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Formatted question description: https://leetcode.ca/all/2358.html

2358. Maximum Number of Groups Entering a Competition

  • Difficulty: Medium.
  • Related Topics: Array, Math, Binary Search, Greedy.
  • Similar Questions: Maximum Height by Stacking Cuboids .

Problem

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

Solution (Java, C++, Python)

  • class Solution {
    public int maximumGroups(int[] grades) {
        int len = grades.length;
        return (int) (-1 + Math.sqrt(1D + 8 * len)) / 2;
    }
}

############

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:
    def maximumGroups(self, grades: List[int]) -> int:
        grades.sort()
        ans = 1
        prev = [1, grades[0]]
        curr = [0, 0]
        for v in grades[1:]:
            curr[0] += 1
            curr[1] += v
            if prev[0] < curr[0] and prev[1] < curr[1]:
                prev = curr
                curr = [0, 0]
                ans += 1
        return ans

############

# 2358. Maximum Number of Groups Entering a Competition
# https://leetcode.com/problems/maximum-number-of-groups-entering-a-competition/

class Solution:
    def maximumGroups(self, grades: List[int]) -> int:
        n = len(grades)
        res = 1
        currSize = 0
        size = 2
        
        for i in range(1, n):
            currSize += 1
            
            if currSize == size:
                res = currSize
                size += 1
                currSize = 0
        
        return res


  • 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;
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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