maximum-score-from-removing-stones

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Algorithms

You are playing a solitaire game with three piles of stones of sizes <code>a</code>​​​​​​, <code>b</code>,​​​​​​ and <code>c</code>​​​​​​ respectively. Each turn you choose two different non-empty piles, take one stone from each, and add <code>1</code> 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 <code>a</code>​​​​​, <code>b</code>,​​​​​ and <code>c</code>​​​​​, return the maximum score you can get.

&nbsp;
Example 1:

<pre>
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.
</pre>

Example 2:

<pre>
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.
</pre>

Example 3:

<pre>
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.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= a, b, c &lt;= 105</code></li>
</ul>

maximumScore

Maximum Score From Removing Stones

Minimum Amount of Time to Fill Cups

minimum-amount-of-time-to-fill-cups

装满杯子需要的最短总时长

Greedy

Math

Heap (Priority Queue)

你正在玩一个单人游戏，面前放置着大小分别为 <code>a</code>​​​​​​、<code>b</code> 和 <code>c</code>​​​​​​ 的 三堆 石子。

每回合你都要从两个 不同的非空堆 中取出一颗石子，并在得分上加 <code>1</code> 分。当存在 两个或更多 的空堆时，游戏停止。

给你三个整数 <code>a</code> 、<code>b</code> 和 <code>c</code> ，返回可以得到的 最大分数 。
 

示例 1：

<pre>
输入：a = 2, b = 4, c = 6
输出：6
解释：石子起始状态是 (2, 4, 6) ，最优的一组操作是：
- 从第一和第三堆取，石子状态现在是 (1, 4, 5)
- 从第一和第三堆取，石子状态现在是 (0, 4, 4)
- 从第二和第三堆取，石子状态现在是 (0, 3, 3)
- 从第二和第三堆取，石子状态现在是 (0, 2, 2)
- 从第二和第三堆取，石子状态现在是 (0, 1, 1)
- 从第二和第三堆取，石子状态现在是 (0, 0, 0)
总分：6 分 。
</pre>

示例 2：

<pre>
输入：a = 4, b = 4, c = 6
输出：7
解释：石子起始状态是 (4, 4, 6) ，最优的一组操作是：
- 从第一和第二堆取，石子状态现在是 (3, 3, 6)
- 从第一和第三堆取，石子状态现在是 (2, 3, 5)
- 从第一和第三堆取，石子状态现在是 (1, 3, 4)
- 从第一和第三堆取，石子状态现在是 (0, 3, 3)
- 从第二和第三堆取，石子状态现在是 (0, 2, 2)
- 从第二和第三堆取，石子状态现在是 (0, 1, 1)
- 从第二和第三堆取，石子状态现在是 (0, 0, 0)
总分：7 分 。
</pre>

示例 3：

<pre>
输入：a = 1, b = 8, c = 8
输出：8
解释：最优的一组操作是连续从第二和第三堆取 8 回合，直到将它们取空。
注意，由于第二和第三堆已经空了，游戏结束，不能继续从第一堆中取石子。
</pre>

 

提示：

<ul>
	<li><code>1 <= a, b, c <= 105</code></li>
</ul>