How do you solve Maximum Points in an Archery Competition on LeetCode?

Maximum Points in an Archery Competition (LeetCode #2212) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Bob should aim to win the sections where he can do so with the least arrows.

What is the brute force solution for Maximum Points in an Archery Competition?

The brute force approach for Maximum Points in an Archery Competition is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we try every possible combination of arrows Bob can shoot to see which sections he can win. Since there are only 12 sections, we can afford to check all combinations, even if it takes time. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Points in an Archery Competition?

Maximum Points in an Archery Competition uses the following concepts: Array, Backtracking, Bit Manipulation, Enumeration. The recommended patterns to study are: Greedy, Backtracking.

What are the interview tips for Maximum Points in an Archery Competition?

Think about edge cases, such as when numArrows is very small or very large. Explain your thought process clearly; it shows your understanding. Practice coding the logic without looking at the solution to reinforce your understanding.

What are common mistakes when solving Maximum Points in an Archery Competition?

Not considering how to allocate remaining arrows after winning sections. Overcomplicating the distribution logic.

#2212

Maximum Points in an Archery Competition

Medium

ArrayBacktrackingBit ManipulationEnumerationGreedyBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a greedy approach to allocate Bob's arrows strategically. By focusing on the sections where Bob can win with the least arrows, we maximize his score efficiently.

⚙️

Algorithm

4 steps

1Step 1: Initialize an array for Bob's arrows and set the maximum points to zero.
2Step 2: Iterate from the highest score section (11) to the lowest (0).
3Step 3: For each section, check if Bob can win by shooting more arrows than Alice. If so, allocate the necessary arrows and adjust the remaining arrows.
4Step 4: If there are arrows left after winning, distribute them to the lowest scoring sections.

solution.py16 lines

1# Full working Python code
2
3def maxPoints(numArrows, aliceArrows):
4    bobArrows = [0] * 12
5    maxPoints = 0
6    arrowsLeft = numArrows
7
8    for score in range(11, -1, -1):
9        if arrowsLeft > aliceArrows[score]:
10            bobArrows[score] = aliceArrows[score] + 1
11            arrowsLeft -= bobArrows[score]
12
13    if arrowsLeft > 0:
14        bobArrows[0] += arrowsLeft
15
16    return bobArrows

ℹ

Complexity note: The optimal solution iterates through the scoring sections once, leading to linear complexity. The space complexity is also linear due to the output array.

1Bob should aim to win the sections where he can do so with the least arrows.
2Distributing remaining arrows to lower scoring sections can help maximize points.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.