# 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 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 <= 105

## 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
}