Problem Description

In this problem, you are provided with two primary inputs: An array nums containing n positive integers and an integer k, which represents the maximum number of operations you can apply. Your goal is to maximize a score, which starts at 1, by performing a series of operations on the nums array.

Each operation consists of the following steps:

• First, you select a non-empty subarray from nums that you haven't chosen before.
• Then, within that subarray, you find the element with the highest prime score. The prime score is defined as the count of distinct prime factors of a number. If multiple elements share the highest prime score, you choose the one with the smallest index within the subarray.
• Finally, you multiply your current score by the selected element.

You can repeat this operation at most k times. The challenge is to calculate the maximum possible score you can achieve after performing up to k operations. Due to the fact that the final score can be quite large, the result should be returned modulo 10^9 + 7.

Intuition

To arrive at the solution, we must prioritize two key observations. First, since the score is a product of selected elements, larger values inherently contribute more to maximizing the score.

Second, the number of times an element can contribute (its "count") depends on how many subarrays it dominates in terms of prime score. The prime score determines eligibility: an element can be selected in all subarrays where it is the first element with the highest prime score. To compute this count efficiently, we determine the boundaries l and r for each element. These boundaries represent the farthest indices where the element’s prime score remains the maximum. The total count of qualifying subarrays is derived from these boundaries, calculated as (i - l) * (r - i).

By combining these insights—greedily selecting larger values first and using prime scores to determine contribution counts—we ensure the optimal balance between maximizing per-operation impact and leveraging eligibility constraints.

With these key observations, the solution employs a greedy approach. It suggests sorting elements in descending order based on their values since higher numbers are likely to have a higher prime score and hence contribute more significantly to maximizing the score. While processing the sorted elements, we calculate the cnt for each and if cnt is less than or equal to k, we incorporate that element's contribution (nums[i]^cnt) into the total score. If cnt exceeds k, we take the contribution as nums[i]^k instead, and then stop, since we've used up all k operations.

A monotonic stack is used to efficiently find the leftmost and rightmost indexes l and r to calculate cnt for each element. This helps to maintain a running track of the prime scores in a way that allows quick look-ups of the needed boundaries.

Finally, the solution applies a fast power algorithm while taking the modulo at each step to ensure the answer stays within bounds, considering the potential size of the number due to repeated multiplication.

Learn more about Stack, Greedy, Math and Monotonic Stack patterns.

Solution Approach

To implement the solution, we use two important techniques: a greedy approach and a monotonic stack.

The prime factors function primeFactors(n) is first defined to calculate the number of distinct prime factors of n to determine the prime score of each number in the given array nums.

Then, in the maximumScore method of Solution, we begin by preparing an array arr which contains tuples for each element in nums. Each tuple consists of the index of the element, the prime score obtained from primeFactors function, and the element's value itself.

To efficiently calculate the leftmost (l) and rightmost (r) boundaries where an element nums[i] has the highest prime score in subarrays, we leverage a monotonic stack. A monotonic stack is a stack where the elements are always in increasing or decreasing order. We maintain two stacks to determine the indices l and r.

• The left array is filled by iterating through arr from the beginning, and we use a stack to track the first previous element with a prime score not less than the current element's prime score.
• The right array is filled by iterating through arr in reverse, maintaining a stack to track the next element with a prime score greater than the current element's prime score.

In both processes, if the current element's prime score is higher than the one on the top of the stack, we pop elements from the stack because we're only interested in the closest boundaries that have a higher or equal prime score.

After determining the left and right bounds for each element, we sort the array arr by the element's value in decreasing order, since we want to use the largest elements first to maximize the score.

We then iterate over this sorted array. For each element nums[i], we calculate the count cnt as the product of the distance from l to i and the distance from i to r. This count represents the number of subarrays where nums[i] is the maximum prime score that hasn't been picked yet.

We check if cnt is less than or equal to k, and if so, we update our score by multiplying it by nums[i]^cnt, and decrement k by cnt. If cnt is greater than k, we multiply our score by nums[i]^k, as we can only perform k more operations, and then we break out of the loop.

Lastly, we return the score modulo 10^9 + 7. The power function in Python (pow(base, exp, mod)) is used to handle the large exponents and perform modulus at the same time for efficiency, reducing the computational complexity.

By following this approach, we ensure a greedy selection of elements translates into a maximum score, bounded by the number of times k we can perform the operation, and make use of efficient data structures and algorithms to handle the constraints of the problem.

