What is the time complexity of Maximum Points in an Archery Competition?

The optimal solution has a time complexity of O(2^n).

What data structures are used to solve Maximum Points in an Archery Competition?

This problem uses Array, Backtracking, Bit Manipulation, Enumeration. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Points in an Archery Competition?

Insight 1: Bob should only focus on target sections where he can score points (i.e., shoot more arrows than Alice). Insight 2: The optimal strategy involves choosing a subset of target sections to win and allocating arrows accordingly.

Which companies ask LeetCode 2212. Maximum Points in an Archery Competition in interviews?

This problem is frequently asked at Kakao.

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Maximum Points in an Archery Competition - LeetCode 2212 Solution

Q: How to solve Maximum Points in an Archery Competition?

The solution uses a brute-force approach combined with bit manipulation to explore all possible subsets of sections Bob could try to win. For each subset, it calculates the total score Bob would achieve and the number of arrows needed. If Bob has enough arrows for a particular subset, the solution checks if the obtained score is greater than the current maximum score. If it is, the current best score and the number of arrows used on each section are updated. Finally, the remaining arrows, if any, are put in section 0.

Maximum Points in an Archery Competition - Complete Solution Guide

Maximum Points in an Archery Competition is LeetCode problem 2212, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Alice and Bob are opponents in an archery competition. The competition has set the following rules: Alice first shoots numArrows arrows and then Bob shoots numArrows arrows. The points are then calculated as follows: The target has integer scoring sections ranging from 0 to 11 inclusive . For each section of the target with score k (in between 0 to 11 ), say Alice and Bob have shot a k and b k arrows on that section respectively. If a k >= b k , then Alice takes k points. If a k < b k , then Bob

Detailed Explanation

The problem describes an archery competition between Alice and Bob. Alice shoots a given number of arrows (numArrows), distributing them across 12 target sections (0-11). We are given the number of arrows Alice shot at each section. Bob also shoots numArrows arrows, and his goal is to maximize his total score. The scoring system is as follows: If Bob shoots more arrows than Alice on a section, Bob gets points equal to the section number. If Alice shoots at least as many arrows as Bob, Alice gets the points. If both shoot zero arrows, nobody gets the points. We must return an array representing the number of arrows Bob shoots on each section to maximize his score, such that the sum of the arrows equals numArrows.

Solution Approach

The solution uses a brute-force approach combined with bit manipulation to explore all possible subsets of sections Bob could try to win. For each subset, it calculates the total score Bob would achieve and the number of arrows needed. If Bob has enough arrows for a particular subset, the solution checks if the obtained score is greater than the current maximum score. If it is, the current best score and the number of arrows used on each section are updated. Finally, the remaining arrows, if any, are put in section 0.

Step-by-Step Algorithm

Step 1: Initialize max_score to -1 and best_bob_arrows to an array of 12 zeros.
Step 2: Iterate through all possible subsets of sections 1 to 11 using a bitmask 'i' from 0 to 2^11 - 1.
Step 3: For each subset, initialize current_score to 0, arrows_used to 0, and current_bob_arrows to an array of 12 zeros.
Step 4: Iterate through sections 1 to 11 (represented by index k from 1 to 11).
Step 5: If the (k-1)-th bit of the bitmask 'i' is set (meaning Bob wins this section), calculate the arrows needed (aliceArrows[k] + 1), add k to current_score, add the arrows needed to arrows_used, and set current_bob_arrows[k] to the number of arrows needed.
Step 6: If arrows_used is less than or equal to numArrows, check if current_score is greater than max_score. If it is, update max_score to current_score, update best_bob_arrows to current_bob_arrows, and put any remaining arrows (numArrows - arrows_used) in section 0.
Step 7: After iterating through all subsets, return best_bob_arrows.

Key Insights

Insight 1: Bob should only focus on target sections where he can score points (i.e., shoot more arrows than Alice).
Insight 2: The optimal strategy involves choosing a subset of target sections to win and allocating arrows accordingly.
Insight 3: Bit manipulation can be used to efficiently iterate through all possible subsets of target sections (from section 1 to 11).
Insight 4: Any remaining arrows after maximizing the score should be placed in section 0 to avoid giving Alice extra points.

Complexity Analysis

Time Complexity: O(2^n)

Space Complexity: O(n)

Topics

This problem involves: Array, Backtracking, Bit Manipulation, Enumeration.

Companies

Asked at: Kakao.

