Minimum Number of Groups to Create a Valid Assignment Solution in Java

Explore Solutions in Other Languages

Problem Description

Ah, the classic dilemma of group assignments! You know, the one where you have to figure out how to split a bunch of people into groups without causing a riot? The problem is simple yet complex, much like trying to explain why your cat knocked over your coffee.

In this problem, you are given an array of integers, nums, where each integer represents a task that needs to be assigned to a group. Your mission, should you choose to accept it, is to determine the minimum number of groups required to ensure that each group can handle the tasks without any overlap.

Imagine you’re at a party, and everyone wants to play a game. But wait! You can’t have two people playing the same game in the same group. So, how do you organize this chaos? That’s where our code comes in!

Code Solution


class Solution {
  public int minGroupsForValidAssignment(int[] nums) {
    Map count = new HashMap<>();
    int minFreq = nums.length;

    for (final int num : nums)
      count.merge(num, 1, Integer::sum);

    for (final int freq : count.values())
      minFreq = Math.min(minFreq, freq);

    for (int groupSize = minFreq; groupSize >= 1; --groupSize) {
      final int numGroups = getNumGroups(count, groupSize);
      if (numGroups > 0)
        return numGroups;
    }

    throw new IllegalArgumentException();
  }

  private int getNumGroups(Map count, int groupSize) {
    int numGroups = 0;
    for (final int freq : count.values()) {
      final int a = freq / (groupSize + 1);
      final int b = freq % (groupSize + 1);
      if (b == 0) {
        numGroups += a;
      } else if (groupSize - b <= a) {
        numGroups += a + 1;
      } else {
        return 0;
      }
    }
    return numGroups;
  }
}

Approach Explanation

The code begins by counting the frequency of each task in the nums array. It then determines the minimum frequency of any task, which helps in deciding the maximum possible size of the groups. The algorithm iterates from this minimum frequency down to 1, checking how many groups can be formed for each potential group size.

The helper function getNumGroups calculates how many groups can be formed based on the current group size. It does this by dividing the frequency of each task by the group size and checking if the remaining tasks can fit into the groups without exceeding the limits.

Time and Space Complexity

Complexity Type	Complexity
Time Complexity	O(n), where n is the number of elements in the `nums` array.
Space Complexity	O(n), due to the storage of frequencies in a HashMap.

Real-World Example

Imagine you’re organizing a cooking class where each participant can only cook one dish at a time. If you have a list of dishes that people want to cook, you need to figure out how many groups you can form so that no one is cooking the same dish in the same group. This problem mirrors the task of assigning groups based on the frequency of dish requests, ensuring that everyone gets to cook without any overlap.