# Minimum number of distinct elements after removing m items

Given an array of items, an i-th index element denotes the item id’s and given a number m, the task is to remove m elements such that there should be minimum distinct id’s left.Print the number of distinct id’s.

Examples:

```Input : arr[] = { 2, 2, 1, 3, 3, 3}
m = 3
Output : 1
Remove 1 and both 2's.So, only 3 will be
left that's why distinct id is 1.

Input : arr[] = { 2, 4, 1, 5, 3, 5, 1, 3}
m = 2
Output : 3
Remove 2 and 4 completely. So, remaining ids
are 1, 3 and 5 i.e. 3
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

1- Count the occurrence of elements and store in the hash.
2- Sort the hash.
3- Start removing elements from hash.
4- Return the number of values left in the hash.

## C++

 `// C++ program for above implementation ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find distintc id's ` `int` `distinctIds(``int` `arr[], ``int` `n, ``int` `mi) ` `{ ` `    ``unordered_map<``int``, ``int``> m; ` `    ``vector > v; ` `    ``int` `count = 0; ` ` `  `    ``// Store the occurrence of ids ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``m[arr[i]]++; ` ` `  `    ``// Store into the vector second as first and vice-versa ` `    ``for` `(``auto` `it = m.begin(); it != m.end(); it++) ` `        ``v.push_back(make_pair(it->second, it->first)); ` ` `  `    ``// Sort the vector ` `    ``sort(v.begin(), v.end()); ` ` `  `    ``int` `size = v.size(); ` ` `  `    ``// Start removing elements from the beginning ` `    ``for` `(``int` `i = 0; i < size; i++) { ` ` `  `        ``// Remove if current value is less than  ` `        ``// or equal to mi ` `        ``if` `(v[i].first <= mi) { ` `            ``mi -= v[i].first; ` `            ``count++; ` `        ``} ` ` `  `        ``// Return the remaining size ` `        ``else` `            ``return` `size - count; ` `    ``} ` `    ``return` `size - count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 3, 1, 2, 3, 3 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``int` `m = 3; ` ` `  `    ``cout << distinctIds(arr, n, m); ` `    ``return` `0; ` `} `

## Java

 `//Java program for Minimum number of ` `//distinct elements after removing m items ` `import` `java.util.HashMap; ` `import` `java.util.Map; ` `import` `java.util.Map.Entry; ` ` `  `public` `class` `DistinctIds ` `{ ` `    ``// Function to find distintc id's ` `    ``static` `int` `distinctIds(``int` `arr[], ``int` `n, ``int` `mi) ` `    ``{ ` ` `  `        ``Map m = ``new` `HashMap(); ` `        ``int` `count = ``0``; ` `        ``int` `size = ``0``; ` ` `  `        ``// Store the occurrence of ids ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` ` `  `            ``// If the key is not add it to map ` `            ``if` `(m.containsKey(arr[i]) == ``false``) ` `            ``{ ` `                ``m.put(arr[i], ``1``); ` `                ``size++; ` `            ``} ` ` `  `            ``// If it is present then increase the value by 1 ` `            ``else` `m.put(arr[i], m.get(arr[i]) + ``1``); ` `        ``} ` ` `  `        ``// Start removing elements from the beginning ` `        ``for` `(Entry mp:m.entrySet()) ` `        ``{ ` `            ``// Remove if current value is less than ` `            ``// or equal to mi ` `            ``if` `(mp.getKey() <= mi) ` `            ``{ ` `                ``mi -= mp.getKey(); ` `                ``count++; ` `            ``} ` `            ``// Return the remaining size ` `            ``else` `return` `size - count; ` `        ``} ` ` `  `        ``return` `size - count; ` `    ``} ` ` `  `    ``//Driver method to test above function ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``// TODO Auto-generated method stub ` `        ``int` `arr[] = {``2``, ``3``, ``1``, ``2``, ``3``, ``3``}; ` `        ``int` `m = ``3``; ` ` `  `        ``System.out.println(distinctIds(arr, arr.length, m)); ` `    ``} ` `} ` `//This code is contributed by Sumit Ghosh `

Output:

```1
```

Time Complexity : O(n log n)

This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.