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2358. Maximum Number of Groups Entering a Competition 👍

  • Time: $O(1)$
  • Space: $O(1)$
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class Solution {
 public:
  int maximumGroups(vector<int>& grades) {
    // Sort grades, then we can seperate the students into groups of sizes 1, 2,
    // 3, ..., k, s.t. the i-th group < the (i + 1)-th group for both sum and
    // size. So, we can rephrase the problem into:
    //   Find the maximum k s.t. 1 + 2 + 3 + ... + k <= n

    //  1 + 2 + 3 + ... + k <= n
    //         k(k + 1) / 2 <= n
    //              k^2 + k <= 2n
    //   (k + 0.5)^2 - 0.25 <= 2n
    //          (k + 0.5)^2 <= 2n + 0.25
    //                    k <= sqrt(2n + 0.25) - 0.5
    return sqrt(grades.size() * 2 + 0.25) - 0.5;
  }
};

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class Solution {
  public int maximumGroups(int[] grades) {
    // Sort grades, then we can seperate the students into groups of sizes 1, 2,
    // 3, ..., k, s.t. the i-th group < the (i + 1)-th group for both sum and
    // size. So, we can rephrase the problem into:
    //   Find the maximum k s.t. 1 + 2 + 3 + ... + k <= n

    //  1 + 2 + 3 + ... + k <= n
    //         k(k + 1) / 2 <= n
    //              k^2 + k <= 2n
    //   (k + 0.5)^2 - 0.25 <= 2n
    //          (k + 0.5)^2 <= 2n + 0.25
    //                    k <= sqrt(2n + 0.25) - 0.5
    return (int) (Math.sqrt(grades.length * 2 + 0.25) - 0.5);
  }
}

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class Solution:
  def maximumGroups(self, grades: list[int]) -> int:
    # Sort grades, then we can seperate the students into groups of sizes 1, 2,
    # 3, ..., k, s.t. the i-th group < the (i + 1)-th group for both sum and
    # size. So, we can rephrase the problem into:
    #   Find the maximum k s.t. 1 + 2 + 3 + ... + k <= n

    #  1 + 2 + 3 + ... + k <= n
    #         k(k + 1) // 2 <= n
    #              k^2 + k <= 2n
    #   (k + 0.5)^2 - 0.25 <= 2n
    #          (k + 0.5)^2 <= 2n + 0.25
    #                    k <= sqrt(2n + 0.25) - 0.5
    return int(math.sqrt(len(grades) * 2 + 0.25) - 0.5)