
Problem Statement
Given two integer arrays nums
and multipliers
, each index starting from zero and of lengths n
and m
respectively (with n
always being greater than or equal to m
), you initiate a game with a score of zero. Your goal is to maximize your score through m
specific operations. For each operation, indexed from 0 to m-1
, you are allowed to perform the following actions:
- Select an integer
x
from either the beginning or the end of thenums
array. - Multiply the selected integer
x
by the multiplier at the corresponding index from themultipliers
array and add the result to your score. - Remove the selected integer
x
from thenums
array.
The challenge is to determine the maximum score possible after carrying out all m
operations as described.
Examples
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
Approach and Intuition
The task is essentially about making optimal choices from two potential candidates (from the start or the end of the nums
array) at each operation step, to maximize the resultant score after all m
operations are completed. Here's how the process and choice-making can break down:
Dynamic Programming Approach: Each decision affects the subsequent ones since removing an element alters the available candidates for the following operations. To systematically evaluate the choices, dynamic programming can be particularly effective.
State Representation: Let's visualize the state by
DP[i][j]
, wherei
represents the number of elements taken from the beginning of thenums
array andj
represents the number taken from the end up to the current operation. Note thati + j
would give the total number of operations performed so far.State Transitions: Each step can involve two choices:
- Take the element from the start of the
nums
array which would affect the state transition as: [ DP[i+1][j] = \max(DP[i+1][j], DP[i][j] + nums[i] \times multipliers[i+j]) ] - Take the element from the end of the
nums
array which translates to: [ DP[i][j+1] = \max(DP[i][j+1], DP[i][j] + nums[n-j-1] \times multipliers[i+j]) ]
These equations help recursively build up the answer by simulating each possible choice and comparing their outcomes to opt for the one that yields the maximum score.
- Take the element from the start of the
Result Computation: Since
m
operations need to be performed, the result lies among various states where the sum of indicesi
andj
equalsm
(i.e.,DP[i][m-i]
fori
from0
tom
). The final answer will be the maximum among these values.Optimal Substructure and Overlapping Subproblems: The problem exhibits the properties of optimal substructure and overlapping subproblems, crucial for the applicability of dynamic programming. Each optimal solution to a state relies on the optimal solutions to its substates.
By following this approach and making use of memory-efficient techniques, the problem not only becomes manageable but also efficient to solve even for large input sizes defined by the constraints. The boundary conditions and specific cases are well-handled by adjusting the range and updates in state transitions ensuring robustness in the solution.
Solutions
- C++
- Java
- C
- Python
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 sizeweightsSize + 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 positionsl
wherel
ranges from 0 tostep
. 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 thenumbers
array and adding the resulting product to the score at positionl+1
. - Multiplying the current weight with the element from the end of the
numbers
array (adjusted for the currentstep
andl
) and adding this product to the score at positionl
.
- Multiplying the current weight with the element at position
- 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:
- Initialize
elementCount
andfactorCount
to store the lengths ofelements
andfactors
respectively. - Create an array
scores
of sizefactorCount + 1
to store intermediate results and ultimately the maximum score. - Iterate through the
factors
array from the last index to the first (reverse order), using variableoperation
. - For each
operation
, iterate through all possible positionsleftIndex
thatelements
can be matched withfactors[operation]
. - Calculate the score by taking the maximum between:
- Multiplying
factors[operation]
with theelements[leftIndex]
and adding the score from the next operation. - Multiplying
factors[operation]
with the element from the right end of theelements
corresponding to the current operation (elementCount - 1 - (operation - leftIndex)
) and adding the existing score atleftIndex
.
- Multiplying
- Update the
scores
array with the computed maximum value for eachleftIndex
. - 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 integersx
andy
.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 arrayselements
andcoefficients
, and their respective sizeselementsCount
andcoefficientsCount
.
Initialization:
- An integer
total_elements
is set toelementsCount
, representing the total number of elements. - An integer
total_operations
is set tocoefficientsCount
, indicating the total number of operations that can be performed. - An array
scores[]
is created with a size oftotal_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.
- An integer
Dynamic Programming Algorithm:
- The outer loop iterates from
total_operations - 1
down to 0, representing each operation that can be performed. - The inner loop iterates through each possible
leftIndex
from 0 to the currentoperation
value. This index helps in selecting the elements from the start (leftIndex
) and from the end (total_elements - 1 - (operation - leftIndex)
) of the array. - 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]
).
- Multiplying the current coefficient with the element at
- The greater of these two calculated values is determined using the
greater
function and assigned toscores[leftIndex]
.
- The outer loop iterates from
Result:
- The function returns
scores[0]
, which after completion of all operations contains the maximum score that can be achieved.
- The function returns
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
andelements
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.
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