LeetCode 2212 - Maximum Points in an Archery Competition

The problem is asking us to determine the optimal way for Bob to distribute his numArrows across 12 scoring sections (from 0 to 11) in an archery competition in order to maximize his total points against Alice. Alice's arrow distribution is given in aliceArrows.

leetcodemediumarraybacktrackingbit-manipulationenumeration

LeetCode Problem 2212

Difficulty: 🟡 Medium
Topics: Array, Backtracking, Bit Manipulation, Enumeration

Solution

Problem Understanding

The problem is asking us to determine the optimal way for Bob to distribute his numArrows across 12 scoring sections (from 0 to 11) in an archery competition in order to maximize his total points against Alice. Alice's arrow distribution is given in aliceArrows. The scoring rules are such that for each section, the player who has more arrows gets the points corresponding to that section. If both players have the same number of arrows, the points go to Alice. If both have zero arrows, the points are not awarded to anyone.

The input consists of numArrows, an integer indicating how many arrows Bob has, and aliceArrows, an array of 12 integers showing how Alice distributed her arrows across the sections. The output should be an array bobArrows of length 12, showing how Bob should distribute his arrows to maximize his points. The sum of Bob's arrows must equal numArrows.

Key edge cases include scenarios where Bob cannot beat Alice in any section due to insufficient arrows, or when multiple distributions yield the same maximum points. The constraints indicate numArrows can be as high as 10^5, so an efficient solution is necessary.

Approaches

A brute-force approach would consider all possible ways Bob can allocate his numArrows across the 12 sections. For each allocation, we calculate the score and select the allocation with the maximum points. However, the number of allocations grows exponentially with numArrows, making this approach infeasible.

The key observation is that for each section i, Bob only needs to beat Alice by placing at least aliceArrows[i] + 1 arrows in that section. This reduces the problem to a subset selection problem, where we consider all combinations of sections to "win" and compute the corresponding arrow usage and points. Since there are only 12 sections, the total number of subsets is 2^12 = 4096, which is computationally feasible. For each subset, we check if Bob has enough arrows to beat Alice in all chosen sections. We also ensure any leftover arrows are assigned to a section arbitrarily to meet the numArrows constraint.

Approach Time Complexity Space Complexity Notes
Brute Force O(numArrows^12) O(1) Try all possible arrow distributions; infeasible due to exponential combinations
Optimal O(2^12 * 12) O(12) Enumerate all subsets of sections to beat Alice; feasible because 2^12 is small

Algorithm Walkthrough

  1. Initialize a variable maxPoints to 0 to track the highest score Bob can achieve, and a list bestDistribution to store the arrow allocation for that score.
  2. Iterate through all subsets of the 12 sections using a bitmask from 0 to 2^12 - 1. Each bit represents whether Bob tries to win that section.
  3. For each subset, initialize arrowsUsed to 0 and points to 0. Also initialize a temporary arrow distribution array tempDistribution of length 12 with zeros.
  4. For each section i (from 0 to 11), check if the subset includes this section (bit is set). If so, compute the arrows needed to win: aliceArrows[i] + 1. If adding these arrows exceeds numArrows, break and discard this subset.
  5. If the subset is valid (arrowsUsed <= numArrows), compute the total points for the subset by summing the section indices included. Assign the required arrows in tempDistribution for these sections.
  6. If arrowsUsed < numArrows, assign the remaining arrows arbitrarily to the highest scoring section in the subset to ensure sum equals numArrows.
  7. If the total points for this subset are greater than maxPoints, update maxPoints and copy tempDistribution to bestDistribution.
  8. After iterating all subsets, return bestDistribution.

Why it works: The algorithm considers all combinations of sections Bob could try to win. By only considering subsets where Bob beats Alice, we maximize points per arrow usage. The leftover arrows allocation ensures that the solution satisfies the numArrows constraint. Since we evaluate every subset, we are guaranteed to find an optimal solution.

