2491. Divide Players Into Teams of Equal Skill
Problem
You are given a positive integer array skill of even length n where skill[i] denotes the skill of the ith player. Divide the players into n / 2 teams of size 2 such that the total skill of each team is equal.
The chemistry of a team is equal to the product of the skills of the players on that team.
Return the sum of the **chemistry of all the teams, or return -1 if there is no way to divide the players into teams such that the total skill of each team is equal.**
Example 1:
Input: skill = [3,2,5,1,3,4]
Output: 22
Explanation:
Divide the players into the following teams: (1, 5), (2, 4), (3, 3), where each team has a total skill of 6.
The sum of the chemistry of all the teams is: 1 * 5 + 2 * 4 + 3 * 3 = 5 + 8 + 9 = 22.
Example 2:
Input: skill = [3,4]
Output: 12
Explanation:
The two players form a team with a total skill of 7.
The chemistry of the team is 3 * 4 = 12.
Example 3:
Input: skill = [1,1,2,3]
Output: -1
Explanation:
There is no way to divide the players into teams such that the total skill of each team is equal.
Constraints:
2 <= skill.length <= 105skill.lengthis even.1 <= skill[i] <= 1000
C++
class Solution { public: long long dividePlayers(vector<int>& skill) { sort(skill.begin(), skill.end()); int n = skill.size(); int t = skill[0] + skill[n - 1]; long long ans = 0; for (int i = 0, j = n - 1; i < j; ++i, --j) { if (skill[i] + skill[j] != t) return -1; ans += 1ll * skill[i] * skill[j]; } return ans; } };
PYTHON
class Solution: def dividePlayers(self, skill: list[int]) -> int: skill.sort() total_product = 0 target_sum = skill[0] + skill[-1] i, j = 0, len(skill) - 1 while i < j: if skill[i] + skill[j] != target_sum: return -1 total_product += skill[i] * skill[j] i += 1 j -= 1 return total_product
JAVA
class Solution { public long dividePlayers(int[] skill) { Arrays.sort(skill); int n = skill.length; int t = skill[0] + skill[n - 1]; long ans = 0; for (int i = 0, j = n - 1; i < j; ++i, --j) { if (skill[i] + skill[j] != t) { return -1; } ans += (long) skill[i] * skill[j]; } return ans; } }
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