Watch 10 video solutions for Maximum Number of Groups Entering a Competition, a medium level problem involving Array, Math, Binary Search. This walkthrough by Bro Coders has 3,086 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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:
ith group is less than the sum of the grades of students in the (i + 1)th group, for all groups (except the last).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 <= 1051 <= grades[i] <= 105Problem Overview: You receive an array grades representing student scores. Students must be divided into groups such that each new group has more members than the previous one and the total grade sum is strictly larger than the previous group. The task is to compute the maximum number of groups you can form.
Approach 1: Sorting and Greedy Approach (O(n log n) time, O(1) extra space)
Sort the array so smaller grades appear first. Greedy grouping works because building groups with the smallest available elements keeps early sums minimal, leaving larger elements for later groups. Iterate through the sorted list while expanding the next group size from 1, 2, 3.... Maintain counters for the current group size and the total number of used students. Each time you have enough remaining students to form the next larger group, commit that group and move forward. Sorting guarantees that each later group naturally has a larger sum because it contains more elements and larger values. This approach uses a classic greedy strategy combined with an array traversal.
Approach 2: Mathematical Incremental Grouping (O(1) time, O(1) space)
The key observation: if group sizes are strictly increasing, they must follow 1, 2, 3, ..., k. The total number of students required is the triangular number 1 + 2 + ... + k = k(k+1)/2. Because grades are positive and the array can be sorted, satisfying the size condition automatically ensures the sum condition. Therefore the task reduces to finding the largest k such that k(k+1)/2 ≤ n, where n is the number of students. Solve this using the quadratic formula or a small iterative check. This turns the problem into a simple math calculation.
Alternative: Binary Search on Group Count (O(log n) time, O(1) space)
You can also treat the number of groups k as the search space. Using binary search, check whether k(k+1)/2 ≤ n. If the condition holds, try a larger k; otherwise reduce it. This approach is useful when the formula is not immediately obvious or when interviewers want to see how you convert a monotonic condition into a search problem.
Recommended for interviews: The mathematical observation is the cleanest and runs in constant time. However, many candidates first discover the greedy grouping after sorting, which demonstrates clear reasoning about constraints. Showing the greedy idea and then simplifying it to the triangular-number formula signals strong problem-solving ability.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Sorting and Greedy | O(n log n) | O(1) | When working directly with the grades array and reasoning about group formation step by step |
| Mathematical Incremental Grouping | O(1) | O(1) | Optimal solution when you recognize the triangular number pattern |
| Binary Search on Group Count | O(log n) | O(1) | Useful when converting the constraint k(k+1)/2 ≤ n into a monotonic search problem |