# Maximum distinct elements after removing k elements

• Difficulty Level : Medium
• Last Updated : 31 Oct, 2022

Given an array arr[] containing n elements. The problem is to find the maximum number of distinct elements (non-repeating) after removing k elements from the array.
Note: 1 <= k <= n.
Examples:

```Input : arr[] = {5, 7, 5, 5, 1, 2, 2}, k = 3
Output : 4
Remove 2 occurrences of element 5 and
1 occurrence of element 2.

Input : arr[] = {1, 2, 3, 4, 5, 6, 7}, k = 5
Output : 2

Input : arr[] = {1, 2, 2, 2}, k = 1
Output : 1```

Approach: Following are the steps:

1. Make a multi set from the given array.

2. During making this multiset check if the current element is present or not in multiset, if it is already present then simply reduce the k value and do not insert in the multiset.

3. If k becomes 0 then simply just put values in multiset.

4. After traversing the whole given array,

a) if k is not equal to zero then it means the multiset is consist of only unique elements and we have to remove any of the k elements from the multiset to make k=0, so in this case the answer will be size of multiset minus k value at that time.

b) if k is equal to zero then it means there may be duplicate values present in the multiset so put all the values in a set and the size of this set will be the number of distinct elements after removing k elements

## C++

 `// CPP implementation of the above approach``#include ``using` `namespace` `std;``  ` `// function to find maximum distinct elements``// after removing k elements``int` `maxDistinctNum(``int` `a[], ``int` `n, ``int` `k)``{``  ``int` `i;``  ``multiset<``int``> s;``  ``// making multiset from given array``        ``for``(i=0;i st;``            ``for``(``auto` `it:s){``                ``st.insert(it);``            ``}``            ``return` `st.size();``        ``}``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 5, 7, 5, 5, 1, 2, 2 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `k = 3;``  ` `    ``// Function Call``    ``cout << ``"Maximum distinct elements = "``         ``<< maxDistinctNum(arr, n, k);``    ``return` `0;``}`

## Python3

 `# Python implementation of the above approach` `# function to find maximum distinct elements after removing k elements``def` `maxDistinctNum(a, n, k):``  ` `   ``# making multiset from given array multisets are like dictionaries ,``   ``# so will initialise a dictionary``    ``s ``=` `{}``    ``for` `i ``in` `range``(n):``        ``if` `a[i] ``not` `in` `s ``or` `k ``=``=` `0``:``            ``s[a[i]] ``=` `s.get(a[i], ``0``)``+``1``        ``else``:``            ``s[a[i]] ``=` `1``            ``k ``-``=` `1``    ``if` `k !``=` `0``:``        ``return` `len``(s)``-``k``    ``else``:` `        ``st ``=` `set``()``        ``for` `i ``in` `s:``            ``st.add(i)``        ``return` `len``(st)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `  ``# Array``    ``arr ``=` `[``5``, ``7``, ``5``, ``5``, ``1``, ``2``, ``2``]``    ``K ``=` `3` `    ``# Size of array``    ``N ``=` `len``(arr)``    ` `    ``# Function Call``    ``print``(``"Maximum distinct elements = "``, maxDistinctNum(arr, N, K))` `# This code is contributed by vivekmaddheshiya205`

## Java

 `// Java implementation of the``// above approach``import` `java.util.