# Maximum number of similar groups of a given size that can be made from a given array

I am given an array of numbers, not necessarily unique, and the size of a group. Let the array be denoted by $$B$$ and the size of the group be $$A$$.

I need to find the maximum number of groups with the exact same contents and of size $$A$$ that can be made from the elements of $$B$$. Once an element is used in a group it is exhausted. A group can have more than one element with the same value as long as the element is still available in $$B$$.

Example:

1. If the input array is $$\{1, 2, 3, 1, 2, 3, 4\}$$, and the group size is $$3$$ the maximum number of groups that can be made is $$2$$, which are $$\{1, 2, 3\}$$ and $$\{1, 2, 3\}$$.
2. If the input array is $$\{1, 3, 3, 3, 6, 3, 10\}$$, and the group size is $$4$$ the maximum number of groups that can be made is $$1$$, which is $$\{1, 3, 3, 3\}$$.

What I have tried so far, is to frame some equations ( given below ) but after that, I am struggling to come up with an algorithm to solve them.

Let $$F_1$$ be the frequency of the element $$B_1$$, $$F_2$$ be the frequency of the element $$B_2$$ and so on till $$B_n$$, where $$B_1 \dots B_n$$ are distinct elements from the array $$B$$.

Now I need to choose $$n_1, n_2, \dots n_i$$ such that

1. $$n_1 + n_2 + \dots + n_i = A$$
2. $$k\cdot n_1 \leq F_1\text{ , } k\cdot n_2 \leq F_2\text{ , }\dots \text{ , }k\cdot n_i \leq F_i$$
3. $$k$$ is the number of groups and we need to maximize it.

Length of $$B$$ can be as large as $$10^5$$ and $$A$$ can also be as large as $$10^5$$.

• Hint: try to reduce the problem to a decisional version of it. A decision version is: given $k$: find if it is possible to find a solution of size $k$ – narek Bojikian Jan 12 at 9:07
• @narekBojikian Okay, I thought about an approach by considering it as a decisional problem. $k$ is given, and I have the group with $A$ places to fill. I maintain a priority queue of the frequency array. To fill a place in all the $k$ groups, I do an extract max, subtract $k$ from the frequency val and insert it back into the queue. I do the same until I fill all the $A$ places in all the groups or I am unable to fill it. Is this approach right? – Debarun Mukherjee Jan 12 at 11:15
The function $$f:\mathbb{N}\rightarrow\{0, 1\}:f(k) = \begin{cases} 1; &\text{if there is a solution of size k,}\\ 0; &\text{otherwise} \end{cases}$$ is monoton, since if there is no solution of size $$k$$ then there is no solution of size $$k+1$$. That means we can binary search the value of $$k$$ in the interval $$[1, |B|]$$, and output the greatest $$k$$ for which there is a solution of size $$k$$. Thereby, we turned the problem into a decision problem with an $$O(\log |B|)$$ factor in the running time.
For a fixed value of $$k$$, a greedy solution suffices. If we have $$F_i$$ copies of some element $$B_i$$, then each set can contain at most $$\left\lfloor\frac{F_i}{k}\right\rfloor$$ of this element. So we have a solution of size $$k$$ if and only if $$\sum\limits_{i=1}^{n} \left\lfloor\frac{F_i}{k}\right\rfloor \geq A.$$
As an exercise, try to prove that the greedy solution is optimal. The running time can be bounded in $$O(|B|\log|B|).$$