Python Solution

from typing import List

class Solution:
    def maximumBobPoints(self, numArrows: int, aliceArrows: List[int]) -> List[int]:
        maxPoints = 0
        bestDistribution = [0] * 12

        for mask in range(1 << 12):
            arrowsUsed = 0
            points = 0
            tempDistribution = [0] * 12

            for i in range(12):
                if mask & (1 << i):
                    needed = aliceArrows[i] + 1
                    arrowsUsed += needed
                    if arrowsUsed > numArrows:
                        break
                    points += i
                    tempDistribution[i] = needed
            else:
                if arrowsUsed < numArrows:
                    for i in reversed(range(12)):
                        if mask & (1 << i):
                            tempDistribution[i] += numArrows - arrowsUsed
                            break
                if points > maxPoints:
                    maxPoints = points
                    bestDistribution = tempDistribution[:]

        return bestDistribution

The implementation iterates through all possible subsets of the 12 sections using a bitmask. For each subset, it calculates the arrows required to win each section and checks if it exceeds numArrows. The total points are computed and, if higher than the previous maximum, the arrow distribution is stored. Any remaining arrows are assigned to the highest section in the subset to meet the total arrow constraint.

Go Solution

func maximumBobPoints(numArrows int, aliceArrows []int) []int {
    maxPoints := 0
    bestDistribution := make([]int, 12)

    for mask := 0; mask < (1 << 12); mask++ {
        arrowsUsed := 0
        points := 0
        tempDistribution := make([]int, 12)

        valid := true
        for i := 0; i < 12; i++ {
            if mask&(1<<i) != 0 {
                needed := aliceArrows[i] + 1
                arrowsUsed += needed
                if arrowsUsed > numArrows {
                    valid = false
                    break
                }
                points += i
                tempDistribution[i] = needed
            }
        }

        if valid {
            if arrowsUsed < numArrows {
                for i := 11; i >= 0; i-- {
                    if mask&(1<<i) != 0 {
                        tempDistribution[i] += numArrows - arrowsUsed
                        break
                    }
                }
            }
            if points > maxPoints {
                maxPoints = points
                copy(bestDistribution, tempDistribution)
            }
        }
    }

    return bestDistribution
}

The Go version follows the same logic as the Python solution. A key difference is explicit handling of slice copying with copy to avoid reference issues.

Worked Examples

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

Subset mask Sections Bob tries Arrows used Points Distribution
101010... 4,5,8,9,10,11 9 47 [0,0,0,0,1,1,0,0,1,2,3,1]

The algorithm evaluates all 4096 subsets. It finds that selecting sections 4,5,8,9,10,11 yields the maximum points 47, using exactly 9 arrows.

Complexity Analysis

Measure Complexity Explanation
Time O(2^12 * 12) Enumerates all subsets of 12 sections, checking up to 12 elements per subset
Space O(12) Stores temporary arrow distributions and the best distribution

The algorithm is feasible because 2^12 = 4096 subsets, and each subset evaluation is O(12). Space is minimal since only fixed-length arrays are stored.

Test Cases

# test cases
sol = Solution()

# Example 1
assert sum(sol.maximumBobPoints(9, [1,1,0,1,0,0,2,1,0,1,2,0])) == 9  # sum of arrows should equal numArrows

# Example 2
assert sum(sol.maximumBobPoints(3, [0,0,1,0,0,0,0,0,0,0,0,2])) == 3

# Minimum arrows
assert sum(sol.maximumBobPoints(1, [0]*12)) == 1

# Bob cannot beat Alice in any section
assert sum(sol.maximumBobPoints(2, [3]*12)) == 2

# All arrows to highest scoring section
assert sol.maximumBobPoints(5, [0]*12)[11] == 5
Test Why
Example 1 Standard problem example, checks correct subset selection
Example 2 Another example, smaller number of arrows
Minimum arrows Edge case with only 1 arrow
Bob cannot beat Alice Edge case