LeetCode #3181 — HARD

Maximum Total Reward Using Operations II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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 <= 5 * 10⁴
1 <= rewardValues[i] <= 5 * 10⁴

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Bit Manipulation

Example 1

[1,1,3,3]

Example 2

[1,6,4,3,2]

Step 02

Core Insight

What unlocks the optimal approach

Sort the rewards array first.
If we decide to apply some rewards, it's always optimal to apply them in order.
The transition is given by: <code>dp[i][j] = dp[i - 1][j − rewardValues[i]]</code> if <code>j − rewardValues[i] < rewardValues[i]</code>.
Note that the dp array is a boolean array. We just need 1 bit per element, so we can use a bitset or something similar. We just need a "stream" of bits and apply bitwise operations to optimize the computations by a constant factor.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3181: Maximum Total Reward Using Operations II
import java.math.BigInteger;

class Solution {
    public int maxTotalReward(int[] rewardValues) {
        int[] nums = Arrays.stream(rewardValues).distinct().sorted().toArray();
        BigInteger f = BigInteger.ONE;
        for (int v : nums) {
            BigInteger mask = BigInteger.ONE.shiftLeft(v).subtract(BigInteger.ONE);
            BigInteger shifted = f.and(mask).shiftLeft(v);
            f = f.or(shifted);
        }
        return f.bitLength() - 1;
    }
}

// Accepted solution for LeetCode #3181: Maximum Total Reward Using Operations II
func maxTotalReward(rewardValues []int) int {
	slices.Sort(rewardValues)
	rewardValues = slices.Compact(rewardValues)
	one := big.NewInt(1)
	f := big.NewInt(1)
	p := new(big.Int)
	for _, v := range rewardValues {
		mask := p.Sub(p.Lsh(one, uint(v)), one)
		f.Or(f, p.Lsh(p.And(f, mask), uint(v)))
	}
	return f.BitLen() - 1
}

# Accepted solution for LeetCode #3181: Maximum Total Reward Using Operations II
class Solution:
    def maxTotalReward(self, rewardValues: List[int]) -> int:
        nums = sorted(set(rewardValues))
        f = 1
        for v in nums:
            f |= (f & ((1 << v) - 1)) << v
        return f.bit_length() - 1

// Accepted solution for LeetCode #3181: Maximum Total Reward Using Operations II
/**
 * [3181] Maximum Total Reward Using Operations II
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn max_total_reward(reward_values: Vec<i32>) -> i32 {
        let mut reward_values: Vec<usize> = reward_values.into_iter().map(|x| x as usize).collect();
        reward_values.sort_unstable();
        reward_values.dedup();

        let max_value = *reward_values.last().unwrap();
        let n = reward_values.len();

        if n >= 2 && max_value - 1 == reward_values[n - 2] {
            return (2 * max_value - 1) as i32;
        }

        let mut dp = vec![false; 2 * max_value + 1];
        dp[0] = true;

        unsafe {
            for v in reward_values {
                for i in (v..v << 1).rev() {
                    *(dp.get_unchecked_mut(i)) |= *dp.get_unchecked(i - v);
                }
            }
        }

        dp.iter().enumerate().rfind(|(_, &x)| x).unwrap().0 as i32
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3181() {
        assert_eq!(4, Solution::max_total_reward(vec![1, 1, 3, 3]));
        assert_eq!(11, Solution::max_total_reward(vec![1, 6, 4, 3, 2]));
    }
}

// Accepted solution for LeetCode #3181: Maximum Total Reward Using Operations II
function maxTotalReward(rewardValues: number[]): number {
    rewardValues.sort((a, b) => a - b);
    rewardValues = [...new Set(rewardValues)];
    let f = 1n;
    for (const x of rewardValues) {
        const mask = (1n << BigInt(x)) - 1n;
        f = f | ((f & mask) << BigInt(x));
    }
    return f.toString(2).length - 1;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × M / w)

Space

O(n + M / w)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.