*;``class` `GFG{``    ` `// Function to find maximum``// distinct elements after``// removing k elements``static` `int` `maxDistinctNum(``int` `arr[],``                          ``int` `n, ``int` `k)``{``  ``HashMap numToFreq = ``new` `HashMap<>();` `  ``// Build frequency map``  ``for``(``int` `i = ``0` `; i < n ; i++)``  ``{``    ``numToFreq.put(arr[i],``    ``numToFreq.getOrDefault(arr[i], ``0``) + ``1``);``  ``}` `  ``int` `result = ``0``;` `  ``// Min-heap``  ``PriorityQueue minHeap =``                ``new` `PriorityQueue();` `  ``// Add all number with freq=1 to``  ``// result and push others to minHeap``  ``for``(Map.Entry p : numToFreq.entrySet())``  ``{``    ``if``(p.getValue() == ``1``)``      ``++result;``    ``else``      ``minHeap.add(p.getValue());``  ``}` `  ``// Perform k operations``  ``while``(k != ``0` `&& !minHeap.isEmpty())``  ``{``    ``// Pop the top() element``    ``Integer t = minHeap.poll();``    ` `    ``// Increment Result``    ``if``(t == ``1``)``    ``{``      ``++result;``    ``}` `    ``// Reduce t and k``    ``// Push it again``    ``else``    ``{``      ``--t;``      ``--k;``      ``minHeap.add(t);``    ``}``  ``}` `  ``// Return result``  ``return` `result;``}` `// Driver code``public` `static` `void` `main(String[] args)``{       ``  ``int` `arr[] = {``5``, ``7``, ``5``, ``5``, ``1``, ``2``, ``2``};``  ``int` `n = arr.length;``  ``int` `k = ``3``;` `  ``// Function Call``  ``System.out.println(``"Maximum distinct elements = "` `+ ``                      ``maxDistinctNum(arr, n, k));``}``}` `// This code is contributed by rutvik_56`

## Javascript

 ``

Output

`Maximum distinct elements = 4`

Time Complexity: O(k*logd), where d is the number of distinct elements in the given array.

Another Approach: Follow the below steps, to solve this problem:

• Find the Number of distinct Toys.
• Sum of number of element except one element form every distinct Toys.
• Check sum if greater than or equal K then Return all distinct element.
• Otherwise decrement number of distinct element and to fill K.
• Return Size of vector.

Below is the implementation of the above approach:

## C++

 `// C++ code for the above approach``#include ``using` `namespace` `std;``// function to return maximum number of distinct Toys``int` `MaxNumber(``int` `arr[], ``int` `N, ``int` `K)``{``    ``// Count Number of distinct Number``    ``unordered_map<``int``, ``int``> mp;``    ``for` `(``int` `i = 0; i < N; i++) {``        ``mp[arr[i]]++;``    ``}``    ``// push them into vector``    ``vector<``int``> v1;``    ``for` `(``auto` `i : mp) {``        ``v1.push_back(i.second);``    ``}``    ``// add number of element except one element from every``    ``// distinct element``    ``int` `temp = 0;``    ``for` `(``int` `i = 0; i < v1.size(); i++) {``        ``temp += v1[i] - 1;``    ``}``    ``// check if it is greater than simply return size of``    ``// vector otherwise decrement size of vector to fill k``    ``if` `(K <= temp) {``        ``return` `v1.size();``    ``}``    ``else` `{``        ``K = K - temp;``        ``int` `ans = v1.size();``        ``while` `(K) {``            ``ans--;``            ``K--;``        ``}``        ``return` `ans;``    ``}``}``// Driver Code``int` `main()``{``    ``// array``    ``int` `arr[] = { 10, 10, 10, 50, 50 };``    ``int` `K = 3;``    ``// size of array``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << MaxNumber(arr, N, K) << endl;``    ``return` `0;``}`

## Java

 `// Java code for the above approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG {` `  ``// Function to return maximum number of distinct Toys``  ``static` `int` `MaxNumber(``int``[] arr, ``int` `N, ``int` `K)``  ``{` `    ``// Count Number of distinct Number``    ``HashMap mp = ``new` `HashMap<>();``    ``for` `(``int` `i = ``0``; i < N; i++) {``      ``mp.put(arr[i], mp.getOrDefault(arr[i], ``0``) + ``1``);``    ``}` `    ``// pust them into arraylist``    ``List v1 = ``new` `ArrayList<>();``    ``for` `(Map.Entry i :``         ``mp.entrySet()) {``      ``v1.add(i.getValue());``    ``}` `    ``// add number of element except one element from``    ``// every distinct element``    ``int` `temp = ``0``;``    ``for` `(``int` `i = ``0``; i < v1.size(); i++) {``      ``temp += v1.get(i) - ``1``;``    ``}` `    ``// check if it is greater than simply return size of``    ``// vector otherwise decrement size of vector to fill``    ``// k``    ``if` `(K <= temp) {``      ``return` `v1.size();``    ``}``    ``else` `{``      ``K = K - temp;``      ``int` `ans = v1.size();``      ``while` `(K != ``0``) {``        ``ans--;``        ``K--;``      ``}``      ``return` `ans;``    ``}``  ``}` `  ``public` `static` `void` `main(String[] args)``  ``{``    ``int` `arr[] = { ``10``, ``10``, ``10``, ``50``, ``50` `};``    ``int` `K = ``3``;``    ``int` `N = arr.length;` `    ``System.out.println(MaxNumber(arr, N, K));``  ``}``}` `// This code is contributed by lokeshmvs21.`

