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Maximum Score from Performing Multiplication Operations

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 ith 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

Solution Explanation

This problem asks to find the maximum score achievable by performing a series of operations on an array nums using the multipliers in the array multipliers. In each operation, we choose a number from either the start or the end of nums, multiply it by the corresponding multiplier, and add the product to the score. The key is to optimally choose between the start and end elements at each step to maximize the final score.

Approach

This problem can be efficiently solved using dynamic programming. We can define a recursive function or iterative approach to explore all possible choices.

State: The state of the dynamic programming solution can be represented by three variables:

  • i: The index indicating the number of elements chosen from the beginning of the nums array.
  • j: The index indicating the number of elements chosen from the end of the nums array.
  • k: The index of the current multiplier in the multipliers array.

Recursive Relation: The recursive relation can be formulated as follows:

dp(i, j, k) = max(
    dp(i + 1, j, k + 1) + nums[i] * multipliers[k],  // Choose from the beginning
    dp(i, j + 1, k + 1) + nums[n - 1 - j] * multipliers[k] // Choose from the end
)

The base case is when k reaches the end of multipliers (all operations are done), where the score is 0.

Optimization: A naive recursive implementation would lead to exponential time complexity due to overlapping subproblems. Therefore, memoization (top-down dynamic programming) or tabulation (bottom-up dynamic programming) is crucial for optimization.

Time and Space Complexity Analysis

Time Complexity: The time complexity of both the recursive (with memoization) and iterative (tabulation) approaches is O(m*m), where 'm' is the length of the multipliers array. This is because we explore all possible combinations of choices from the beginning and end of nums.

Space Complexity:

  • Recursive (with Memoization): The space complexity is O(m*m) due to the memoization table.
  • Iterative (Tabulation): The space complexity is also O(m*m) because of the DP table.

Code Implementation (Python, Java, C++, Go, and Typescript)

The code implementations in different programming languages (Python, Java, C++, Go, and TypeScript) demonstrate the dynamic programming approach using tabulation (bottom-up) to avoid redundant computations. Memoization (top-down) is another efficient approach. Both achieve the same time and space complexity.

Note: The provided code solutions illustrate both a recursive (with memoization) approach and an iterative (tabulation) approach in most of the languages. Both approaches solve the problem efficiently with the same time complexity. The choice depends on personal preference and the specifics of the problem context. The iterative (tabulation) approach might be slightly faster in some cases.