LeetCode 1753. Maximum Score From Removing Stones Solution in Java, C++, Python & Go | Explanation + Code

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

PythonJavaC++Go

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

Solution 2

PythonJavaC++Go

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)