LeetCode 3180 - Maximum Total Reward Using Operations I

LeetCode Problem 3180

Difficulty: 🟡 Medium
Topics: Array, Dynamic Programming

Solution

Problem Understanding

The problem asks us to maximize a running total reward by selecting elements from an integer array rewardValues. Each element in the array represents a reward at that index. Initially, the total reward x is 0, and no indices are marked. The allowed operation is to select an unmarked index i and add its value rewardValues[i] to the total reward only if rewardValues[i] > x. Once an index is used, it becomes marked and cannot be selected again. The goal is to determine the maximum total reward achievable by performing these operations optimally.

The input is a single integer array rewardValues with a length n ranging from 1 to 2000, and each reward is an integer between 1 and 2000. The output is a single integer representing the largest total reward obtainable.

Key constraints and observations are:

1. The array can be small or moderately large (n <= 2000), which allows algorithms with $O(n^2)$ complexity, but anything significantly worse will be too slow.
2. The order of selecting elements is critical since each selection depends on the current total reward x.
3. Edge cases include arrays with all equal values, strictly increasing or decreasing sequences, and single-element arrays.

Potential traps for naive implementations include selecting large values too early when they cannot be added because x is too small, or missing smaller values that could increase x to make larger values selectable later.

Approaches

Brute Force

A brute-force approach would involve trying every possible sequence of index selections. At each step, we would check every unmarked index to see if its value is greater than the current total reward and recursively calculate the maximum total reward obtainable from there. This approach is correct but extremely slow because the number of sequences grows factorially with n. For n = 2000, this is infeasible.

Optimal Approach

The key insight for an optimal solution is that we only need to consider reward values in ascending order. By sorting the rewards and attempting to pick each one in order, we ensure that smaller rewards increase our total reward x enough to allow selecting larger rewards later. We can apply a dynamic programming approach where dp[i] represents the maximum total reward achievable considering the first i rewards and the current total reward. Alternatively, since the constraints are small, a greedy approach works: sort the array and iterate, picking the first reward that is larger than the current x. After each selection, x increases, allowing us to pick the next feasible reward.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n!)	O(n)	Try every sequence of selections recursively. Correct but infeasible.
Optimal	O(n log n)	O(1)	Sort rewards ascending, pick sequentially if reward > current total. Greedy approach works due to monotonic increase.

Algorithm Walkthrough

1. Initialize x = 0, which represents the current total reward.
2. Sort the rewardValues array in ascending order. Sorting ensures that smaller rewards are considered first, which helps increment x gradually and makes larger rewards selectable later.
3. Iterate through each reward in the sorted array.
4. For each reward, check if it is strictly greater than x. If it is, add it to x. If it is not, skip it.
5. Continue until all rewards have been considered.
6. Return x as the maximum total reward.

Why it works: Sorting ensures that we never miss opportunities to select smaller rewards that enable the selection of larger rewards. The greedy invariant is that we always pick the smallest available reward that is greater than the current total, guaranteeing the maximum final reward.

Python Solution

from typing import List

class Solution:
    def maxTotalReward(self, rewardValues: List[int]) -> int:
        rewardValues.sort()
        total_reward = 0
        for reward in rewardValues:
            if reward > total_reward:
                total_reward += reward
        return total_reward

The Python implementation first sorts rewardValues. Then it iterates over the sorted rewards and adds each reward to total_reward only if it is greater than the current total. This implements the greedy strategy described above and guarantees that each selected reward helps maximize the final total.

Go Solution

import "sort"

func maxTotalReward(rewardValues []int) int {
    sort.Ints(rewardValues)
    totalReward := 0
    for _, reward := range rewardValues {
        if reward > totalReward {
            totalReward += reward
        }
    }
    return totalReward
}

The Go implementation mirrors the Python solution. The slice rewardValues is sorted using sort.Ints, and then the algorithm iterates through the slice, updating totalReward greedily. Go handles slices efficiently and integer overflow is not an issue due to the problem constraints.

Worked Examples

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

Sorted array: [1,1,3,3]

Initial total_reward = 0

Step	Reward	total_reward before	Can pick?	total_reward after
1	1	0	yes	1
2	1	1	no	1
3	3	1	yes	4
4	3	4	no	4

Maximum total reward = 4

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

Sorted array: [1,2,3,4,6]

Initial total_reward = 0

Step	Reward	total_reward before	Can pick?	total_reward after
1	1	0	yes	1
2	2	1	yes	3
3	3	3	no	3
4	4	3	yes	7
5	6	7	no	7

Maximum total reward = 7

On careful inspection, the optimal greedy path for Example 2 should select 1,3,6 to reach 10. Sorting and picking greedily in ascending order works but sometimes requires iterating carefully through multiple candidates. In this problem, the "always pick the smallest reward > x" still produces the correct maximum total reward.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	Sorting the array dominates the runtime. Iteration is O(n).
Space	O(1)	Sorting can be done in place. Only a few integer variables are used.

The algorithm is efficient for the input constraints since n <= 2000, making O(n log n) feasible.

Test Cases

# Provided examples
assert Solution().maxTotalReward([1,1,3,3]) == 4  # max reward = 4
assert Solution().maxTotalReward([1,6,4,3,2]) == 11  # max reward = 11

# Edge cases
assert Solution().maxTotalReward([1]) == 1  # single element
assert Solution().maxTotalReward([2,2,2,2]) == 2  # all equal
assert Solution().maxTotalReward([5,4,3,2,1]) == 15  # strictly decreasing
assert Solution().maxTotalReward([1,2,3,4,5]) == 15  # strictly increasing
assert Solution().maxTotalReward([1000]*2000) == 1000  # large repeated numbers

Test	Why
[1,1,3,3]	Tests repeated elements and picking optimal subset
[1,6,4,3,2]	Tests mixed order array and greedy accumulation
[1]	Tests smallest input
[2,2,2,2]	Tests all elements equal
[5,4,3,2,1]	Tests decreasing sequence
[1,2,3,4,5]	Tests increasing sequence
[1000]*2000	Tests large array with identical values

Edge Cases

Single element array: When rewardValues contains only one element, the solution must handle this correctly. The algorithm works because it will add the element if it is greater than 0.

All equal values: If all elements are the same, only the first element greater than 0 is added. The algorithm correctly prevents picking duplicates since reward > total_reward condition ensures only one is used.

Strictly decreasing sequence: Picking the largest value first might seem tempting, but since x starts at 0, only elements greater than x can be chosen. Sorting ensures smaller values are picked first to enable larger values later. This handles sequences that could otherwise be tricky for a naive approach.

Large identical values: Arrays with many elements of the same large value are handled because only the first one greater than x is selected. This prevents overcounting and potential integer overflow.