# Split an array containing N elements into K sets of distinct elements

Given two integers N and K and an array arr[] consisting of duplicate elements, the task is to split N elements into K sets of distinct elements.

Examples:

Input: N = 5, K = 2, arr[] = {3, 2, 1, 2, 3}
Output:
( 3 2 1 )
( 2 3 )

Input: N = 5, K = 2, arr[] = {2, 1, 1, 2, 1}
Output: -1
Explanation:
It is not possible to split all the elements into K sets of distinct elements as 1 appears more than K times in the array.

Approach: In order to solve the problem, we are using a map to store the frequency of every element. If the frequency of any element exceeds K, print -1. Maintain another map to store the sets for every respective frequencies. If no element has a frequency greater than K, print the sets for all corresponding frequencies as the required set.

Below is the implementation of the above approach:

## C++

 `// C++ Program to split N elements ` `// into exactly K sets consisting ` `// of no distinct elements ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to check if possible to ` `// split N elements into exactly K ` `// sets consisting of no distinct elements ` `void` `splitSets(``int` `a[], ``int` `n, ``int` `k) ` `{ ` `    ``// Store the frequency ` `    ``// of each element ` `    ``map<``int``, ``int``> freq; ` ` `  `    ``// Store the required sets ` `    ``map<``int``, vector<``int``> > ans; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``// If frequency of a ` `        ``// particular element ` `        ``// exceeds K ` `        ``if` `(freq[a[i]] + 1 > k) { ` `            ``// Not possible ` `            ``cout << -1 << endl; ` `            ``return``; ` `        ``} ` ` `  `        ``// Increase the frequency ` `        ``freq[a[i]] += 1; ` `        ``// Store the element for the ` `        ``// respective set ` `        ``ans[freq[a[i]]].push_back(a[i]); ` `    ``} ` ` `  `    ``// Display the sets ` `    ``for` `(``auto` `it : ans) { ` `        ``cout << ``"( "``; ` `        ``for` `(``int` `i : it.second) { ` `            ``cout << i << ``" "``; ` `        ``} ` `        ``cout << ``")\n"``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `arr[] = { 2, 1, 2, 3, 1, ` `                  ``4, 1, 3, 1, 4 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` ` `  `    ``int` `k = 4; ` ` `  `    ``splitSets(arr, n, k); ` ` `  `    ``return` `0; ` `} `

Output:

```( 2 1 3 4 )
( 2 1 3 4 )
( 1 )
( 1 )
```

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