2212. Maximum Points in an Archery Competition

Problem Description

Alice and Bob participate in an archery competition with a unique scoring system. They each shoot numArrows arrows at a target divided into sections numbered from 0 to 11. For each section, a player earns points equal to the section number if they shoot strictly more arrows into that section than the opponent, and no points are awarded if both competitors shoot an equal number of arrows and that number is zero.

The challenge is to create a strategy for Bob to shoot his arrows to maximize his points, given the fixed number of arrows he can shoot (numArrows) and an array aliceArrows detailing how many arrows Alice shot in each section.

The goal is to return the array bobArrows representing the number of arrows Bob should shoot at each scoring section to maximize his points total. Bob's strategy doesn't have to be unique; any optimal configuration satisfying the points maximization condition is a valid answer as long as the sum of bobArrows equals numArrows.

Intuition

The problem is essentially about making strategic decisions to outscore Alice while working with a limited resource: a fixed number of arrows. Since higher-scoring sections contribute more to the total score, it's generally advantageous for Bob to focus on winning these sections, provided the cost (number of arrows needed) does not exceed the potential gain in points.

However, we must balance this approach with the constraint on the total number of arrows Bob has. The strategy thus becomes a variant of the knapsack problem, where each section is like an item with a 'weight' (number of arrows needed to win the section) and 'value' (the score of the section), and the total number of arrows Bob has is like the capacity of the knapsack.

To arrive at the optimal solution, we can use a bit mask to represent whether or not Bob should compete in each individual section (1 for compete, 0 for not compete). We iterate over all possible combinations (states) Bob could take, and for each state, calculate the total points earned and the number of arrows used. We keep track of the state that yields the highest score without exceeding the total number of arrows Bob can use.

The solution code then converts the optimal state into a specific distribution of arrows, ensuring that the sections Bob won are those indicated by the state, and that all leftover arrows (if any) are assigned to the 0 section, which has no score and hence will not change Bob's total points.

Learn more about Backtracking patterns.

Solution Approach

The solution involves a bit manipulation and brute-force approach, iterating over all possible states that represent shooting strategies for Bob. For each state, we check if it is possible for Bob to use his arrows to gain more points than Alice with the numArrows available.

Here's the step-by-step breakdown of the implementation, referencing the supplied Python code:

1. Initialization: We know there are 12 sections on the target, so n = len(aliceArrows) ensures we work with all sections. We initialize state to 0 to track the best combination of movements for Bob to beat or tie Alice's score, and mx to -1 to keep track of the maximum points Bob can score.

2. Bitmask Enumeration: We loop through all 2^n (from 1 to 1 << n, exclusive of 1 << n) possible combinations of sections using a bitmask mask. Each bit in the mask indicates whether Bob should try to win that section (bit is 1) or not (bit is 0).

3. Scoring Calculation: Inside the loop, we initialize cnt to track the number of arrows Bob is shooting under this state and points to track the score acquired. We iterate through all sections, and if the bitmask indicates Bob should compete in a section i (using (mask >> i) & 1), we add the number of arrows Alice shot in this section plus one to cnt and increment points with the section number i.

4. State and Score Updating: If the number of arrows cnt is less than or equal to numArrows (meaning Bob can actually apply this strategy without running out of arrows), and the accumulated points of this state are higher than the maximum points mx seen so far, we update state with the current bitmask and mx with the current points.

5. Construct the Result Array: After finding the optimal state, we initialize an array ans of zeros with n slots to represent the number of arrows Bob will shoot in each section. We go through each section again and if the state indicates that Bob should win that section, we set ans[i] with the number Alice shot plus one (alice + 1), and subtract that amount from numArrows. We assign all remaining arrows to section 0, which does not contribute any points (ans[0] = numArrows).

6. Returning the Result: Finally, we return the array ans, which represents the strategy that Bob should follow to shoot his arrows to maximize his score with respect to Alice's recorded shots.

Through this approach, we have ensured Bob's possible strategies are exhaustively considered and the most beneficial one is chosen, in compliance with the constraints provided by the number of arrows Bob has (numArrows) and the result of Alice's shots (aliceArrows).

Example Walkthrough

Let's use a small example to illustrate the solution approach described. Let's say Bob has numArrows = 5 arrows he can shoot and Alice's shots in each section are given by the array aliceArrows = [0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0].

We're looking to maximize Bob's score by deciding how many arrows Bob should shoot in each section.

1. Initialization: We have n = 12 sections on the target. Initially, state = 0 and mx = -1.

2. Bitmask Enumeration: We go through all possible 2^12 combinations of sections. For illustration, we'll consider two example states (bitmasks): 0b000000000101 where Bob chooses to compete in sections 2 and 0; and 0b000000100001, where Bob chooses to compete in sections 0 and 3.

3. Scoring Calculation:

   • For the state 0b000000000101, Bob would need to shoot aliceArrows[2]+1 = 2 arrows to win section 2 and no arrows in section 0 since Alice shot 0 arrows there. This means he uses 2 arrows and scores 2 points.
   • For the state 0b000000100001, Bob needs aliceArrows[3]+1 = 3 arrows to win section 3. This uses 3 arrows and gets 3 points.

4. State and Score Updating:

   • With the first state 0b000000000101, Bob uses 2 arrows out of 5, scoring 2 points. This is our current maximum.
   • With the second state 0b000000100001, Bob uses 3 arrows scoring 3 points. This is now our new maximum, so state = 0b000000100001 and mx = 3.

5. Construct the Result Array: Based on the state 0b000000100001, the array ans would start as [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]. We update it to have 3 arrows in section 3 (ans[3] = 3), and since Bob has 2 arrows remaining, we put them in section 0 (ans[0] = 2) for no additional points.

6. Returning the Result: The final ans array representing Bob's optimal strategy is [2, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0].

Through this example, we followed the solution approach to maximize Bob's score by distributing his 5 arrows across the sections, taking into account Alice's shots. The outcome showed that competing in sections with a higher score (i.e., section 3 worth 3 points) is generally more beneficial than competing in several lower-scoring sections.

Solution Implementation

Time and Space Complexity

The given code aims to maximize the points Bob can score over Alice by choosing specific targets to shoot given a limited number of arrows. The total number of targets is denoted by n.

Time Complexity:

The time complexity of the code is determined by the main loop iterating over all the subsets of the targets that could be shot by Bob (2^n possible states), and the inner loop which sums the arrows needed and the points scored for each subset.

• Main loop: iterates over 2^n subsets (2^n times)
• Inner loop: iterates over n targets to calculate cnt and points (takes O(n) for each subset)

Multiplying the number of iterations of both loops gives us the overall time complexity of the code, which will be O(n * 2^n).

Space Complexity:

The space complexity of the code involves storing the best state (state), maximum points (mx), the final answer (ans array), and the variables used in loop iterations.

• The ans array uses O(n) space.
• The variables state, mx, mask, cnt, and points use O(1) space each.

Since ans is the most significant space consumption, the overall space complexity is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.