How do you solve Maximum Score From Removing Stones on LeetCode?

Maximum Score From Removing Stones (LeetCode #1753) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Always prioritize removing stones from the two largest piles to maximize score.

What is the time complexity of Maximum Score From Removing Stones?

The optimal solution for Maximum Score From Removing Stones runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) due to sorting the piles, but since we only have three piles, this is effectively constant time.

What is the brute force solution for Maximum Score From Removing Stones?

The brute force approach for Maximum Score From Removing Stones is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we simulate the game by repeatedly choosing the two largest piles until we can no longer make a move. This is straightforward but inefficient as it checks all possible combinations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Score From Removing Stones?

Maximum Score From Removing Stones uses the following concepts: Math, Greedy, Heap (Priority Queue). The recommended patterns to study are: Greedy Algorithm, Sorting.

What are the interview tips for Maximum Score From Removing Stones?

Explain your thought process clearly as you work through the problem. Consider edge cases and discuss them with the interviewer. Practice similar problems to recognize patterns and improve your problem-solving speed.

What are common mistakes when solving Maximum Score From Removing Stones?

Not considering edge cases where one or more piles may be empty initially. Failing to optimize the approach and relying on brute force even when a better solution exists.

#1753

Maximum Score From Removing Stones

Medium

MathGreedyHeap (Priority Queue)Greedy AlgorithmSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(1)

The optimal solution uses a greedy approach where we always remove stones from the two largest piles. This ensures that we maximize the score by making the most effective moves possible.

⚙️

Algorithm

3 steps

1Step 1: Sort the piles to easily access the two largest ones.
2Step 2: While the two largest piles are non-empty, remove one stone from each and increment the score.
3Step 3: Return the score when fewer than two piles are left.

solution.py9 lines

1def maxScore(a, b, c):
2    piles = sorted([a, b, c])
3    score = 0
4    while piles[1] > 0:
5        piles[1] -= 1
6        piles[2] -= 1
7        score += 1
8        piles.sort()
9    return score

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the piles, but since we only have three piles, this is effectively constant time.

1Always prioritize removing stones from the two largest piles to maximize score.
2Sorting helps easily identify the largest piles, but for three piles, it can be done with simple comparisons.

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