LeetCode 2358 - Maximum Number of Groups Entering a Competition

The problem asks us to partition a list of student grades into multiple ordered groups under two strict conditions. First, the total sum of grades in the i-th group must be less than that of the (i+1)-th group.

leetcodemediumarraymathbinary-searchgreedy

LeetCode Problem 2358

Difficulty: 🟡 Medium
Topics: Array, Math, Binary Search, Greedy

Solution

Problem Understanding

The problem asks us to partition a list of student grades into multiple ordered groups under two strict conditions. First, the total sum of grades in the i-th group must be less than that of the (i+1)-th group. Second, the number of students in each group must strictly increase as we progress. Essentially, both the size of the group and the sum of grades should grow from group to group.

The input is an array of positive integers grades, representing individual student grades. The output is a single integer representing the maximum number of groups that can be formed while satisfying the constraints.

The constraints indicate that grades can have up to 10^5 elements, and each grade can be up to 10^5. This rules out approaches that require enumerating all possible group partitions since that would be computationally infeasible. The problem guarantees that grades is non-empty, so there is at least one student.

Important edge cases include:

  • A single student: only one group can exist.
  • All students having the same grade: only group size differentiation matters.
  • The number of students is small but the grades are large or identical.

The key insight is that because both the group sizes and sum of grades must strictly increase, we do not need to examine the exact values of the grades, only the total number of students to determine the maximum groups. This transforms the problem into a mathematical one: how many groups can be formed such that 1 + 2 + ... + k <= n, where n is the number of students.

Approaches

The brute-force approach would try every possible partition of the grades array and check both the sum and size constraints. This is correct in principle, but the number of partitions grows exponentially with n, making it impractical for n = 10^5.

The optimal approach leverages the observation that for k groups with increasing sizes, the minimum number of students required is the sum of the first k natural numbers: 1 + 2 + ... + k = k*(k+1)/2. Therefore, the problem reduces to finding the largest k such that k*(k+1)/2 <= len(grades). This can be solved using a simple iterative approach or by solving the quadratic equation directly.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n) Enumerates all possible partitions and checks sum/size constraints
Optimal O(√n) O(1) Uses the sum of first k integers to determine maximum number of groups

Algorithm Walkthrough

  1. Compute the total number of students n = len(grades).
  2. Initialize a counter k = 0 to represent the number of groups.
  3. While the sum of the first k + 1 natural numbers (k+1)*(k+2)/2 is less than or equal to n, increment k.
  4. Return k as the maximum number of groups.

Why it works: By forming groups with sizes 1, 2, 3, ..., k, we guarantee that the number of students in each group strictly increases. Since the grades can be arbitrarily arranged, the sum of grades in these increasing-sized groups can also be made to strictly increase by appropriate assignment. Therefore, the limiting factor is the total number of students, not the actual grades.

Python Solution

from typing import List

class Solution:
    def maximumGroups(self, grades: List[int]) -> int:
        n = len(grades)
        k = 0
        while (k + 1) * (k + 2) // 2 <= n:
            k += 1
        return k

The implementation counts the maximum number of groups by iteratively testing k until the triangular number k*(k+1)/2 exceeds n. This ensures the group size condition is satisfied, and the sum condition can always be met by sorting the grades appropriately.

Go Solution

func maximumGroups(grades []int) int {
    n := len(grades)
    k := 0
    for (k+1)*(k+2)/2 <= n {
        k++
    }
    return k
}

The Go solution mirrors the Python logic. We handle integer division explicitly and rely on Go's integer arithmetic, which naturally floors the result of division, consistent with the Python integer division operator //.

Worked Examples

Example 1: grades = [10,6,12,7,3,5]

Number of students n = 6.

  • Try k = 1: sum = 1 ≤ 6
  • Try k = 2: sum = 3 ≤ 6
  • Try k = 3: sum = 6 ≤ 6
  • Try k = 4: sum = 10 > 6

Maximum groups = 3

Example 2: grades = [8,8]

Number of students n = 2.

  • Try k = 1: sum = 1 ≤ 2
  • Try k = 2: sum = 3 > 2

Maximum groups = 1

Complexity Analysis

Measure Complexity Explanation
Time O(√n) The while loop runs until k(k+1)/2 > n, which is approximately √(2n) iterations
Space O(1) Only counters are used; no extra space proportional to n

The time complexity is very efficient for n ≤ 10^5 and the space complexity is constant.

Test Cases

# Basic provided examples
assert Solution().maximumGroups([10,6,12,7,3,5]) == 3  # Example 1
assert Solution().maximumGroups([8,8]) == 1            # Example 2

# Single student
assert Solution().maximumGroups([42]) == 1

# All grades equal
assert Solution().maximumGroups([5,5,5,5,5]) == 2

# Large number of students
assert Solution().maximumGroups(list(range(1, 101))) == 13  # 1+2+...+13=91 <=100, 1+2+...+14=105>100

# Edge case: maximum number of students all equal
assert Solution().maximumGroups([100000]*105) == 14
Test Why
[10,6,12,7,3,5] Standard example with mixed grades
[8,8] Only one group possible due to size constraint
[42] Single student, smallest non-empty input
[5,5,5,5,5] Equal grades test, size differentiation is key
range(1,101) Tests larger input, triangular number logic
[100000]*105 Maximum value edge case for grades and length

Edge Cases

Single Student: When the array has only one student, only one group can be formed. The implementation handles this naturally because the loop will never increment k beyond 1.

All Equal Grades: When all grades are equal, the sum condition does not block group formation, but group sizes still must strictly increase. The algorithm relies solely on student count and thus correctly computes the maximum groups.

Maximum Size Input: For n = 10^5 and maximum grades, the approach still works efficiently due to the O(√n) loop. There is no risk of integer overflow in Python, and in Go, the multiplication (k+1)*(k+2) stays well below int limits for n = 10^5.