Example Walkthrough

Let's take an example to illustrate the solution approach. Suppose we have an array nums with values [6, 12, 5], and k equals 2. The prime factors counts for these numbers are as follows: The number 6 has two distinct prime factors (2 and 3), 12 also has two distinct prime factors (2 and 3 as well), and 5 has one prime factor (5 itself).

Let's walk through the solution step by step:

1. Determine the prime score for each element:

   • 6 (index 0): 2 prime factors
   • 12 (index 1): 2 prime factors
   • 5 (index 2): 1 prime factor

2. Create arr with tuples (index, prime score, value):

   • [(0, 2, 6), (1, 2, 12), (2, 1, 5)]

3. Calculate left and right boundaries using monotonic stacks:

   • Left boundaries (farthest left index with prime score ≥ current):

     • For 6 (index 0): No elements to the left → left[0] = -1
     • For 12 (index 1): Previous element (index 0) has same prime score → left[1] = 0
     • For 5 (index 2): Previous elements (index 1) has higher prime score → left[2] = 1
     • Result: left = [-1, 0, 1]

   • Right boundaries (farthest right index with prime score > current):

     • For 6 (index 0): No higher scores to the right → right[0] = 3
     • For 12 (index 1): No higher scores to the right → right[1] = 3
     • For 5 (index 2): No higher scores to the right → right[2] = 3
     • Result: right = [3, 3, 3]

4. Sort arr by value in descending order:

   • [(1, 2, 12), (0, 2, 6), (2, 1, 5)]

5. Calculate contribution counts (cnt) for each element:

   • For 12 (index 1): cnt = (1 - 0) * (3 - 1) = 1 * 2 = 2
   • For 6 (index 0): cnt = (0 - (-1)) * (3 - 0) = 1 * 3 = 3
   • For 5 (index 2): cnt = (2 - 1) * (3 - 2) = 1 * 1 = 1

6. Apply operations greedily:

   • Initial score: 1
   • Process 12 (cnt=2 ≤ k=2):
     • Multiply score by 12^2 = 144
     • Update k = 2 - 2 = 0
   • All operations are used. Final score: 144

This demonstrates how the corrected boundary calculations ensure accurate subarray counts, leading to the optimal selection of elements to maximize the score within k operations.

Keep in mind that the actual implementation would handle much larger numbers and array sizes, efficiently managing the modulo operation and arithmetic, but the core steps would follow this logical sequence.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code consists of several parts:

1. Calculating prime factors for each number in the nums list:

   • For each number x, the primeFactors function can go up to sqrt(x) which is the worst-case scenario. However, since the prime factors are only up to the square root of x, this part is typically less than O(n * sqrt(x)).
   • Across all numbers, since we can't assume the average value of x, if we treat the input array elements as having value up to m in the worst case, this part can be seen as O(n * sqrt(m)), where m is the maximum number in nums.

2. Building the left and right arrays using a monotonically decreasing stack approach:

   • In usual stack operations (push/pop), each element is processed once, making this O(n).

3. Sorting the transformed arr list based on the third element (the value of the number itself):

   • Sorting is typically O(n * log(n)).

4. The final loop to calculate the answer:

   • The loop runs at most n times, and in each iteration, pow function is called which is generally O(log(x)) for each x. Since there can be up to n such calls, in the worst case, this part is O(n * log(m)) where m is the maximum number in nums (note that the exponentiation is modulo mod, which doesn't affect time complexity).

Combining these parts, the dominant term is from sorting and the pow operations, resulting in O(n * log(n) + n * log(m)). If m (the maximum element in nums) is on the same order as n or less, it simplifies to O(n * log(n)).

Space Complexity

The space complexity of the code consists of the following parts:

1. The ans set inside primeFactors function has the space complexity that depends on the number of distinct prime factors. In the worst case, all numbers are distinct primes, so the set size can be O(sqrt(m)). However, since only one set is maintained at a time and is rewritten for each number, this doesn't scale with n.

2. The arr list would store a tuple for each element in nums, resulting in O(n).

3. The left and right arrays which are both of size n, together contribute O(2n).

4. The stack stk which is used twice could size up to n elements in the worst case, contributing another O(n).

Combining these, the space complexity would be O(n) as the dominant term, conforming with the reference provided.

Learn more about how to find time and space complexity quickly using problem constraints.