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 1: Greedy
Take two stones (one each) from the largest two piles, until one is empty.
Time complexity: O(n)
Space complexity: O(1)
C++
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class Solution { public: int maximumScore(int a, int b, int c) { array<int, 3> s{a,b,c}; sort(begin(s), end(s)); int ans = 0; while (s[1]-- && s[2]--) { ++ans; sort(begin(s), end(s)); } return ans; } }; |
Solution 2: Math
First, let’s assuming a <= b <= c.
There are two conditions:
1. a + b <= c, we can pair c with a first and then b. Total pairs is (a + b + (a + b)) / 2
2. a + b > c, we can pair c with a, b “evenly”, and then pair a with b, total pairs is (a + b + c) / 2
ans = (a + b + min(a + b, c)) / 2
Time complexity: O(1)
Space complexity: O(1)
C++
1 2 3 4 5 6 7 8 |
class Solution { public: int maximumScore(int a, int b, int c) { array<int, 3> s{a,b,c}; sort(begin(s), end(s)); return (s[0] + s[1] + min(s[0] + s[1], s[2])) / 2; } }; |
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