Maximum Number of Groups With Increasing Length

Maximum Number of Groups With Increasing Length - Problem

Imagine you're organizing a tournament where participants are grouped into teams, and each team must have a strictly increasing number of members. You have n different types of participants (numbered 0 to n-1), and each type i can only be used up to usageLimits[i] times across all teams.

Your challenge: Create the maximum number of teams possible while following these rules:

Each team must contain distinct participant types (no duplicates within a team)
Each team (except the first) must be strictly larger than the previous team
Don't exceed the usage limit for any participant type

Goal: Return the maximum number of teams you can create.

Example: If usageLimits = [1,2,5], you could create teams of sizes 1, 2, 3... as long as you don't run out of participants!

Input & Output

example_1.py — Basic Case

$ Input: usageLimits = [1,2,5]

› Output: 3

💡 Note: We can form 3 teams: Team 1 (size 1): use participant 0 once. Team 2 (size 2): use participants 1 and 2 once each. Team 3 (size 3): use participant 2 four more times (total 5) and participant 1 once more (total 2). This uses exactly the limits: [1,2,5].

example_2.py — Small Limits

$ Input: usageLimits = [2,1,2]

› Output: 2

💡 Note: We can form 2 teams: Team 1 (size 1): use participant 0 once. Team 2 (size 2): use participants 0 and 2 once each. We cannot form a third team of size 3 because we would need 3 distinct participants but only have participant 1 (limit 1) and participant 2 (limit 1) remaining.

example_3.py — Single Element

$ Input: usageLimits = [10]

› Output: 1

💡 Note: With only one type of participant, we can form at most 1 team of size 1, since each team must consist of distinct participant types.

Constraints

1 ≤ usageLimits.length ≤ 10⁵
1 ≤ usageLimits[i] ≤ 10⁹
Each group must have distinct elements
Groups must have strictly increasing lengths

Visualization

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Understanding the Visualization

Binary Search Setup

We search for the maximum number of teams between 0 and the total sum of all usage limits

Greedy Verification

For each candidate answer, we verify by greedily forming teams using participants with highest availability

Team Formation

Form teams of increasing sizes (1, 2, 3, ..., k) using distinct participants

Optimization

Always use participants with the most remaining availability to maximize future team formation potential

Key Takeaway

🎯 Key Insight: Use binary search on the answer combined with greedy verification. Always assign participants with highest availability first to maximize the potential for forming larger teams later.

Asked in

G Google 42 a Amazon 38 ⊞ Microsoft 29 f Meta 25

The optimal solution uses binary search on the answer combined with greedy verification. Since if we can form k teams we can also form k-1 teams (monotonic property), we can binary search for the maximum k. The verification step sorts participants by availability and greedily assigns them to teams, always using the most available participants first.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Possibilities)	O(n! × 2^n)	O(n)	Try forming teams one by one until impossible
Binary Search + Greedy Verification (Optimal)	O(n log(sum) log n)	O(n)	Binary search on answer with greedy verification

Brute Force (Try All Possibilities) — Algorithm Steps

Start with an empty list of teams and current team size = 1
For each team size, try all combinations of participants
Check if the combination uses within usage limits
If valid, form the team and increment team size
Continue until no valid team can be formed
Return the number of teams formed

Visualization

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Step-by-Step Walkthrough

Try Team Size 1

Check all possible single-member teams

Try Team Size 2

Check all possible two-member teams

Continue Until Impossible

Keep increasing team size until no valid team can be formed

Code -

solution.c — C

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

int canFormTeam(int* limits, int n, int size, int startIdx, int* currentUsage) {
    if (size == 0) return 1;
    if (startIdx >= n) return 0;
    
    // Try including current participant
    if (currentUsage[startIdx] < limits[startIdx]) {
        currentUsage[startIdx]++;
        if (canFormTeam(limits, n, size - 1, startIdx + 1, currentUsage)) {
            return 1;
        }
        currentUsage[startIdx]--;
    }
    
    // Try not including current participant
    return canFormTeam(limits, n, size, startIdx + 1, currentUsage);
}

int maximumGroups(int* usageLimits, int n) {
    int* usage = (int*)calloc(n, sizeof(int));
    int teams = 0;
    int teamSize = 1;
    
    while (1) {
        int* tempUsage = (int*)malloc(n * sizeof(int));
        memcpy(tempUsage, usage, n * sizeof(int));
        
        if (canFormTeam(usageLimits, n, teamSize, 0, tempUsage)) {
            memcpy(usage, tempUsage, n * sizeof(int));
            teams++;
            teamSize++;
        } else {
            free(tempUsage);
            break;
        }
        free(tempUsage);
    }
    
    free(usage);
    return teams;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n! × 2^n)

For each team size k, we check all C(n,k) combinations, leading to exponential time

⚠ Quadratic Growth

Space Complexity

O(n)

Space to store current team formation and usage tracking

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Possibilities) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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