inner loop, scores[leftIndex] is updated to the maximum value obtained by:
     • Multiplying the current coefficient with the element at leftIndex and adding the next possible score (scores[leftIndex + 1]).
     • Multiplying the current coefficient with an element from the end of the array and adding the current score (scores[leftIndex]).
  4. The greater of these two calculated values is determined using the greater function and assigned to scores[leftIndex].

• Result:

  • The function returns scores[0], which after completion of all operations contains the maximum score that can be achieved.

A key aspect of this approach is that it optimizes the computation by systematically breaking down the problem using dynamic programming, making efficient use of space and time by storing interim results and only recalculating necessary values. This solution is highly efficient for cases where there are a large number of operations and elements.

class Solution:
    def maxScore(self, elements: List[int], factors: List[int]) -> int:
    
        count_factors = len(factors)
        count_elements = len(elements)
    
        score_table = [0] * (count_factors + 1)
    
        for operation in range(count_factors - 1, -1, -1):
            for start in range(0, operation + 1, 1):
                score_table[start] = max(factors[operation] * elements[start] + score_table[start + 1],
                                         factors[operation] * elements[count_elements - 1 - (operation - start)] + score_table[start])
    
        return score_table[0]

This Python solution implements a function to compute the maximum score from performing multiplication operations given two lists: elements and factors. The function maxScore accepts elements and factors as arguments, representing the integer lists for the elements and multiplication factors, respectively.

The approach employs dynamic programming to solve the problem efficiently:

• Initializations are carried out to determine the lengths of the factors and elements lists. A score table is configured with zeros and an additional space to accommodate the boundary of operations, using length of factors plus one.

• Iterating begins from the last factor to the first, and for each factor, a nested loop iterates through possible start points for multiplication.

• Inside the loop, the maximum score is determined by comparing two potential scores:

  • Multiplying the current factor with the element corresponding to the start of the range and adding it with the pre-computed value in the score table at the next position.
  • Multiplying the current factor with an element from the end of the elements list adjusted for the current depth in nested iteration, adding this to the pre-computed value in the score table at the current position.

• The result is stored back in the score table at the current starting index, progressively building up the maximum score achievable from the zeroth starting index, taking into account various multiplication and addition operations.

• Ultimately, the value at the first index of the score table (score_table[0]) contains the maximum score possible with the given list of elements and factors as the computation completes.

This approach ensures all possible combinations of multiplications are considered through dynamic programming, thus ensuring that the maximum score calculation is comprehensive and efficient.