# Minimum number of subsets with distinct elements

You are given an array of n-element. You have to make subsets from the array such that no subset contain duplicate elements. Find out minimum number of subset possible.

Examples :

```Input : arr[] = {1, 2, 3, 4}
Output :1
Explanation : A single subset can contains all
values and all values are distinct

Input : arr[] = {1, 2, 3, 3}
Output : 2
Explanation : We need to create two subsets
{1, 2, 3} and {3} [or {1, 3} and {2, 3}] such
that both subsets have distinct elements.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We basically need to find the most frequent element in the array. The result is equal to the frequency of the most frequent element.

A simple solution is to run two nested loops to count frequency of every element and return the frequency of the most frequent element. Time complexity of this solution is O(n2).

A better solution is to first sort the array and then start count number of repetitions of elements in an iterative manner as all repetition of any number lie beside the number itself. By this method you can find the maximum frequency or repetition by simply traversing the sorted array. This approach will cost O(nlogn) time complexity

## C++

 `// A sorting based solution to find the  ` `// minimum number of subsets of a set  ` `// such that every subset contains distinct ` `// elements. ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count subsets such that all ` `// subsets have distinct elements. ` `int` `subset(``int` `ar[], ``int` `n) ` `{ ` `    ``// Take input and initialize res = 0 ` `    ``int` `res = 0; ` ` `  `    ``// Sort the array ` `    ``sort(ar, ar + n); ` ` `  `    ``// Traverse the input array and ` `    ``// find maximum frequency ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``int` `count = 1; ` ` `  `        ``// For each number find its repetition / frequency ` `        ``for` `(; i < n - 1; i++) { ` `            ``if` `(ar[i] == ar[i + 1]) ` `                ``count++; ` `            ``else` `                ``break``; ` `        ``} ` ` `  `        ``// Update res ` `        ``res = max(res, count); ` `    ``} ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 5, 6, 9, 3, 4, 3, 4 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << subset(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// A sorting based solution to find the  ` `// minimum number of subsets of a set  ` `// such that every subset contains distinct ` `// elements. ` `import` `java.util.*; ` `import` `java.lang.*; ` ` `  `public` `class` `GfG{ ` `     `  `    ``// Function to count subsets such that all ` `    ``// subsets have distinct elements. ` `    ``public` `static` `int` `subset(``int` `ar[], ``int` `n) ` `    ``{ ` `        ``// Take input and initialize res = 0 ` `        ``int` `res = ``0``; ` ` `  `        ``// Sort the array ` `        ``Arrays.sort(ar); ` ` `  `        ``// Traverse the input array and ` `        ``// find maximum frequency ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``int` `count = ``1``; ` ` `  `            ``// For each number find its repetition / frequency ` `            ``for` `(; i < n - ``1``; i++) { ` `                ``if` `(ar[i] == ar[i + ``1``]) ` `                    ``count++; ` `                ``else` `                    ``break``; ` `            ``} ` ` `  `            ``// Update res ` `            ``res = Math.max(res, count); ` `        ``} ` ` `  `        ``return` `res; ` `    ``} ` `     `  `    ``// Driver function ` `    ``public` `static` `void` `main(String argc[]) ` `    ``{ ` `        ``int` `arr[] = { ``5``, ``6``, ``9``, ``3``, ``4``, ``3``, ``4` `}; ` `        ``int` `n = ``7``; ` `        ``System.out.println(subset(arr, n)); ` `    ``} ` `     `  `} ` ` `  `/* This code is contributed by Sagar Shukla */`

## Python3

 `# A sorting based solution to find the  ` `# minimum number of subsets of a set  ` `# such that every subset contains distinct ` `# elements. ` ` `  `# function to count subsets such that all ` `# subsets have distinct elements. ` `def` `subset(ar, n): ` ` `  `    ``# take input and initialize res = 0 ` `    ``res ``=` `0` ` `  `    ``# sort the array ` `    ``ar.sort() ` ` `  `    ``# traverse the input array and ` `    ``# find maximum frequency ` `    ``for` `i ``in` `range``(``0``, n) : ` `        ``count ``=` `1` ` `  `        ``# for each number find its repetition / frequency ` `        ``for` `i ``in` `range``(n ``-` `1``): ` `            ``if` `ar[i] ``=``=` `ar[i ``+` `1``]: ` `                ``count``+``=``1` `            ``else``: ` `                ``break` `             `  `        ``# update res ` `        ``res ``=` `max``(res, count) ` `     `  `    ``return` `res ` ` `  ` `  `# Driver code ` `ar ``=` `[ ``5``, ``6``, ``9``, ``3``, ``4``, ``3``, ``4` `] ` `n ``=` `len``(ar) ` `print``(subset(ar, n)) ` ` `  `# This code is contributed by ` `# Smitha Dinesh Semwal `

## C#

 `// A sorting based solution to find the  ` `// minimum number of subsets of a set  ` `// such that every subset contains distinct ` `// elements. ` `using` `System; ` ` `  `public` `class` `GfG { ` `     `  `    ``// Function to count subsets such that all ` `    ``// subsets have distinct elements. ` `    ``public` `static` `int` `subset(``int` `[]ar, ``int` `n) ` `    ``{ ` `        ``// Take input and initialize res = 0 ` `        ``int` `res = 0; ` ` `  `        ``// Sort the array ` `        ``Array.Sort(ar); ` ` `  `        ``// Traverse the input array and ` `        ``// find maximum frequency ` `        ``for` `(``int` `i = 0; i < n; i++) { ` `            ``int` `count = 1; ` ` `  `            ``// For each number find its  ` `            ``// repetition / frequency ` `            ``for` `( ; i < n - 1; i++) { ` `                ``if` `(ar[i] == ar[i + 1]) ` `                    ``count++; ` `                ``else` `                    ``break``; ` `            ``} ` ` `  `            ``// Update res ` `            ``res = Math.Max(res, count); ` `        ``} ` ` `  `        ``return` `res; ` `    ``} ` `     `  `    ``// Driver function ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[]arr = { 5, 6, 9, 3, 4, 3, 4 }; ` `        ``int` `n = 7; ` `         `  `        ``Console.WriteLine(subset(arr, n)); ` `    ``} ` `     `  `} ` ` `  `/* This code is contributed by Vt_m */`

## PHP

 ` `

Output:

`2`

An efficient solution is to use hashing. We count frequencies of all elements in a hash table. Finally we return the key with maximum value in hash table.

