What is the time complexity of Maximum Total Reward Using Operations II?

The optimal solution has a time complexity of O(N * R).

What data structures are used to solve Maximum Total Reward Using Operations II?

This problem uses Array, Dynamic Programming, Bit Manipulation. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Total Reward Using Operations II?

Insight 1: Sorting the unique reward values helps in processing them in ascending order, which is crucial for building the DP bitmask. Insight 2: Using a bitmask (dynamic programming) allows us to efficiently track which reward sums are achievable at each step.

Which companies ask LeetCode 3181. Maximum Total Reward Using Operations II in interviews?

This problem is frequently asked at Mitsogo.

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Maximum Total Reward Using Operations II - LeetCode 3181 Solution

Maximum Total Reward Using Operations II - Complete Solution Guide

Maximum Total Reward Using Operations II is LeetCode problem 3181, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The problem asks us to find the maximum total reward we can obtain from an array of reward values. We start with a total reward of 0, and we can iteratively choose unmarked indices from the array. If the value at the chosen index is greater than our current total reward, we add that value to our total reward and mark the index. The goal is to maximize the final total reward.

Solution Approach

The solution uses a dynamic programming approach with a bitmask to represent the achievable reward sums. The algorithm iterates through the sorted unique reward values. For each reward value 'r', it calculates all previously achievable sums that are less than 'r'. Then, it adds 'r' to each of those sums, effectively creating new achievable sums. The bitmask is updated to include these new sums. Finally, the algorithm finds the largest achievable sum, which is the maximum total reward.

Step-by-Step Algorithm

Step 1: Extract the unique reward values from the input array and sort them in ascending order.
Step 2: Initialize a bitmask 'dp' to 1. This represents that a sum of 0 is initially achievable.
Step 3: Iterate through the sorted unique reward values.
Step 4: For each reward 'r', create a mask of 'r' ones. This mask is used to isolate the achievable sums that are less than 'r'.
Step 5: Find the achievable sums less than 'r' by performing a bitwise AND operation between 'dp' and the mask.
Step 6: Create new achievable sums by shifting the achievable sums less than 'r' to the left by 'r' bits. This effectively adds 'r' to each of those sums.
Step 7: Update the bitmask 'dp' by performing a bitwise OR operation with the new sums. This adds the new achievable sums to the bitmask.
Step 8: After processing all reward values, find the position of the most significant bit (MSB) that is set in 'dp'. This position represents the maximum total reward.
Step 9: Return MSB - 1, as dp also accounts for a reward of 0

Key Insights

Insight 1: Sorting the unique reward values helps in processing them in ascending order, which is crucial for building the DP bitmask.
Insight 2: Using a bitmask (dynamic programming) allows us to efficiently track which reward sums are achievable at each step.
Insight 3: The problem inherently limits the maximum possible sum by the values in the array. This makes a bitmask approach feasible.

Complexity Analysis

Time Complexity: O(N * R)

Space Complexity: O(R)

Topics

This problem involves: Array, Dynamic Programming, Bit Manipulation.

Companies

Asked at: Mitsogo.

Maximum Total Reward Using Operations II | LeetCode 3181

Hard

Array Dynamic Programming Bit Manipulation

Time: O(N * R)Space: O(R)

Solution Code

Available in:

import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
import java.util.stream.Collectors;

class Solution {
    public int maxTotalReward(int[] rewardValues) {
        Set<Integer> rewardSet = new HashSet<>();
        for (int reward : rewardValues) {
            rewardSet.add(reward);
        }
        List<Integer> rewardsList = rewardSet.stream().sorted().collect(Collectors.toList());
        int[] rewards = rewardsList.stream().mapToInt(Integer::intValue).toArray();

        long dp = 1;

        for (int r : rewards) {
            // We need to find all achievable rewards x such that x < r.
            // This corresponds to the lower r bits of our bitmask dp.
            // We can isolate these bits by creating a mask of r ones.
            long mask = (1L << r) - 1;
            long achievable_less_than_r = dp & mask;

            // For each sum x that is achievable and less than r,
            // we can now achieve a new sum of x + r.
            // In terms of bitmasks, this is equivalent to shifting
            // the mask of lower sums by r.
            long new_sums = achievable_less_than_r << r;

            // Update the dp bitmask with these newly achievable sums.
            dp |= new_sums;
        }

