3040. Maximum Number of Operations With the Same Score II

Problem Description

In this problem, we're asked to focus on an array of integers called nums. We can perform an operation on this array as long as it has at least two elements left. An operation consists of choosing and deleting any two elements from the array based on the following three possibilities:

1. The first two elements.
2. The last two elements.
3. The first and the last elements.

The score for an operation is calculated by summing up the deleted elements. We need to determine the maximum number of operations we can perform where each operation yields the same score. The goal is to return this maximum number of operations that all have the same score.

To understand how to approach this problem, one must recognize the constraint that every operation must yield the same score. This significantly narrows down the possibilities and directs us towards considering only specific pairs that sum up to a uniform score.

Intuition

The solution to this problem involves memorization and depth-first search (DFS) to explore all valid operations. Given the constraint that all operations must yield the same score, the initial observation is that there are only three potential scores, based on pairs that can be taken from either the beginning or the end of the array. These scores are:

1. The sum of the first two elements.
2. The sum of the last two elements.
3. The sum of the first and the last elements.

Using these scores, we use DFS to determine the maximum number of pairs that can be created with the same sum. A recursive dfs function is employed to attempt to form pairs starting from different indices of the array for each of the three scores. The arguments i and j represent the current subarray's bounds we're searching, and s represents the targeted score. The function returns the maximum number of valid operations between positions i and j.

To avoid redundant calculations, we use memorization (caching the results of previous computations). This optimization is particularly useful when overlapping subproblems occur, which is a common characteristic in this kind of recursive search.

In summary, the intuition is to exhaustively search for the maximum number of pairs that can be formed for each of the three possible scores and to use caching to optimize the process. Finally, we compare the results of these three exhaustive searches to find and return the maximum achievable number of operations with the same score.

Learn more about Memoization and Dynamic Programming patterns.

Solution Approach

The solution provided utilizes a recursive DFS (Depth-First Search) strategy, coupled with memoization to significantly reduce the complexity that would typically be associated with brute force recursive algorithms due to repeated computations.

DFS with Memoization

The core algorithm is wrapped within the dfs(i, j, s) function, where i and j represent the indices defining the current subarray of nums, and s is the constant score that all operations must match.

Here's a deep dive into the dfs function steps:

• Base Case: If the current subarray size defined by i and j is less than 2 (j - i < 1), we cannot perform any more operations, so we return 0.

• Recursive Step: The function attempts to delete a pair of elements from either end of the current subarray or from both ends, provided the sum of the chosen elements equals s. It explores three possibilities:

  • If nums[i] + nums[i + 1] == s, deletes these elements and proceeds recursively with dfs(i + 2, j, s).
  • If nums[i] + nums[j] == s, deletes these elements and proceeds recursively with dfs(i + 1, j - 1, s).
  • If nums[j - 1] + nums[j] == s, deletes these elements and proceeds recursively with dfs(i, j - 2, s).

Each recursive call returns the maximum number of operations starting from that subsequence with indices i and j. The function keeps track of the maximum possible operations (ans) that respect the equal score rule in each subtree, essentially performing a depth-first traversal of all the valid operation sequences.

Memoization:

To avoid recalculating the results for the same subarray, the dfs function uses the @cache decorator for memoization, which stores the result of each unique (i, j, s) state. This optimization prevents duplicate calculations, which can quickly explode in a recursive algorithm.

Main Solution Workflow:

The solution block calculates the maximum possible operations for each of the three valid scores separately. For each score, we adjust the subarray and call the dfs function:

1. The score is the sum of the first two elements: dfs(2, n - 1, nums[0] + nums[1])
2. The score is the sum of the last two elements: dfs(0, n - 3, nums[-1] + nums[-2])
3. The score is the sum of the first and last elements: dfs(1, n - 2, nums[0] + nums[-1])

The algorithm returns the maximum number of operations for a single score plus one (to account for the initial operation that determined the score).

This approach avoids unnecessary computations while effectively exploring all meaningful paths, leading to an efficient solution to the problem.

Example Walkthrough

Let's consider a simple example where nums = [5, 3, 1, 2, 4, 6]. The task is to find the maximum number of operations with equal scores.

According to our problem description, we have three initial scores to consider based on the first two elements, the last two elements, and the first and last elements. These scores are:

1. nums[0] + nums[1] = 5 + 3 = 8
2. nums[5] + nums[4] = 6 + 4 = 10
3. nums[0] + nums[5] = 5 + 6 = 11

Each score will lead to a different path of operations and we have to explore each through DFS.

Let's pick the first score (8) and apply our DFS approach:

• Start with dfs(2, n - 1, nums[0] + nums[1]) which is dfs(2, 5, 8). We are looking to find pairs that sum up to 8 in the subarray nums[2:6] which is [1, 2, 4, 6].

• For the current subarray, neither the first two 1 + 2 nor the last two 4 + 6 sum up to 8, but the first and last 1 + 6 do. We can perform one operation here bringing our subarray to [2, 4].

• With the new subarray [2, 4], we notice that 2 + 4 equals 6, not 8. Hence, we cannot perform any further operations to match our score of 8. The dfs function would return 0 for this path because we have no more valid operations.

• The result for the first score would then be one initial operation that determined the score, plus dfs(2, 5, 8) which returns 0. So, for a score of 8, we get 1 + 0 = 1 operation.

Applying the same steps for the second and third scores (10 and 11 respectively), we would eventually find the maximum number of operations for each score and then return the largest among them. This way, we will have recursively explored all possible operations respecting the constraint that they should all yield the same score and guaranteed to have found the maximum number of such operations.

Suppose the second starting score 10 led to the maximum of 2 operations. Then, the result for our nums array would be 2 + 1 = 3 operations including the operation that yielded the initial score of 10.

This example simplifies the DFS process but illustrates the recursive nature and decision-making involved in arriving at the solution using the described approach with memoization aiding efficiency.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n^2) where n is the length of the input list nums. This is because the dfs function that is used to compute the answer is called recursively, and at each step, it considers either incrementing the left pointer by 2, incrementing the left pointer and decrementing the right pointer, or decrementing the right pointer by 2. Therefore, at most n/2 recursive calls can be made at each level of recursion before base cases are reached, and there are n possible positions for the left and right pointers, which results in O(n^2) recursive calls in the worst case.

The space complexity of the code is also O(n^2) due to the memoization (@cache) used in the recursive dfs function. This memoization stores results for each unique pair of indices (i, j) to avoid redundant calculations. Since there can be n choices for i and n choices for j, there can be at most O(n^2) unique states to store, which determines the space complexity.

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