# Minimum number of distinct elements after removing m items

• Difficulty Level : Medium
• Last Updated : 11 Nov, 2022

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```

1. Count the occurrence of elements and store them in the hash.
2. Sort the hash.
3. Start removing elements from the hash whose frequency count is less than m.
4. Return the number of values left in the hash.

Implementation:

## C++

 `// C++ program for above implementation` `#include ` `using` `namespace` `std;`   `// Function to find distinct 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[0]);`   `    ``int` `m = 3;`   `    ``cout << distinctIds(arr, n, m);` `    ``return` `0;` `}`

## Java

 `import` `java.util.*;`   `class` `Solution {` `    ``static` `int` `distinctIds(``int` `arr[], ``int` `n, ``int` `m)` `    ``{` `        ``// Creating HashMap to store frequency count` `        ``HashMap h = ``new` `HashMap<>();`   `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``if` `(h.containsKey(arr[i])) {` `                ``h.put(arr[i], h.get(arr[i]) + ``1``);` `            ``}` `            ``else` `{` `                ``h.put(arr[i], ``1``);` `            ``}` `        ``}`   `        ``// Creating a list to sort HashMap according to` `        ``// values` `        ``List > l` `            ``= ``new` `LinkedList >(` `                ``h.entrySet());`   `        ``// sorting using Comparator` `        ``Collections.sort(` `            ``l,` `            ``new` `Comparator >() {` `                ``public` `int` `compare(` `                    ``Map.Entry o1,` `                    ``Map.Entry o2)` `                ``{` `                    ``return` `o1.getValue() - o2.getValue();` `                ``}` `            ``});`   `        ``// Creating new map after sorting and also` `        ``// maintaining insertion order` `        ``LinkedHashMap lh` `            ``= ``new` `LinkedHashMap<>();` `        ``for` `(Map.Entry e : l) {` `            ``lh.put(e.getKey(), e.getValue());` `        ``}`   `        ``for` `(Integer i : lh.keySet()) {` `            ``// removing element from whose frequency count is` `            ``// less than m ,Sorted manner to get minimum` `            ``// distinct ids` `            ``if` `(h.get(i) <= m) {` `                ``m -= h.get(i);` `                ``h.remove(i);` `            ``}` `        ``}`   `        ``return` `h.size();` `    ``}`   `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``// TODO Auto-generated method stub` `        ``int` `arr[] = { ``2``, ``4``, ``1``, ``5``, ``3``, ``5``, ``1``, ``3` `};` `        ``int` `m = ``2``;`   `        ``System.out.println(distinctIds(arr, arr.length, m));` `    ``}` `}`

## Python3

 `# Python program for above implementation`   `# Function to find distinct id's` `def` `distinctIds(arr, n, mi):` `  ``m ``=` `{}` `  ``v ``=` `[]` `  ``count ``=` `0`   `  ``# Store the occurrence of ids` `  ``for` `i ``in` `range``(n):` `    ``if` `arr[i] ``in` `m:` `      ``m[arr[i]] ``+``=` `1` `    ``else``:` `      ``m[arr[i]] ``=` `1`   `  ``# Store into the list value as key and vice-versa` `  ``for` `i ``in` `m:` `    ``v.append([m[i],i])`   `  ``v.sort()` `  ``size ``=` `len``(v)`   `  ``# Start removing elements from the beginning` `  ``for` `i ``in` `range``(size):` `    `  `    ``# Remove if current value is less than ` `    ``# or equal to mi` `    ``if` `(v[i][``0``] <``=` `mi):` `      ``mi ``-``=` `v[i][``0``]` `      ``count ``+``=` `1` `        `  `    ``else``:   ``# Return the remaining size` `      ``return` `size ``-` `count` `  ``return` `size ``-` `count`   `# Driver code` `arr ``=` `[ ``2``, ``3``, ``1``, ``2``, ``3``, ``3` `]` `n ``=` `len``(arr)`   `m ``=` `3`   `# To display the result` `print``(distinctIds(arr, n, m))`   `# This code is contributed by rohitsingh07052`

## C#

 `// C# program for Minimum number of` `// distinct elements after removing m items` `using` `System;` `using` `System.Collections.Generic;  ` `class` `GFG` `{`   `  ``// Function to find distinct id's` `  ``static` `int` `distinctIds(``int``[] arr, ``int` `n, ``int` `mi)` `  ``{`   `    ``Dictionary<``int``, ``int``> m = ``new` `Dictionary<``int``, ``int``>();  ` `    ``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[arr[i]] = 1;` `        ``size++;` `      ``}`   `      ``// If it is present then increase the value by 1` `      ``else` `      ``{` `        ``m[arr[i]]++;` `      ``}` `    ``}`   `    ``// Start removing elements from the beginning` `    ``foreach``(KeyValuePair<``int``, ``int``> mp ``in` `m) ` `    ``{`   `      ``// Remove if current value is less than` `      ``// or equal to mi` `      ``if` `(mp.Key <= mi)` `      ``{` `        ``mi -= mp.Key;` `        ``count++;` `      ``}` `    ``}` `    ``return` `size - count;` `  ``}`   `  ``// Driver code` `  ``static` `void` `Main() ` `  ``{`   `    ``// TODO Auto-generated method stub` `    ``int``[] arr = {2, 3, 1, 2, 3, 3};` `    ``int` `m = 3;`   `    ``Console.WriteLine(distinctIds(arr, arr.Length, m));` `  ``}` `}`   `// This code is contributed by divyeshrabadiya07`

## Javascript

 ``

Output

`1`

Time Complexity: O(n log n)

Space Complexity: O(n), As we  are using map to store elements

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