| comments | difficulty | edit_url | rating | source | tags | |||
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true |
Medium |
1487 |
Weekly Contest 227 Q2 |
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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
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 ansclass 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;
}
};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
}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) >> 1class 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());
if (s[0] + s[1] < s[2]) return s[0] + s[1];
return (a + b + c) >> 1;
}
};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
}