[LeetCode] 1753. Maximum Score From Removing Stones

- Problem

1753. Maximum Score From Removing Stones
You are playing a solitaire game with three piles of stones of sizes a, b, and c respectively. Each turn you choose two different non-empty piles, take one stone from each, and add 1 point to your score. The game stops when there are fewer than two non-empty piles (meaning there are no more available moves).

Given three integers a, b, and c, return the maximum score you can get.

Example 1:

Input: a = 2, b = 4, c = 6
Output: 6
Explanation: The starting state is (2, 4, 6). One optimal set of moves is:
- Take from 1st and 3rd piles, state is now (1, 4, 5)
- Take from 1st and 3rd piles, state is now (0, 4, 4)
- Take from 2nd and 3rd piles, state is now (0, 3, 3)
- Take from 2nd and 3rd piles, state is now (0, 2, 2)
- Take from 2nd and 3rd piles, state is now (0, 1, 1)
- Take from 2nd and 3rd piles, state is now (0, 0, 0)
There are fewer than two non-empty piles, so the game ends. Total: 6 points.

Example 2:

Input: a = 4, b = 4, c = 6
Output: 7
Explanation: The starting state is (4, 4, 6). One optimal set of moves is:
- Take from 1st and 2nd piles, state is now (3, 3, 6)
- Take from 1st and 3rd piles, state is now (2, 3, 5)
- Take from 1st and 3rd piles, state is now (1, 3, 4)
- Take from 1st and 3rd piles, state is now (0, 3, 3)
- Take from 2nd and 3rd piles, state is now (0, 2, 2)
- Take from 2nd and 3rd piles, state is now (0, 1, 1)
- Take from 2nd and 3rd piles, state is now (0, 0, 0)
There are fewer than two non-empty piles, so the game ends. Total: 7 points.

Example 3:

Input: a = 1, b = 8, c = 8
Output: 8
Explanation: One optimal set of moves is to take from the 2nd and 3rd piles for 8 turns until they are empty.
After that, there are fewer than two non-empty piles, so the game ends.

Constraints:

• 1 <= a, b, c <= 10^5

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- 내 풀이 1 (Heap)

class Solution:
    def maximumScore(self, a: int, b: int, c: int) -> int:
        answer, heap = 0, [-a,-b,-c]
        
        while True:
            x, y = heapq.heappop(heap), heapq.heappop(heap)

            if x == 0 or y == 0:
                return answer
            
            heapq.heappush(heap, x+1)
            heapq.heappush(heap, y+1)
            answer += 1

- 결과

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- 내 풀이 2 (Math)

class Solution:
    def maximumScore(self, a: int, b: int, c: int) -> int:
        sum_piles = a+b+c
        return min(sum_piles // 2, sum_piles - max(a,b,c))

- 결과