Maximum Number of Groups Entering a Competition

Maximum Number of Groups Entering a Competition - Problem

You're organizing a university programming competition and need to divide students into groups based on their grades. The challenge is to create the maximum number of groups possible while following two strict ordering rules:

Grade Rule: Each group must have a higher total grade sum than the previous group
Size Rule: Each group must have more students than the previous group

Given an array grades representing student grades, determine the maximum number of valid groups you can form.

Example: If you have grades [10, 6, 12, 7, 3, 5], you could form 3 groups:

Group 1: [3] → 1 student, sum = 3
Group 2: [6, 5] → 2 students, sum = 11
Group 3: [10, 12, 7] → 3 students, sum = 29

Input & Output

example_1.py — Basic Case

$ Input: [10, 6, 12, 7, 3, 5]

› Output: 3

💡 Note: We can form 3 groups: [3] (1 student, sum=3), [5,6] (2 students, sum=11), [7,10,12] (3 students, sum=29). Each group satisfies both constraints.

example_2.py — Small Array

$ Input: [8, 8]

› Output: 1

💡 Note: With only 2 students, we can form at most 1 group. We cannot form 2 groups because the first would need 1 student and second would need 2 students, but we only have 2 total.

example_3.py — Larger Array

$ Input: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

› Output: 4

💡 Note: With 10 students, we can form 4 groups: [1] (1 student), [2,3] (2 students), [4,5,6] (3 students), [7,8,9,10] (4 students). Total: 1+2+3+4=10 students used.

Constraints

1 ≤ grades.length ≤ 10⁵
1 ≤ grades[i] ≤ 10⁶
The grades array contains positive integers only

Visualization

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Understanding the Visualization

Gather All Players

Start with all student grades representing player skills

Sort by Skill

Arrange players from weakest to strongest to optimize assignments

Build Level 1

Assign 1 weakest player to the bottom level

Build Level 2

Assign next 2 weakest players to second level

Continue Building

Keep adding levels with k players until you run out of players

Count Levels

The number of complete levels is your answer

Key Takeaway

🎯 Key Insight: To maximize the number of groups, sort the grades and greedily assign the minimum required students (1, 2, 3, ...) to each group using the smallest available grades. This is equivalent to finding the largest triangular number ≤ n.

Asked in

G Google 25 ⊞ Microsoft 18 a Amazon 15 f Meta 12

The optimal approach uses a greedy algorithm with sorting. After sorting grades in ascending order, we assign the minimum required students to each group (1, 2, 3, ...) using the smallest available grades. This maximizes the number of groups by ensuring we can form as many valid groups as possible. The key insight is that the actual grade values don't matter - only the count of students matters for determining the maximum groups.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Partitions)	O(2^n × n)	O(n)	Generate all possible ways to partition students into groups and check validity
Greedy Algorithm (Optimal)	O(n log n)	O(1)	Sort grades and greedily assign minimum students to each group using smallest available grades

Brute Force (Try All Partitions) — Algorithm Steps

Generate all possible partitions of the array
For each partition, check if group sizes are strictly increasing
Check if group sums are strictly increasing
Track the maximum number of valid groups found

Visualization

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Step-by-Step Walkthrough

Generate Partitions

Generate all possible ways to group the students

Check Size Constraint

Verify each group has more students than the previous

Check Sum Constraint

Verify each group has higher total grades

Track Maximum

Keep track of the maximum valid group count

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int maxGroups = 0;

int sum(int* arr, int size) {
    int total = 0;
    for (int i = 0; i < size; i++) {
        total += arr[i];
    }
    return total;
}

int isValidPartition(int groups[][100], int groupSizes[], int numGroups) {
    if (numGroups == 0) return 0;
    
    int prevSize = 0, prevSum = 0;
    for (int i = 0; i < numGroups; i++) {
        int currSize = groupSizes[i];
        int currSum = sum(groups[i], currSize);
        
        if (currSize <= prevSize || currSum <= prevSum) {
            return 0;
        }
        
        prevSize = currSize;
        prevSum = currSum;
    }
    
    return 1;
}

void backtrack(int index, int* grades, int n, int groups[][100], int groupSizes[], int numGroups) {
    if (index == n) {
        if (isValidPartition(groups, groupSizes, numGroups)) {
            if (numGroups > maxGroups) {
                maxGroups = numGroups;
            }
        }
        return;
    }
    
    // Try adding to existing groups
    for (int i = 0; i < numGroups; i++) {
        groups[i][groupSizes[i]] = grades[index];
        groupSizes[i]++;
        backtrack(index + 1, grades, n, groups, groupSizes, numGroups);
        groupSizes[i]--;
    }
    
    // Try creating new group
    if (numGroups < 100) {
        groups[numGroups][0] = grades[index];
        groupSizes[numGroups] = 1;
        backtrack(index + 1, grades, n, groups, groupSizes, numGroups + 1);
        groupSizes[numGroups] = 0;
    }
}

int maximumGroups(int* grades, int n) {
    maxGroups = 0;
    int groups[100][100];
    int groupSizes[100] = {0};
    backtrack(0, grades, n, groups, groupSizes, 0);
    return maxGroups;
}

int main() {
    char input[1000];
    fgets(input, sizeof(input), stdin);
    
    int grades[100];
    int n = 0;
    char* token = strtok(input, "[],\n");
    
    while (token != NULL && n < 100) {
        grades[n++] = atoi(token);
        token = strtok(NULL, "[],\n");
    }
    
    printf("%d\n", maximumGroups(grades, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n × n)

2^n possible partitions, each taking O(n) time to validate

⚠ Quadratic Growth

Space Complexity

O(n)

Space for recursion stack and storing current partition

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Partitions) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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