Question

2358. Maximum Number of Groups Entering a Competition

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Description

You are given a positive integer array grades which represents the grades of students in a university. You would like to enter all these students into a competition in ordered non-empty groups, such that the ordering meets the following conditions:

• The sum of the grades of students in the ith group is less than the sum of the grades of students in the (i + 1)th group, for all groups (except the last).
• The total number of students in the ith group is less than the total number of students in the (i + 1)th group, for all groups (except the last).

Return _the maximum number of groups that can be formed_.

 

Example 1:

Input: grades = [10,6,12,7,3,5]
Output: 3
Explanation: The following is a possible way to form 3 groups of students:
- 1st group has the students with grades = [12]. Sum of grades: 12. Student count: 1
- 2nd group has the students with grades = [6,7]. Sum of grades: 6 + 7 = 13. Student count: 2
- 3rd group has the students with grades = [10,3,5]. Sum of grades: 10 + 3 + 5 = 18. Student count: 3
It can be shown that it is not possible to form more than 3 groups.

Example 2:

Input: grades = [8,8]
Output: 1
Explanation: We can only form 1 group, since forming 2 groups would lead to an equal number of students in both groups.

 

Constraints:

• 1 <= grades.length <= 105
• 1 <= grades[i] <= 105

Solutions

Solution 1

class Solution:
  def maximumGroups(self, grades: List[int]) -> int:
    n = len(grades)
    return bisect_right(range(n + 1), n * 2, key=lambda x: x * x + x) - 1

class Solution {
  public int maximumGroups(int[] grades) {
    int n = grades.length;
    int l = 0, r = n;
    while (l < r) {
      int mid = (l + r + 1) >> 1;
      if (1L * mid * mid + mid > n * 2L) {
        r = mid - 1;
      } else {
        l = mid;
      }
    }
    return l;
  }
}

class Solution {
public:
  int maximumGroups(vector<int>& grades) {
    int n = grades.size();
    int l = 0, r = n;
    while (l < r) {
      int mid = (l + r + 1) >> 1;
      if (1LL * mid * mid + mid > n * 2LL) {
        r = mid - 1;
      } else {
        l = mid;
      }
    }
    return l;
  }
};

func maximumGroups(grades []int) int {
  n := len(grades)
  return sort.Search(n, func(k int) bool {
    k++
    return k*k+k > n*2
  })
}

function maximumGroups(grades: number[]): number {
  const n = grades.length;
  let l = 1;
  let r = n;
  while (l < r) {
    const mid = (l + r + 1) >> 1;
    if (mid * mid + mid > n * 2) {
      r = mid - 1;
    } else {
      l = mid;
    }
  }
  return l;
}