LeetCode #2358 — MEDIUM

Maximum Number of Groups Entering a Competition

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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 i^th 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 i^th 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:
- 1^st group has the students with grades = [12]. Sum of grades: 12. Student count: 1
- 2^nd group has the students with grades = [6,7]. Sum of grades: 6 + 7 = 13. Student count: 2
- 3^rd 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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Binary Search · Greedy

Example 1

[10,6,12,7,3,5]

Example 2

[8,8]

Core Insight

What unlocks the optimal approach

Would it be easier to place the students into valid groups after sorting them based on their grades in ascending order?
Notice that, after sorting, we can separate them into groups of sizes 1, 2, 3, and so on.
If the last group is invalid, we can merge it with the previous group.
This creates the maximum number of groups because we always greedily form the smallest possible group.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2358: Maximum Number of Groups Entering a Competition
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;
    }
}

// Accepted solution for LeetCode #2358: Maximum Number of Groups Entering a Competition
func maximumGroups(grades []int) int {
	n := len(grades)
	return sort.Search(n, func(k int) bool {
		k++
		return k*k+k > n*2
	})
}

# Accepted solution for LeetCode #2358: Maximum Number of Groups Entering a Competition
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

// Accepted solution for LeetCode #2358: Maximum Number of Groups Entering a Competition
impl Solution {
    pub fn maximum_groups(grades: Vec<i32>) -> i32 {
        let n = grades.len() as i64;
        let (mut l, mut r) = (0i64, n);
        while l < r {
            let mid = (l + r + 1) / 2;
            if mid * mid + mid > 2 * n {
                r = mid - 1;
            } else {
                l = mid;
            }
        }
        l as i32
    }
}

// Accepted solution for LeetCode #2358: Maximum Number of Groups Entering a Competition
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.