Problem Description

The problem deals with an integer array nums and an integer k. We're tasked with partitioning the array into at most k non-empty adjacent subarrays. Our goal is to maximize the 'score' of the partition, where the score is defined as the sum of the averages of each subarray. Important to note is that we must use every integer in nums and that the partitions we create will affect the score. We need to find the best possible way to split the array to achieve the maximum score.

Intuition

The core of this problem lies in dynamic programming, particularly in finding the best partition at each stage to maximize the score. A naïve approach might try every possible partition, but this would be prohibitively slow. Instead, we need an efficient way to solve smaller subproblems and combine their solutions to solve the larger problem.

The intuition behind the dynamic programming solution is that if we know the best score we can get from the first i elements with k subarrays, we can use this to compute the best score for the first i+1 elements with k subarrays, and so on. The recursive function dfs represents this notion, where dfs(i, k) returns the best score we can get starting from index i with k subarrays left to form.

The solution uses a top-down approach with memoization (caching) to avoid re-computing the scores for the same i and k values more than once. It also uses a prefix sum array s to quickly compute the sum of elements for any given subarray, which is used to calculate the averages needed to compute the scores for the partitions.

By caching results and avoiding repetition of work, we arrive at the best solution in a much faster and more efficient way than brute-force methods.

Learn more about Dynamic Programming and Prefix Sum patterns.

Solution Approach

The solution uses several algorithms and concepts:

Dynamic Programming:

Dynamic programming is utilized to break the problem down into smaller, manageable subproblems. A function dfs recursively computes the maximum score obtainable from a given starting index i and a certain number of partitions k. The essence of dynamic programming is apparent as dfs computes the scores based on previously solved smaller problems.

Memoization:

To optimize the dynamic programming solution, memoization is used. The @cache decorator in Python automatically memorizes the result of the dfs function calls with particular arguments, so when the function is called again with the same arguments, the result is retrieved from the cache instead of re-computing it.

Prefix Sums:

To efficiently calculate the average of any subarray within nums, a prefix sum array s is created using the accumulate function from Python's itertools module, with initial=0 to include the starting point. This data structure allows constant-time retrieval of the sum of elements between any two indices.

Implementation Details:

The main function largestSumOfAverages first computes the cumulative sums of nums. Then it defines the dfs function with parameters i (the starting index for considering partitions) and k (the number of remaining partitions).

The dfs function works as follows:

• If i == n, where n is the length of nums, it means we've considered all elements and there is no score to be added, so it returns 0.
• If k == 1, we can only make one more partition, so the best score is the average of all remaining elements, calculated by (s[-1] - s[i]) / (n - i).
• For other cases, the function iterates through the elements starting from i up to the last element, creating a subarray from i to j and calculating its average. It then recursively calls dfs(j + 1, k - 1) to compute the score of the remaining array with one less partition.
• The max function keeps track of the highest score found during the iteration.

At the end of the dfs calls, dfs(0, k) provides the maximum score for the entire array with k partitions.

The implementation takes advantage of Python's concise syntax and powerful standard library functions like itertools.accumulate and functools.cache to create an elegant and efficient solution.

Example Walkthrough

Let's assume we have nums = [9, 1, 2, 3, 9] and k = 3. We want to find the maximum score by partitioning the array into at most k adjacent subarrays.

First, let's calculate the prefix sums to efficiently compute the sums of subarrays:

nums = [9, 1, 2, 3, 9] prefix_sums = [0, 9, 10, 12, 15, 24] (include a 0 at the beginning for easy calculation)

Here's the Python function dfs(i, k) that will compute the maximum score starting from index i with k partitions left.

Now, let's walk through the solution:

Step 1: Call dfs(0, 3) for the full array with 3 partitions allowed.

Step 2: dfs(0, 3) explores partitioning the array from index 0. It tries partitioning after every index to find the maximum score:

• Partition after index 0: average of first part [9] is 9, now call dfs(1, 2).
• Partition after index 1: average of first part [9, 1] is 5, now call dfs(2, 2).
• Partition after index 2: average of first part [9, 1, 2] is 4, now call dfs(3, 2).
• Partition after index 3: average of first part [9, 1, 2, 3] is 3.75, now call dfs(4, 2).

Step 3: For each of the above calls to dfs(i, 2), it again divides the remaining part of the array and calls dfs(j, 1). When k == 1, it simply takes the average of the remaining elements since it has to be one partition.

Step 4: Each time the score is computed, we take the average of the current partition plus the result of dfs for the remaining partitions. The maximum value from these recursive calls is the answer for the current dfs.

Step 5: Following the steps recursively will lead us to find the maximum score. For instance, one of the optimal solutions is to partition the array into [9], [1, 2, 3], [9] with scores 9 + 2 + 9 = 20.

Step 6: Once all possibilities for dfs(0, 3) are evaluated, the maximum score that can be returned is cached to ensure that the same state is not recomputed.

Step 7: Finally, dfs(0, k) returns the maximum score for the entire array with k partitions.

The recursive and memoizing nature of the dfs function allows us to efficiently explore all possibilities, while the prefix sums provide a quick way to calculate averages as needed without repeated summation. By combining these techniques, the algorithm efficiently finds the solution.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given recursive algorithm involves analyzing the number of subproblems solved and the time it takes to solve each subproblem. Here, n represents the length of the nums array, and k represents the number of partitions.

• There are at most n * k subproblems because for each starting index i in nums, there are at most k partitions possible.
• For each subproblem defined by a starting index i and a remaining partition count k, the algorithm iterates from i to n-1 to consider all possible partition points.

The time taken for each subproblem is O(n) because of the for-loop from i to n-1. Therefore, the overall time complexity is O(n^2 * k).

Space Complexity

The space complexity consists of the space required by the recursion stack and the caching of subproblem results.

• The maximum depth of the recursion stack is k, because the algorithm makes a recursive call with k-1 whenever it makes a partition.
• The cache stores results for the n * k subproblems.

Therefore, the space complexity is O(n * k) due to caching, plus O(k) for the recursion stack, which simplifies to O(n * k).

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