Maximum Points in an Archery Competition | LeetCode 2212

Medium

Array Backtracking Bit Manipulation Enumeration

Time: O(2^n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int[] maximumBobPoints(int numArrows, int[] aliceArrows) {
        int max_score = -1;
        int[] best_bob_arrows = new int[12];

        // Iterate through all 2^11 subsets of sections {1, ..., 11} that Bob can win.
        // A bitmask `i` represents a subset.
        for (int i = 0; i < (1 << 11); i++) {
            int current_score = 0;
            int arrows_spent = 0;
            int[] current_bob_arrows = new int[12];

            // For each section k from 1 to 11
            for (int k = 1; k < 12; k++) {
                // Check if we decide to win section k.
                // Section k corresponds to the (k-1)-th bit of i.
                if (((i >> (k - 1)) & 1) == 1) {
                    int cost = aliceArrows[k] + 1;
                    arrows_spent += cost;
                    current_score += k;
                    current_bob_arrows[k] = cost;
                }
            }

            // If this is a valid strategy (Bob has enough arrows)
            if (arrows_spent <= numArrows) {
                // If this strategy yields a better score
                if (current_score > max_score) {
                    max_score = current_score;
                    // Any remaining arrows are assigned to section 0
                    int remaining_arrows = numArrows - arrows_spent;
                    current_bob_arrows[0] = remaining_arrows;
                    best_bob_arrows = current_bob_arrows.clone(); // Important: create a copy to avoid modification
                }
            }
        }

        return best_bob_arrows;
    }
}

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Problem Info

Description

Problem 2212: Maximum Points in an Archery Competition - Detailed explanation and solution approaches.

Topics

Array Backtracking Bit Manipulation Enumeration

Companies

Kakao

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2212

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

7 steps

1
Step 1: Initialize max_score to -1 and best_bob_arrows to an array of 12 zeros.
2
Step 2: Iterate through all possible subsets of sections 1 to 11 using a bitmask 'i' from 0 to 2^11 - 1.
3
Step 3: For each subset, initialize current_score to 0, arrows_used to 0, and current_bob_arrows to an array of 12 zeros.
4
Step 4: Iterate through sections 1 to 11 (represented by index k from 1 to 11).
5
Step 5: If the (k-1)-th bit of the bitmask 'i' is set (meaning Bob wins this section), calculate the arrows needed (aliceArrows[k] + 1), add k to current_score, add the arrows needed to arrows_used, and set current_bob_arrows[k] to the number of arrows needed.
6
Step 6: If arrows_used is less than or equal to numArrows, check if current_score is greater than max_score. If it is, update max_score to current_score, update best_bob_arrows to current_bob_arrows, and put any remaining arrows (numArrows - arrows_used) in section 0.
7
Step 7: After iterating through all subsets, return best_bob_arrows.

💡

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(2^n)

🐌 Exponential - Very slow! Often needs optimization.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is exponential because we are exploring all possible subsets of sections. The space complexity is linear because we use an array of size 12 to store the results.

📖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

Dynamic Programming

Dynamic programming could be used to solve this problem. We can create a DP table to store the maximum score that can be achieved with a given number of arrows and a given set of sections. However, since n is fixed at 11, the benefits of DP may not outweigh the increased complexity in implementation.

Time: O(n * numArrows), where n is 11.Space: O(n * numArrows)

Trade-offs: DP may be slightly more efficient if numArrows is very large and close to the constraint limit. However, for the given constraints, the brute-force approach is simpler and faster due to its constant factor advantage.

✨

Key Insights

1Insight 1: Bob should only focus on target sections where he can score points (i.e., shoot more arrows than Alice).
2Insight 2: The optimal strategy involves choosing a subset of target sections to win and allocating arrows accordingly.
3Insight 3: Bit manipulation can be used to efficiently iterate through all possible subsets of target sections (from section 1 to 11).
4Insight 4: Any remaining arrows after maximizing the score should be placed in section 0 to avoid giving Alice extra points.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: numArrows is very small. The algorithm correctly handles this by potentially choosing an empty subset, or a small subset based on what Alice has. Remaining arrows are put in section 0.
Edge case 2: numArrows is just enough to win all sections. The algorithm naturally finds the subset that provides the highest score without exceeding numArrows.
Edge case 3: aliceArrows is all zeros. Bob can win all rounds between 1-11, and the code will find this combination.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not adding 1 to aliceArrows[k] to determine the number of arrows needed to win section k. If Bob shoots the same number of arrows as Alice, Alice gets the points.

✗Common mistake 2: Not assigning the remaining arrows to section 0. If the remaining arrows are not used, Alice can potentially get points if she has arrows in those sections.

✗Common mistake 3: Not creating a copy of the current_bob_arrows array when updating the best_bob_arrows (especially in languages like Java). If you directly assign references, modifications to current_bob_arrows will affect best_bob_arrows.

✗Common mistake 4: Incorrectly handling the bit manipulation, causing incorrect subsets to be considered.

✅

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. Suppose we have a limited budget (numArrows) and want to invest in different projects (sections) to maximize returns (score). Each project has a certain minimum investment requirement (aliceArrows).

🏦Real-world scenario 2: Game strategy optimization. In games where you must allocate resources across different actions to maximize score, this problem represents how to allocate resources optimally, considering constraints and opponent's moves.

🎮Real-world scenario 3: Portfolio Optimization: You have a limited amount of capital (numArrows) to invest across different assets (sections). Each asset has a certain risk profile (represented by Alice's investment). You want to maximize your expected return (score) while ensuring you invest enough to beat the risk.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: July 23, 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What is the difference between backtracking and brute force?

Backtracking prunes invalid paths early using constraints, while brute force explores all possibilities. In backtracking, you "choose" an option, "explore" it recursively, then "unchoose" (backtrack) if it doesn't lead to a solution. This eliminates entire branches of the search tree.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Points in an Archery Competition"?

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

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