Minimum Number of Groups to Create a Valid Assignment Solution in C++

Links to Other Language Solutions

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, “Minimum Number of Groups to Create a Valid Assignment,” is like trying to organize a family reunion where everyone wants to sit next to their favorite cousin but not next to that one uncle who tells the same joke every year.

In this LeetCode problem, you’re given an array of integers, where each integer represents a person. Your task is to determine the minimum number of groups you can create such that no two people in the same group have the same integer. Think of it as trying to assign seats at a wedding without putting the bride and groom’s exes at the same table.

Code Solution


class Solution {
 public:
  int minGroupsForValidAssignment(vector& nums) {
    unordered_map count;
    int minFreq = nums.size();

    for (const int num : nums)
      ++count[num];

    for (const auto& [_, freq] : count)
      minFreq = min(minFreq, freq);

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

    throw;
  }

 private:
  int getNumGroups(unordered_map& count, int groupSize) {
    int numGroups = 0;
    for (const auto& [_, freq] : count) {
      const int a = freq / (groupSize + 1);
      const int b = freq % (groupSize + 1);
      if (b == 0) {
        numGroups += a;
      } else if (groupSize - b <= a) {
        numGroups += a + 1;
      } else {
        return 0;
      }
    }
    return numGroups;
  }
};

Approach

The approach taken in this solution is quite clever. First, it counts the frequency of each integer in the input array. Then, it determines the minimum frequency of any integer, which sets the upper limit for group sizes. The algorithm then 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 the number of groups that can be formed based on the current group size, ensuring that no two identical integers end up in the same group.

Time and Space Complexity

Complexity Type	Complexity
Time Complexity	O(n)
Space Complexity	O(n)

Real-World Example

Imagine you're a teacher trying to group students for a project. Each student has a unique skill set (represented by integers), and you want to ensure that no two students with the same skill set are in the same group. Using the approach from this problem, you can efficiently determine the minimum number of groups needed to accommodate all students without causing any skill overlap.