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1753. Maximum Score From Removing Stones
Description
You are playing a solitaire game with three piles of stones of sizes a
, b
, and c
respectively. Each turn you choose two different nonempty piles, take one stone from each, and add 1
point to your score. The game stops when there are fewer than two nonempty 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 nonempty 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 nonempty 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 nonempty piles, so the game ends.
Constraints:
1 <= a, b, c <= 10^{5}
Solutions

class Solution { public int maximumScore(int a, int b, int c) { int[] s = new int[] {a, b, c}; Arrays.sort(s); int ans = 0; while (s[1] > 0) { ++ans; s[1]; s[2]; Arrays.sort(s); } return ans; } }

class Solution { public: int maximumScore(int a, int b, int c) { vector<int> s = {a, b, c}; sort(s.begin(), s.end()); int ans = 0; while (s[1]) { ++ans; s[1]; s[2]; sort(s.begin(), s.end()); } return ans; } };

class Solution: def maximumScore(self, a: int, b: int, c: int) > int: s = sorted([a, b, c]) ans = 0 while s[1]: ans += 1 s[1] = 1 s[2] = 1 s.sort() return ans

func maximumScore(a int, b int, c int) (ans int) { s := []int{a, b, c} sort.Ints(s) for s[1] > 0 { ans++ s[1] s[2] sort.Ints(s) } return }