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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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

✓ Binary Search + Greedy Verification (Optimal)

⏱️ Time: O(n log(sum) log n) Space: O(n)

The key insight is that if we can form k teams, we can also form k-1 teams. This monotonic property allows us to binary search on the answer. For verification, we use a greedy approach: sort participants by usage limits and always use the most available ones first.

Brute Force (Try All Possibilities)

⏱️ Time: O(n! × 2^n) Space: O(n)

Start with team size 1 and keep trying to form larger teams by checking all possible combinations of participants. This approach simulates the actual team formation process but is extremely inefficient.

Binary Search + Greedy Verification (Optimal) — Algorithm Steps

Binary search on the number of teams (left=0, right=sum of all limits)
For each candidate k, check if k teams can be formed
Verification: sort usage limits in descending order
Greedily assign participants to teams, using most available first
If verification succeeds, try larger k; otherwise try smaller k
Return the maximum valid k

Visualization

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Step-by-Step Walkthrough

Binary Search Setup

Search space: 0 to sum of all usage limits

Verification Step

For each candidate k, check if k teams can be formed greedily

Greedy Assignment

Always use participants with highest availability first

Converge to Answer

Binary search converges to the maximum possible teams

Code -

solution.c — C

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

int compare(const void* a, const void* b) {
    return (*(int*)b - *(int*)a);
}

int canFormKTeams(int* usageLimits, int n, int k) {
    // Check if we have enough total participants
    long long needed = (long long)k * (k + 1) / 2;
    long long total = 0;
    for (int i = 0; i < n; i++) {
        total += usageLimits[i];
    }
    if (total < needed) return 0;
    
    // Check if we have enough distinct participant types
    if (k > n) return 0;
    
    // Sort in descending order for greedy approach
    int* limits = (int*)malloc(n * sizeof(int));
    memcpy(limits, usageLimits, n * sizeof(int));
    qsort(limits, n, sizeof(int), compare);
    
    long long* usage = (long long*)calloc(n, sizeof(long long));
    
    // Try to form k teams with sizes 1, 2, 3, ..., k
    for (int teamSize = 1; teamSize <= k; teamSize++) {
        int participantsNeeded = teamSize;
        
        // Greedily use participants with highest remaining availability
        for (int i = 0; i < n && participantsNeeded > 0; i++) {
            long long canUse = (participantsNeeded < limits[i] - usage[i]) ? 
                         participantsNeeded : limits[i] - usage[i];
            usage[i] += canUse;
            participantsNeeded -= canUse;
        }
        
        // If we couldn't form this team, return false
        if (participantsNeeded > 0) {
            free(limits);
            free(usage);
            return 0;
        }
    }
    
    free(limits);
    free(usage);
    return 1;
}

int maximumGroups(int* usageLimits, int n) {
    int left = 0;
    // Upper bound: min of participant types and max k where k*(k+1)/2 <= total
    long long totalParticipants = 0;
    for (int i = 0; i < n; i++) {
        totalParticipants += usageLimits[i];
    }
    int right = n;  // Can't have more teams than participant types
    int tempRight = 0;
    while ((long long)tempRight * (tempRight + 1) / 2 <= totalParticipants) {
        tempRight++;
    }
    tempRight--;
    if (tempRight < right) right = tempRight;
    
    int result = 0;
    
    while (left <= right) {
        int mid = left + (right - left) / 2;
        
        if (canFormKTeams(usageLimits, n, mid)) {
            result = mid;
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    
    return result;
}

int main() {
    char input[10000];
    fgets(input, sizeof(input), stdin);
    
    int usageLimits[100000];
    int n = 0;
    
    // Remove brackets if present
    char* start = input;
    if (*start == '[') start++;
    
    char* token = strtok(start, ",]\n");
    while (token != NULL) {
        // Skip empty tokens
        while (*token == ' ') token++;
        if (*token != '\0') {
            usageLimits[n++] = atoi(token);
        }
        token = strtok(NULL, ",]\n");
    }
    
    printf("%d\n", maximumGroups(usageLimits, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log(sum) log n)

Binary search takes O(log(sum)) iterations, each verification takes O(n log n) for sorting

⚠ Quadratic Growth

Space Complexity

O(n)

Space for sorting and storing usage limits

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Binary Search + Greedy Verification (Optimal) — Algorithm Steps

Visualization

Code -

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