LeetCode-in-Java
1753. Maximum Score From Removing Stones
Medium
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
Solution
import java.util.Arrays;
public class Solution {
public int maximumScore(int a, int b, int c) {
int[] nums = new int[] {a, b, c};
Arrays.sort(nums);
if (nums[0] + nums[1] < nums[2]) {
return nums[0] + nums[1];
} else {
return (nums[0] + nums[1] + nums[2]) / 2;
}
}
}