Minimum Number of Groups to Create a Valid Assignment

Minimum Number of Groups to Create a Valid Assignment - Problem

Minimum Number of Groups to Create a Valid Assignment

You're organizing a tournament and need to distribute numbered balls into boxes (groups) following strict fairness rules. Each box can only contain balls with the same number, but balls with the same number can be split across different boxes if needed.

The key constraint is balance: the difference between the largest group size and smallest group size cannot exceed 1. In other words, if the smallest group has k balls, the largest group can have at most k+1 balls.

Goal: Find the minimum number of groups needed to distribute all balls while maintaining this balance.

Input: An array of integers representing the numbered balls
Output: The minimum number of groups required

Input & Output

example_1.py — Basic Case

$ Input: [3, 2, 3, 2, 3]

› Output: 2

💡 Note: We have 3 balls of value 3 and 2 balls of value 2. We can create 2 groups: one with [3,3,3] (size 3) and one with [2,2] (size 2). The size difference is 3-2=1, which satisfies our constraint.

example_2.py — Equal Distribution

$ Input: [1, 1, 2, 2, 3, 3]

› Output: 3

💡 Note: Each number appears twice. We can create 3 groups: [1,1], [2,2], [3,3], each with size 2. All groups have equal size, so the difference is 0 ≤ 1.

example_3.py — Single Type

$ Input: [1, 1, 1, 1, 1]

› Output: 2

💡 Note: All balls have the same value. We can split them into 2 groups: [1,1,1] (size 3) and [1,1] (size 2). The difference is 1, which is the maximum allowed.

Constraints

1 ≤ balls.length ≤ 10⁵
1 ≤ balls[i] ≤ 10⁹
The difference between largest and smallest group sizes must be at most 1

Visualization

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Understanding the Visualization

Count Team Sizes

First, count how many teams we have of each strength level

Determine Bracket Sizes

Try different minimum bracket sizes and see if we can fit all teams

Distribute Teams

For each team type, calculate minimum brackets needed while respecting size constraints

Find Optimal Solution

Return the configuration that uses the fewest total brackets

Key Takeaway

🎯 Key Insight: Instead of trying all possible group arrangements, we mathematically determine the minimum groups needed by testing each possible minimum group size and using the ceiling formula to calculate required groups for each frequency.

Asked in

G Google 42 a Amazon 35 f Meta 28 ⊞ Microsoft 31

The optimal solution uses mathematical optimization by trying each possible minimum group size and calculating the required number of groups using the formula ⌈frequency/(minSize+1)⌉. This approach runs in O(n + m²) time where m is the number of unique values, which is much faster than brute force approaches.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Possible Group Counts)	O(n³)	O(n)	Try every possible number of groups from 1 to n and check if valid assignment exists
Optimized Mathematical Solution	O(n + m²)	O(m)	Use mathematical properties to directly determine minimum groups needed

Brute Force (Try All Possible Group Counts) — Algorithm Steps

Count frequency of each ball number
For each possible group count k from 1 to total balls:
Try to assign ball types to groups maintaining size balance
Check if assignment is valid (size difference ≤ 1)
Return the first k that works

Visualization

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Step-by-Step Walkthrough

Count Frequencies

Count how many times each ball number appears

Try Group Counts

For each possible number of groups k from 1 to n

Attempt Distribution

Try to distribute ball types into k groups with size constraint

Check Validity

Verify that max group size - min group size ≤ 1

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 canDistribute(int* frequencies, int freqSize, int numGroups, int minSize, int maxSize) {
    qsort(frequencies, freqSize, sizeof(int), compare);
    int groupsUsed = 0;
    
    for (int i = 0; i < freqSize; i++) {
        int groupsNeeded = (frequencies[i] + maxSize - 1) / maxSize;
        groupsUsed += groupsNeeded;
        
        if (groupsUsed > numGroups) {
            return 0;
        }
    }
    
    int totalBalls = 0;
    for (int i = 0; i < freqSize; i++) {
        totalBalls += frequencies[i];
    }
    return totalBalls >= groupsUsed * minSize && totalBalls <= numGroups * maxSize;
}

int canAssignToGroups(int* frequencies, int freqSize, int numGroups) {
    int totalBalls = 0;
    for (int i = 0; i < freqSize; i++) {
        totalBalls += frequencies[i];
    }
    
    for (int minSize = 1; minSize <= totalBalls / numGroups + 1; minSize++) {
        int maxSize = minSize + 1;
        int* freqCopy = malloc(freqSize * sizeof(int));
        memcpy(freqCopy, frequencies, freqSize * sizeof(int));
        
        if (canDistribute(freqCopy, freqSize, numGroups, minSize, maxSize)) {
            free(freqCopy);
            return 1;
        }
        free(freqCopy);
    }
    return 0;
}

int minGroupsForValidAssignment(int* balls, int ballsSize) {
    int freqMap[1001] = {0}; // Assuming ball values are small
    for (int i = 0; i < ballsSize; i++) {
        freqMap[balls[i]]++;
    }
    
    int frequencies[1001];
    int freqSize = 0;
    for (int i = 0; i < 1001; i++) {
        if (freqMap[i] > 0) {
            frequencies[freqSize++] = freqMap[i];
        }
    }
    
    for (int numGroups = 1; numGroups <= ballsSize; numGroups++) {
        if (canAssignToGroups(frequencies, freqSize, numGroups)) {
            return numGroups;
        }
    }
    
    return ballsSize;
}

int main() {
    int balls[] = {3, 2, 3, 2, 3};
    int ballsSize = sizeof(balls) / sizeof(balls[0]);
    printf("%d\n", minGroupsForValidAssignment(balls, ballsSize));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

For each of n possible group counts, we try to distribute ball types which can take O(n²) in worst case

⚠ Quadratic Growth

Space Complexity

O(n)

Space for frequency map and group assignment tracking

⚡ Linearithmic Space

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Medium-High Frequency

~25 min Avg. Time

1.2K Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Possible Group Counts) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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