## C++

 `// A hashing based solution to find the  ` `// minimum number of subsets of a set  ` `// such that every subset contains distinct ` `// elements. ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count subsets such that all ` `// subsets have distinct elements. ` `int` `subset(``int` `arr[], ``int` `n) ` `{    ` `    ``// Traverse the input array and ` `    ``// store frequencies of elements ` `    ``unordered_map<``int``, ``int``> mp;     ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `        ``mp[arr[i]]++; ` `     `  `    ``// Find the maximum value in map. ` `    ``int` `res = 0; ` `    ``for` `(``auto` `x : mp) ` `       ``res = max(res, x.second); ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 5, 6, 9, 3, 4, 3, 4 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << subset(arr, n); ` `    ``return` `0; ` `} `

## Java

 `import` `java.util.HashMap; ` `import` `java.util.Map; ` ` `  `// A hashing based solution to find the  ` `// minimum number of subsets of a set  ` `// such that every subset contains distinct ` `// elements. ` `class` `GFG  ` `{  ` `     `  `// Function to count subsets such that all ` `// subsets have distinct elements. ` `static` `int` `subset(``int` `arr[], ``int` `n) ` `{  ` `    ``// Traverse the input array and ` `    ``// store frequencies of elements ` `    ``HashMap mp = ``new` `HashMap<>();  ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``mp.put(arr[i],mp.get(arr[i]) == ``null``?``1``:mp.get(arr[i])+``1``); ` `     `  `    ``// Find the maximum value in map. ` `    ``int` `res = ``0``; ` `    ``for` `(Map.Entry entry : mp.entrySet())  ` `    ``res = Math.max(res, entry.getValue()); ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `arr[] = { ``5``, ``6``, ``9``, ``3``, ``4``, ``3``, ``4` `}; ` `    ``int` `n = arr.length; ` `    ``System.out.println( subset(arr, n)); ` ` `  `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# A hashing based solution to find the  ` `# minimum number of subsets of a set such   ` `# that every subset contains distinct ` `# elements. ` ` `  `# Function to count subsets such that  ` `# all subsets have distinct elements. ` `def` `subset(arr, n): ` `     `  `    ``# Traverse the input array and ` `    ``# store frequencies of elements ` `    ``mp ``=` `{i:``0` `for` `i ``in` `range``(``10``)} ` `    ``for` `i ``in` `range``(n): ` `        ``mp[arr[i]] ``+``=` `1` `     `  `    ``# Find the maximum value in map. ` `    ``res ``=` `0` `    ``for` `key, value ``in` `mp.items(): ` `        ``res ``=` `max``(res, value) ` ` `  `    ``return` `res ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[``5``, ``6``, ``9``, ``3``, ``4``, ``3``, ``4``] ` `    ``n ``=` `len``(arr) ` `    ``print``(subset(arr, n)) ` ` `  `# This code is contributed by ` `# Surendra_Gangwar `

## C#

 `// A hashing based solution to find the  ` `// minimum number of subsets of a set  ` `// such that every subset contains distinct ` `// elements. ` `using` `System; ` `using` `System.Collections.Generic;  ` ` `  `class` `GFG  ` `{  ` `     `  `// Function to count subsets such that all ` `// subsets have distinct elements. ` `static` `int` `subset(``int` `[]arr, ``int` `n) ` `{  ` `    ``// Traverse the input array and ` `    ``// store frequencies of elements ` `    ``Dictionary<``int``,  ` `               ``int``> mp = ``new` `Dictionary<``int``,  ` `                                        ``int``>(); ` `    ``for` `(``int` `i = 0 ; i < n; i++) ` `    ``{ ` `        ``if``(mp.ContainsKey(arr[i])) ` `        ``{ ` `            ``var` `val = mp[arr[i]]; ` `            ``mp.Remove(arr[i]); ` `            ``mp.Add(arr[i], val + 1);  ` `        ``} ` `        ``else` `        ``{ ` `            ``mp.Add(arr[i], 1); ` `        ``} ` `    ``} ` `     `  `    ``// Find the maximum value in map. ` `    ``int` `res = 0; ` `    ``foreach``(KeyValuePair<``int``, ``int``> entry ``in` `mp) ` `        ``res = Math.Max(res, entry.Value); ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``int` `[]arr = { 5, 6, 9, 3, 4, 3, 4 }; ` `    ``int` `n = arr.Length; ` `    ``Console.WriteLine(subset(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output :

`2`

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