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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 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: Sorting

Greedily take stones from the two largest piles of stones each time until at least two piles of stones are empty.

The time complexity is $O(n)$, where $n$ is the total number of stones.

Solution 2

We might as well set $a \le b \le c$, then:

• When $a + b \le c$, we can first take stones from the two piles $a$ and $c$ to get the score $a$; then take stones from the two piles $b$ and $c$ to get the score $b$, and the total score is $a + b$;
• When $a + b \gt c$, then we will take stones from the larger pile of $c$ and $a$ and $b$ each time, and eventually $c$ will be empty. At this time, the size difference between $a$ and $b$ is at most $1$. We then take stones from the two piles $a$ and $b$ until we can no longer take them. The total score is $\left \lfloor \frac{a + b + c}{2} \right \rfloor$.

Time complexity $O(1)$.

• Java
• C++
• Python
• Go

• 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;
      }
  }
  
  
  // Solution 2
  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;
      }
  }
  
  

• 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;
      }
  };
  
  
  // Solution 2
  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;
      }
  };
  
  

• 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
  
  
  # Solution 2
  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
  
  

• 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
  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
  }
  
  

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