Skip to content

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

Solution 1

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
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
}

Solution 2

1
2
3
4
5
6
class Solution:
    def maximumScore(self, a: int, b: int, c: int) -> int:
        a, b, c = sorted([a, b, c])
        if a + b < c:
            return a + b
        return (a + b + c) >> 1

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
class Solution {
    public int maximumScore(int a, int b, int c) {
        int[] s = new int[] {a, b, c};
        Arrays.sort(s);
        if (s[0] + s[1] < s[2]) {
            return s[0] + s[1];
        }
        return (a + b + c) >> 1;
    }
}

1
2
3
4
5
6
7
8
9
class Solution {
public:
    int maximumScore(int a, int b, int c) {
        vector<int> s = {a, b, c};
        sort(s.begin(), s.end());
        if (s[0] + s[1] < s[2]) return s[0] + s[1];
        return (a + b + c) >> 1;
    }
};

1
2
3
4
5
6
7
8
func maximumScore(a int, b int, c int) int {
    s := []int{a, b, c}
    sort.Ints(s)
    if s[0]+s[1] < s[2] {
        return s[0] + s[1]
    }
    return (a + b + c) >> 1
}

Comments