## Python3

 `# Python3 code for the above approach` `# function to return maximum number of distinct Toys``def` `MaxNumber(arr, N, K):``  ` `    ``# Count Number of distinct Number``    ``mp ``=` `{}``    ``for` `i ``in` `range``(N):``        ``if` `arr[i] ``not` `in` `mp:``            ``mp[arr[i]] ``=` `0``        ``mp[arr[i]] ``+``=` `1``        ` `        ``# push them into vector``    ``v1 ``=` `[]``    ``for` `i ``in` `mp:``        ``v1.append(mp[i])` `     ``# add number of element except one element from every``    ``# distinct element``    ``temp ``=` `0``    ``for` `i ``in` `range``(``len``(v1)):``        ``temp ``+``=` `v1[i]``-``1``        ` `     ``# check if it is greater than simply return size of``    ``# vector otherwise decrement size of vector to fill k``    ``if` `K <``=` `temp:``        ``return` `len``(v1)``    ``else``:``        ``K ``=` `K``-``temp``        ``ans ``=` `len``(v1)``        ``while` `K:``            ``ans ``-``=` `1``            ``K ``-``=` `1``        ``return` `ans` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``  ` `  ``# Array``    ``arr ``=` `[``10``, ``10``, ``10``, ``50``, ``50``]``    ``K ``=` `3``    ` `    ``# Size of array``    ``N ``=` `len``(arr)``    ``print``(MaxNumber(arr, N, K))` `    ``# This code is contributed by vivekmaddheshiya205`

## C#

 `// C# code for the above approach` `using` `System;``using` `System.Collections;``using` `System.Collections.Generic;` `public` `class` `GFG {` `    ``// Function to return maximum number of distinct Toys``    ``static` `int` `MaxNumber(``int``[] arr, ``int` `N, ``int` `K)``    ``{``        ``// Count Number of distinct Number``        ``Dictionary<``int``, ``int``> mp``            ``= ``new` `Dictionary<``int``, ``int``>();` `        ``for` `(``int` `i = 0; i < N; i++) {``            ``if` `(mp.ContainsKey(arr[i])) {``                ``mp[arr[i]]++;``            ``}``            ``else` `{``                ``mp.Add(arr[i], 1);``            ``}``        ``}` `        ``// put them into arraylist``        ``ArrayList v1 = ``new` `ArrayList();``        ``foreach``(KeyValuePair<``int``, ``int``> i ``in` `mp)``        ``{``            ``v1.Add(i.Value);``        ``}` `        ``// add number of element except one element from``        ``// every distinct element``        ``int` `temp = 0;``        ``for` `(``int` `i = 0; i < v1.Count; i++) {``            ``temp += (``int``)v1[i] - 1;``        ``}` `        ``// check if it is greater than simply return size of``        ``// vector otherwise decrement size of vector to fill``        ``// k``        ``if` `(K <= temp) {``            ``return` `v1.Count;``        ``}``        ``else` `{``            ``K = K - temp;``            ``int` `ans = v1.Count;``            ``while` `(K != 0) {``                ``ans--;``                ``K--;``            ``}``            ``return` `ans;``        ``}``    ``}` `    ``static` `public` `void` `Main()``    ``{` `        ``// Code``        ``int``[] arr = { 10, 10, 10, 50, 50 };``        ``int` `K = 3;``        ``int` `N = arr.Length;` `        ``Console.WriteLine(MaxNumber(arr, N, K));``    ``}``}` `// This code is contributed by lokesh`

Output

`2`

Time Complexity: O(N)

Space Complexity: O(N)

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