#2212

Medium

auto_awesomeBacktracking search with pruning

LeetCode Problem Workspace

Maximum Points in an Archery Competition

In the "Maximum Points in an Archery Competition" problem, Bob aims to maximize his score while respecting Alice's arrow counts.

array backtracking bit manipulation enumeration

Problem Statement

Alice and Bob are participating in an archery competition. You are given an integer numArrows, representing the total number of arrows they can shoot, and an integer array aliceArrows of size 12, where each element corresponds to the number of arrows Alice shot in each scoring section of the competition. Bob wants to maximize his total score while ensuring the sum of his arrows across all sections equals numArrows.

Return an array bobArrows of size 12, representing the number of arrows Bob should shoot in each section, such that the sum of bobArrows equals numArrows. The goal is to find a distribution of arrows for Bob that maximizes his total score, which is the sum of the scores he earns by beating Alice's shots in each section.

Examples

Example 1

Input: numArrows = 9, aliceArrows = [1,1,0,1,0,0,2,1,0,1,2,0]

Output: [0,0,0,0,1,1,0,0,1,2,3,1]

The table above shows how the competition is scored. Bob earns a total point of 4 + 5 + 8 + 9 + 10 + 11 = 47. It can be shown that Bob cannot obtain a score higher than 47 points.

Example 2

Input: numArrows = 3, aliceArrows = [0,0,1,0,0,0,0,0,0,0,0,2]

Output: [0,0,0,0,0,0,0,0,1,1,1,0]

The table above shows how the competition is scored. Bob earns a total point of 8 + 9 + 10 = 27. It can be shown that Bob cannot obtain a score higher than 27 points.

Constraints

1 <= numArrows <= 105
aliceArrows.length == bobArrows.length == 12
0 <= aliceArrows[i], bobArrows[i] <= numArrows
sum(aliceArrows[i]) == numArrows

Solution Approach

Backtracking with Pruning

This problem is solved through a backtracking approach with pruning. You need to consider different distributions of arrows Bob could use while ensuring the total is equal to numArrows. For each distribution, check if Bob's score is greater than Alice's in any given section and prune invalid paths early.

Bit Manipulation for Efficiency

In some cases, bit manipulation can be used to optimize the solution, especially when trying to represent the state of Bob's arrow distribution efficiently and evaluate different possibilities quickly. This can reduce the complexity in certain backtracking scenarios.

Greedy Selection of Sections

While backtracking is central, it is helpful to focus on sections where Bob can outscore Alice with the fewest arrows possible. This greedy strategy can reduce the number of computations by selecting promising sections first and pruning other possibilities.

Complexity Analysis

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

The time and space complexity depends on the final approach used. In the worst case, it could be exponential due to the number of possible arrow distributions. However, by using pruning techniques and bit manipulation, the complexity can be reduced significantly.

What Interviewers Usually Probe

Candidate suggests using backtracking with pruning to explore different arrow distributions.
Candidate attempts to optimize the solution using greedy selection or bit manipulation techniques.
Candidate demonstrates an understanding of how pruning reduces unnecessary calculations in this problem.

Common Pitfalls or Variants

Common pitfalls

Failing to prune invalid paths early can lead to excessive computation and timeouts.
Not considering edge cases where numArrows is small or very large can lead to incorrect solutions.
Misunderstanding the problem's goal, such as incorrectly distributing arrows without considering the competition's scoring system.

Follow-up variants

Alter the number of sections (reduce or increase the size of the aliceArrows array).
Introduce additional constraints on the number of arrows Bob can shoot in each section.
Consider allowing a greater number of arrows or a limit on how many sections Bob must outscore Alice.

FAQ

How does backtracking with pruning work in the "Maximum Points in an Archery Competition" problem?

Backtracking with pruning helps by exploring all possible arrow distributions, while pruning cuts off branches where Bob cannot beat Alice’s score, thus reducing unnecessary computations.

What role does bit manipulation play in this problem?

Bit manipulation can be used to efficiently manage and track arrow distributions, speeding up the backtracking process and reducing the space complexity.

Can a greedy approach be used to solve the "Maximum Points in an Archery Competition" problem?

Yes, a greedy approach can be helpful in selecting sections where Bob can beat Alice with the fewest arrows, which can optimize the solution by focusing on the most promising sections first.

What are the major pitfalls to avoid in this problem?

Key pitfalls include failing to prune invalid paths early, not considering all edge cases, and misunderstanding how to distribute arrows based on Alice's scores.

How does GhostInterview help with this problem?

GhostInterview provides a structured approach to implementing the backtracking algorithm, optimizes the process using pruning, and guides you through common pitfalls in solving this problem.

terminal

Solution

Solution 1: Binary Enumeration

Since there are only $12$ regions, we use binary enumeration to determine in which regions $\textit{Bob}$ scores. We use a variable $\textit{st}$ to represent the scheme in which $\textit{Bob}$ obtains the maximum score, and $\textit{mx}$ to represent the maximum score $\textit{Bob}$ obtains.

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class Solution:
    def maximumBobPoints(self, numArrows: int, aliceArrows: List[int]) -> List[int]:
        st = mx = 0
        m = len(aliceArrows)
        for mask in range(1, 1 << m):
            cnt = s = 0
            for i, x in enumerate(aliceArrows):
                if mask >> i & 1:
                    s += i
                    cnt += x + 1
            if cnt <= numArrows and s > mx:
                mx = s
                st = mask
        ans = [0] * m
        for i, x in enumerate(aliceArrows):
            if st >> i & 1:
                ans[i] = x + 1
                numArrows -= ans[i]
        ans[0] += numArrows
        return ans

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