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3180. Maximum Total Reward Using Operations I

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.

 

Example 1:

Input: rewardValues = [1,1,3,3]

Output: 4

Explanation:

During the operations, we can choose to mark the indices 0 and 2 in order, and the total reward will be 4, which is the maximum.

Example 2:

Input: rewardValues = [1,6,4,3,2]

Output: 11

Explanation:

Mark the indices 0, 2, and 1 in order. The total reward will then be 11, which is the maximum.

 

Constraints:

  • 1 <= rewardValues.length <= 2000
  • 1 <= rewardValues[i] <= 2000

Solutions

We can sort the rewardValues array and then use memoization to solve for the maximum total reward.

We define a function $\textit{dfs}(x)$, representing the maximum total reward that can be obtained when the current total reward is $x$. Thus, the answer is $\textit{dfs}(0)$.

The execution process of the function $\textit{dfs}(x)$ is as follows:

  1. Perform a binary search in the rewardValues array for the index $i$ of the first element greater than $x$;
  2. Iterate over the elements in the rewardValues array starting from index $i$, and for each element $v$, calculate the maximum value of $v + \textit{dfs}(x + v)$.
  3. Return the result.

To avoid repeated calculations, we use a memoization array f to record the results that have already been computed.

The time complexity is $O(n \times (\log n + M))$, and the space complexity is $O(M)$. Where $n$ is the length of the rewardValues array, and $M$ is twice the maximum value in the rewardValues array.

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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)

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class Solution {
    private int[] nums;
    private Integer[] f;

    public int maxTotalReward(int[] rewardValues) {
        nums = rewardValues;
        Arrays.sort(nums);
        int n = nums.length;
        f = new Integer[nums[n - 1] << 1];
        return dfs(0);
    }

    private int dfs(int x) {
        if (f[x] != null) {
            return f[x];
        }
        int i = Arrays.binarySearch(nums, x + 1);
        i = i < 0 ? -i - 1 : i;
        int ans = 0;
        for (; i < nums.length; ++i) {
            ans = Math.max(ans, nums[i] + dfs(x + nums[i]));
        }
        return f[x] = ans;
    }
}

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class Solution {
public:
    int maxTotalReward(vector<int>& rewardValues) {
        sort(rewardValues.begin(), rewardValues.end());
        int n = rewardValues.size();
        int f[rewardValues.back() << 1];
        memset(f, -1, sizeof(f));
        function<int(int)> dfs = [&](int x) {
            if (f[x] != -1) {
                return f[x];
            }
            auto it = upper_bound(rewardValues.begin(), rewardValues.end(), x);
            int ans = 0;
            for (; it != rewardValues.end(); ++it) {
                ans = max(ans, rewardValues[it - rewardValues.begin()] + dfs(x + *it));
            }
            return f[x] = ans;
        };
        return dfs(0);
    }
};

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func maxTotalReward(rewardValues []int) int {
    sort.Ints(rewardValues)
    n := len(rewardValues)
    f := make([]int, rewardValues[n-1]<<1)
    for i := range f {
        f[i] = -1
    }
    var dfs func(int) int
    dfs = func(x int) int {
        if f[x] != -1 {
            return f[x]
        }
        i := sort.SearchInts(rewardValues, x+1)
        f[x] = 0
        for