# Minimum number of distinct elements after removing M items | Set 2

• Difficulty Level : Medium
• Last Updated : 19 May, 2021

Given an array of items, an ith 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
Explanation:
Remove 1 and both 2’s.
So, only 3 will be left, hence distinct number of element is 1.

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

For O(N*log N) approach please refer to the previous article.

Efficient Approach: The idea is to store the occurrence of each element in a Hash and count the occurrence of each frequency in a hash again. Below are the steps:

1. Traverse the given array elements and count the occurrences of each number and store it into the hash.
2. Now instead of sorting the frequency, count the occurrences of the frequency into another array say fre_arr[].
3. Calculate the total unique numbers(say ans) as the number of distinct id’s.
4. Now, traverse the freq_arr[] array and if freq_arr[i] > 0 then remove the minimum of m/i and freq_arr[i](say t) from ans and subtract i*t from m to remove the occurrence of any number i from m.

Below is the implementation of the above approach.

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to return minimum distinct``// character after M removals``int` `distinctNumbers(``int` `arr[], ``int` `m,``                    ``int` `n)``{``    ``unordered_map<``int``, ``int``> count;` `    ``// Count the occurences of number``    ``// and store in count``    ``for` `(``int` `i = 0; i < n; i++)``        ``count[arr[i]]++;` `    ``// Count the occurences of the``    ``// frequencies``    ``vector<``int``> fre_arr(n + 1, 0);``    ``for` `(``auto` `it : count) {``        ``fre_arr[it.second]++;``    ``}` `    ``// Take answer as total unique numbers``    ``// and remove the frequency and``    ``// subtract the answer``    ``int` `ans = count.size();` `    ``for` `(``int` `i = 1; i <= n; i++) {``        ``int` `temp = fre_arr[i];``        ``if` `(temp == 0)``            ``continue``;` `        ``// Remove the minimum number``        ``// of times``        ``int` `t = min(temp, m / i);``        ``ans -= t;``        ``m -= i * t;``    ``}` `    ``// Return the answer``    ``return` `ans;``}` `// Driver Code``int` `main()``{``    ``// Initialize array``    ``int` `arr[] = { 2, 4, 1, 5, 3, 5, 1, 3 };` `    ``// Size of array``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `m = 2;` `    ``// Function call``    ``cout << distinctNumbers(arr, m, n);``    ``return` `0;``}`

## Java

 `// Java program to implement the``// above approach``import` `java.util.*;` `class` `GFG{` `// Function to return minimum distinct``// character after M removals``static` `int` `distinctNumbers(``int` `arr[], ``int` `m,``                                      ``int` `n)``{``    ``Map count = ``new` `HashMap();` `    ``// Count the occurences of number``    ``// and store in count``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``count.put(arr[i],``              ``count.getOrDefault(arr[i], ``0``) + ``1``);``    ` `    ``// Count the occurences of the``    ``// frequencies``    ``int``[] fre_arr = ``new` `int``[n + ``1``];``    ``for``(Integer it : count.values())``    ``{``        ``fre_arr[it]++;``    ``}` `    ``// Take answer as total unique numbers``    ``// and remove the frequency and``    ``// subtract the answer``    ``int` `ans = count.size();` `    ``for``(``int` `i = ``1``; i <= n; i++)``    ``{``        ``int` `temp = fre_arr[i];``        ``if` `(temp == ``0``)``            ``continue``;` `        ``// Remove the minimum number``        ``// of times``        ``int` `t = Math.min(temp, m / i);``        ``ans -= t;``        ``m -= i * t;``    ``}` `    ``// Return the answer``    ``return` `ans;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ` `    ``// Initialize array``    ``int` `arr[] = { ``2``, ``4``, ``1``, ``5``, ``3``, ``5``, ``1``, ``3` `};``    ` `    ``// Size of array``    ``int` `n = arr.length;``    ``int` `m = ``2``;``    ` `    ``// Function call``    ``System.out.println(distinctNumbers(arr, m, n));``}``}` `// This code is contributed by offbeat`

## Python3

 `# Python3 program for the above approach` `# Function to return minimum distinct``# character after M removals``def` `distinctNumbers(arr, m, n):` `    ``count ``=` `{}` `    ``# Count the occurences of number``    ``# and store in count``    ``for` `i ``in` `range``(n):``        ``count[arr[i]] ``=` `count.get(arr[i], ``0``) ``+` `1` `    ``# Count the occurences of the``    ``# frequencies``    ``fre_arr ``=` `[``0``] ``*` `(n ``+` `1``)``    ``for` `it ``in` `count:``        ``fre_arr[count[it]] ``+``=` `1` `    ``# Take answer as total unique numbers``    ``# and remove the frequency and``    ``# subtract the answer``    ``ans ``=` `len``(count)` `    ``for` `i ``in` `range``(``1``, n ``+` `1``):``        ``temp ``=` `fre_arr[i]``        ``if` `(temp ``=``=` `0``):``            ``continue``            ` `        ``# Remove the minimum number``        ``# of times``        ``t ``=` `min``(temp, m ``/``/` `i)``        ``ans ``-``=` `t``        ``m ``-``=` `i ``*` `t` `    ``# Return the answer``    ``return` `ans` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:` `    ``# Initialize array``    ``arr ``=` `[ ``2``, ``4``, ``1``, ``5``, ``3``, ``5``, ``1``, ``3` `]` `    ``# Size of array``    ``n ``=` `len``(arr)``    ``m ``=` `2` `    ``# Function call``    ``print``(distinctNumbers(arr, m, n))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program to implement the``// above approach``using` `System;``using` `System.Collections.Generic;``class` `GFG{` `// Function to return minimum distinct``// character after M removals``static` `int` `distinctNumbers(``int` `[]arr,``                           ``int` `m, ``int` `n)``{``  ``Dictionary<``int``,``             ``int``> count = ``new` `Dictionary<``int``,``                                         ``int``>();``  ` `  ``// Count the occurences of number``  ``// and store in count``  ``for` `(``int` `i = 0; i < n; i++)``    ``if``(count.ContainsKey(arr[i]))``    ``{``      ``count[arr[i]] = count[arr[i]] + 1;``    ``}``  ``else``  ``{``    ``count.Add(arr[i], 1);``  ``}` `  ``// Count the occurences of the``  ``// frequencies``  ``int``[] fre_arr = ``new` `int``[n + 1];``  ``foreach``(``int` `it ``in` `count.Values)``  ``{``    ``fre_arr[it]++;``  ``}` `  ``// Take answer as total unique numbers``  ``// and remove the frequency and``  ``// subtract the answer``  ``int` `ans = count.Count;` `  ``for``(``int` `i = 1; i <= n; i++)``  ``{``    ``int` `temp = fre_arr[i];``    ``if` `(temp == 0)``      ``continue``;` `    ``// Remove the minimum number``    ``// of times``    ``int` `t = Math.Min(temp, m / i);``    ``ans -= t;``    ``m -= i * t;``  ``}` `  ``// Return the answer``  ``return` `ans;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``  ``// Initialize array``  ``int` `[]arr = {2, 4, 1, 5, 3, 5, 1, 3};` `  ``// Size of array``  ``int` `n = arr.Length;``  ``int` `m = 2;` `  ``// Function call``  ``Console.WriteLine(distinctNumbers(arr, m, n));``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``
Output:
`3`

Time Complexity: O(N)
Auxiliary Space: O(N)

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