        // The maximum total reward is the position of the most significant bit
        // that is set in our final dp bitmask.
        return Long.SIZE - Long.numberOfLeadingZeros(dp) - 1;
    }
}

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Problem Info

Description

Problem 3181: Maximum Total Reward Using Operations II - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Bit Manipulation

Companies

Mitsogo

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3181

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Extract the unique reward values from the input array and sort them in ascending order.
2
Step 2: Initialize a bitmask 'dp' to 1. This represents that a sum of 0 is initially achievable.
3
Step 3: Iterate through the sorted unique reward values.
4
Step 4: For each reward 'r', create a mask of 'r' ones. This mask is used to isolate the achievable sums that are less than 'r'.
5
Step 5: Find the achievable sums less than 'r' by performing a bitwise AND operation between 'dp' and the mask.
6
Step 6: Create new achievable sums by shifting the achievable sums less than 'r' to the left by 'r' bits. This effectively adds 'r' to each of those sums.
7
Step 7: Update the bitmask 'dp' by performing a bitwise OR operation with the new sums. This adds the new achievable sums to the bitmask.
8
Step 8: After processing all reward values, find the position of the most significant bit (MSB) that is set in 'dp'. This position represents the maximum total reward.
9
Step 9: Return MSB - 1, as dp also accounts for a reward of 0

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(N * R)

💾

Space Complexity

O(R)

📝

Why?

The time complexity is dominated by the nested loop structure. The outer loop iterates over the unique reward values, and the inner operations involve bitwise manipulations which, in essence, contribute O(R) time since the number of bits is proportional to the largest reward value. The space complexity is determined by the size of the bitmask, which is proportional to the maximum reward value.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Dynamic Programming with Memoization

A recursive solution with memoization can also be used. We can explore all possible combinations of including or excluding each reward value, keeping track of the current total reward.

Time: O(2^N), where N is the number of reward values. Each reward value has two choices, so the upper bound is 2^N. With memoization, the time complexity can be reduced, but in worst case, it won't beat bitmaskingSpace: O(N), for the recursion stack and memoization table.

Trade-offs: Recursive DP is easier to understand, but can be slower for larger inputs due to the recursive calls and potential for stack overflow. The bitmask approach is generally faster and more memory-efficient for this specific problem's constraints because it compacts the state representation into a bitmask. It is better to avoid for values exceeding the max long / integer.

✨

Key Insights

1Insight 1: Sorting the unique reward values helps in processing them in ascending order, which is crucial for building the DP bitmask.
2Insight 2: Using a bitmask (dynamic programming) allows us to efficiently track which reward sums are achievable at each step.
3Insight 3: The problem inherently limits the maximum possible sum by the values in the array. This makes a bitmask approach feasible.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty input array. Although not explicitly tested in the constraints, an empty array would likely return 0, and the initial dp = 1 and the algorithm should still work fine.
Edge case 2: Input array with only one unique reward value. The algorithm correctly calculates the maximum reward as the value itself.
Edge case 3: A large number of small rewards. The bitmask can become very large if there are numerous small rewards, potentially leading to memory issues if R is excessively large (although the constraint limits rewards[i] <= 5*10^4 which prevents the bitmask from becoming truly enormous).
Edge case 4: Input array where all values are 1. The algorithm will correctly mark all values as 1, since they are greater or equal to the total reward value starting from 0, resulting in the number of 1s being returned.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the unique reward values. Sorting is essential to ensure that the DP bitmask is built up correctly, allowing the algorithm to track all achievable sums.

✗Common mistake 2: Using the original array instead of the unique values. Using the duplicates will lead to suboptimal solutions, because you are not optimizing with the ability to pick from a limited set of sums < r.

✗Common mistake 3: Incorrectly implementing the bitmask operations. The bitwise AND, OR, and shift operations must be performed correctly to ensure that the bitmask accurately represents the achievable reward sums.

✗Common mistake 4: Not handling the edge case of the starting reward value of 0 correctly. Initializing the DP array appropriately (dp = 1) is crucial.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation where resources have different costs and adding resources increases the budget. The goal is to maximize the usage of resources while respecting the budget constraint.

🏦Real-world scenario 2: Optimization problems where certain choices can be made only if a certain threshold is met. For example, investing in projects if the current profit is above a certain level to maximize profit over time.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 30, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Total Reward Using Operations II"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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