#3180

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Maximum Total Reward Using Operations I

Optimize the total reward by choosing operations on array indices using state transition dynamic programming techniques efficiently.

array dynamic programming

Problem Statement

You are given an integer array rewardValues of length n, representing reward values for each index. You start with a total reward of 0 and all indices unmarked. You can repeatedly perform an operation that marks an index and adds its reward to your total, following specific rules that may depend on previously marked indices.

Return the maximum total reward achievable by performing the operations optimally. Carefully plan the sequence of index selections to ensure each state transition contributes to the highest possible total, considering overlapping choices and dynamic programming updates.

Examples

Example 1

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

Output: 4

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

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

Solution Approach

Sort and Prepare DP Table

Start by sorting the rewardValues array if necessary and initialize a DP array where dp[i] represents the maximum reward achievable up to index i. Sorting helps to simplify state transitions and ensures that higher rewards can be prioritized systematically.

Define State Transitions

For each index, consider marking it or skipping it. Update dp[i] based on the maximum reward achievable from previous states, reflecting whether the operation can be performed at this index. Each transition captures the effect of choosing or not choosing an index while tracking cumulative rewards.

Compute Maximum Reward

Iterate through the DP table, updating each state based on previous computations. The final maximum reward is the highest value in the DP array after processing all indices. This ensures all valid sequences of operations are considered efficiently without redundant recalculation.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on how the state transitions are computed, typically O(n^2) in naive DP or O(n log n) with optimizations. Space complexity depends on the DP table size, usually O(n) for a 1D table, reflecting the reward states at each index.

What Interviewers Usually Probe

Watch for proper DP initialization and edge cases when indices are at the array boundaries.
Prioritize understanding how each operation affects subsequent states to avoid overcounting rewards.
Explain your reasoning for sorting or not sorting the array as it directly impacts transition correctness.

Common Pitfalls or Variants

Common pitfalls

Ignoring the effect of previous operations on future state transitions, leading to suboptimal total rewards.
Incorrectly updating DP states, which can either double-count rewards or miss valid sequences.
Not handling edge indices properly, which can cause array out-of-bounds errors or wrong maximum computation.

Follow-up variants

Limit the number of operations allowed and compute maximum reward under operation constraints.
Include negative reward values to test state transition handling with mixed contributions.
Extend to multi-dimensional reward arrays, requiring more complex DP state tracking.

FAQ

What is the main pattern used in Maximum Total Reward Using Operations I?

The problem primarily relies on state transition dynamic programming, tracking rewards accumulated as indices are marked.

Should I sort rewardValues before applying dynamic programming?

Sorting can simplify state transitions and ensure higher rewards are considered in an optimal sequence.

What is the time complexity of the DP solution?

A naive implementation is O(n^2) due to nested state transitions, but optimizations can reduce it depending on the DP update strategy.

How do I avoid double-counting rewards in this problem?

Carefully define DP states to track whether an index is already marked and only update states that represent valid sequences.

Can this DP approach handle large arrays efficiently?

Yes, with proper state management and possible optimizations like prefix sums or reduced state representation, it scales to arrays of length up to 2000.

terminal

Solution

Solution 1: Sorting + Memoization + Binary Search

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

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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)

Solution 2: Dynamic Programming

We define $f[i][j]$ as whether it is possible to obtain a total reward of $j$ using the first $i$ reward values. Initially, $f[0][0] = \textit{True}$, and all other values are $\textit{False}$.

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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)

Solution 3: Dynamic Programming + Bit Manipulation

We can optimize Solution 2 by defining a binary number $f$ to save the current state, where the $i$-th bit of $f$ being $1$ indicates that a total reward of $i$ is reachable.

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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)

Continue Practicing

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