Problem Description

You are given an integer array rewardValues of length n, representing the values of rewards.

Initially, your total reward x is 0, and all indices are unmarked. You are allowed to perform the following operation any number of times:

• Choose an unmarked index i from the range [0, n - 1].
• If rewardValues[i] is greater than your current total reward x, then add rewardValues[i] to x (i.e., x = x + rewardValues[i]), and mark the index i.

Return an integer denoting the maximum total reward you can collect by performing the operations optimally.

Intuition

To solve this problem efficiently, we need to figure out a way to maximize the total reward x by leveraging the constraints given. The core idea is to exploit the fact that we can only add reward values greater than the current total reward x.

1. Sort the Rewards: First, sorting rewardValues allows us to later find, using binary search, the smallest value that can be added to x.

2. Use Dynamic Programming (DFS + Memoization): We define a function dfs(x) that returns the maximal reward that can be obtained when the current total reward is x. Using memoization reduces redundant calculations.

3. Binary Search Efficiency: For each current x, perform a binary search to quickly locate the smallest index i in rewardValues satisfying rewardValues[i] > x. This operation allows focus solely on feasible reward candidates.

4. Recursive Solution Exploration: Recursively calculate the potential maximal reward starting from index i. For each v selected, update x to x + v and calculate v + dfs(x + v) to explore further sums, maintaining an optimal approach throughout.

By taking advantage of sorting and efficient searching, combined with recursive decision making and memoization, the solution maximally accumulates rewards.

Learn more about Dynamic Programming patterns.

Solution Approach

Sorting + Memoization + Binary Search

The solution uses a combination of sorting, memoization, and binary search to efficiently collect the maximum reward.

1. Sorting: Begin by sorting the rewardValues array. Sorting allows us to quickly identify the next feasible reward value using binary search, thereby optimizing our decision-making process.

2. Memoization: We implement memoization using a cache to store already computed results of the function dfs(x). This avoids redundant calculations and significantly enhances performance, especially for overlapping subproblems characteristic of dynamic programming.

3. Binary Search: For each current reward x, use bisect_right to perform a binary search on the sorted rewardValues array. This search finds the smallest index i where rewardValues[i] is greater than x.

4. DFS (Depth-First Search) Recursion: Define the recursive function dfs(x):

   • Perform a binary search to find index i where rewardValues[i] > x.
   • Iterate over all reward values starting from index i.
   • For each reward value v that can be added, calculate v + dfs(x + v) and determine the maximum possible reward.
   • Return this maximum calculated reward, storing results in the cache for future access.

Here's how the implementation ties everything together:

class Solution:
    def maxTotalReward(self, rewardValues: List[int]) -> int:
        @cache
        def dfs(x: int) -> int:
            i = bisect_right(rewardValues, x)
            ans = 0
            for v in rewardValues[i:]:
                ans = max(ans, v + dfs(x + v))
            return ans

        rewardValues.sort()
        return dfs(0)

The use of sorting allows for an ordered approach to selecting rewards, memoization avoids recalculations, and the binary search accelerates the identification of optimal choices, effectively combining to solve the problem.

Example Walkthrough

Let's go through the solution approach using a small example to illustrate the steps clearly.

Example

Consider the rewardValues array: [2, 9, 5, 3].

1. Sorting:

   • First, sort the array to get [2, 3, 5, 9]. This helps in efficiently identifying the next potential reward value to add to our total reward x.

2. Initial Setup:

   • Start with x = 0. This is our initial reward, and we need to maximize it by choosing unmarked indices with values greater than x.

3. Memoization:

   • Implement a memoization function dfs(x) to avoid redundant calculations. We'll store results for specific x values to reuse them when needed.

4. Binary Search:

   • Use binary search (bisect_right) to find the smallest index i where the reward value is greater than the current x. For x = 0, the smallest reward value greater than 0 is 2.

5. Recursive DFS Calculation:

   • Starting from index i = 0 (value 2):
     • Add 2 to x: Now, x = 0 + 2 = 2.
     • Calculate potential reward by calling dfs(x). The next potential values are 3, 5, and 9.
     • Next, check adding 3:
       • For x = 2, increase x to 2 + 3 = 5.
       • Continue recursively, exploring further increases, such as adding 5 or 9.
     • Ultimately credit the reward by evaluating each combination and choose the optimal sum path using memoized results.

6. Maximize Reward:

   • Repeat the process recursively for all viable starting indices, using memoized results to quickly obtain previously calculated paths.
   • The maximum reward obtainable will be returned by dfs(0).

This structured approach allows for efficiently reaching the optimal solution through a combination of sorting, binary search, and recursive exploration with memoization.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n \times (\log n + M)), where n is the length of the rewardValues array, and M is twice the maximum value in the rewardValues array. This complexity arises due to the following:

1. Sorting the rewardValues array takes O(n \log n).
2. For each recursive call in the depth-first search (dfs), finding the index with bisect_right takes O(\log n).
3. The number of possible states for dfs(x) is proportional to M, as the maximum possible sum that can be accumulated depends on the values in rewardValues.

The space complexity is O(M), primarily for the cache used to store the results of dfs computations, where M is again related to the largest possible value of the accumulated rewards.

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