Minimum removals required to make frequency of each array element equal to its value

Given an array arr[] of size N, the task is to find the minimum count of array elements required to be removed such that frequency of each array element is equal to its value

Examples:

Input: arr[] = { 2, 4, 1, 4, 2 }
Output:
Explanation:
Removing arr[1] from the array modifies arr[] to { 2, 1, 4, 2 }
Removing arr[2] from the array modifies arr[] to { 2, 1, 2 }
Distinct elements in the array are: { 1, 2 } with frequencies 1 and 2 respectively.
Therefore, the required output is 2.

Input: arr[] = { 2, 7, 1, 8, 2, 8, 1, 8 }
Output: 5

Approach: The problem can be solved using Greedy technique. Follow the steps below to solve the problem:

• Initialize a map say, mp to store the frequency of each distinct element of the array.
• Traverse the array and store the frequency of each distinct element of the array.
• Initialize a variable say, cntMinRem to store the minimum count of array elements required to be removed such that the frequency of arr[i] is equal to arr[i].
• Traverse the map using key value of the map as i and check the following conditions:
• If mp[i] < i, then update the value of cntMinRem += mp[i].
• If mp[i] > i, then update the value of cntMinRem += (mp[i] – i)

• Finally, print the value of cntMinRem.

Below is the implementation of the above approach:

C++14

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;` `// Function to find the minimum count of``// elements required to be removed such``// that frequency of arr[i] equal to arr[i]``int` `min_elements(``int` `arr[], ``int` `N)``{``    ``// Stores frequency of each``    ``// element of the array``    ``unordered_map<``int``, ``int``> mp;` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// Update frequency``        ``// of arr[i]``        ``mp[arr[i]]++;``    ``}` `    ``// Stores minimum count of removals``    ``int` `cntMinRem = 0;` `    ``// Traverse the map``    ``for` `(``auto` `it : mp) {` `        ``// Stores key value``        ``// of the map``        ``int` `i = it.first;` `        ``// If frequency of i is``        ``// less than i``        ``if` `(mp[i] < i) {` `            ``// Update cntMinRem``            ``cntMinRem += mp[i];``        ``}` `        ``// If frequency of i is``        ``// greater than i``        ``else` `if` `(mp[i] > i) {` `            ``// Update cntMinRem``            ``cntMinRem += (mp[i] - i);``        ``}``    ``}` `    ``return` `cntMinRem;``}` `// Driver Code``int` `main()``{` `    ``int` `arr[] = { 2, 4, 1, 4, 2 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << min_elements(arr, N);``    ``return` `0;``}`

Java

 `// Java program to implement``// the above approach``import` `java.util.*;` `class` `GFG{``    ` `// Function to find the minimum count of``// elements required to be removed such``// that frequency of arr[i] equal to arr[i]``public` `static` `int` `min_elements(``int` `arr[], ``int` `N)``{``    ` `    ``// Stores frequency of each``    ``// element of the array``    ``Map mp = ``new` `HashMap();` `    ``// Traverse the array``    ``for``(``int` `i = ``0``; i < N; i++) ``    ``{``        ` `        ``// Update frequency``        ``// of arr[i]``        ``mp.put(arr[i],``               ``mp.getOrDefault(arr[i], ``0``) + ``1``);``    ``}` `    ``// Stores minimum count of removals``    ``int` `cntMinRem = ``0``;` `    ``// Traverse the map``    ``for``(``int` `key : mp.keySet()) ``    ``{``        ` `        ``// Stores key value``        ``// of the map``        ``int` `i = key;``        ``int` `val = mp.get(i);` `        ``// If frequency of i is``        ``// less than i``        ``if` `(val < i)``        ``{``            ` `            ``// Update cntMinRem``            ``cntMinRem += val;``        ``}` `        ``// If frequency of i is``        ``// greater than i``        ``else` `if` `(val > i) ``        ``{``            ` `            ``// Update cntMinRem``            ``cntMinRem += (val - i);``        ``}``    ``}``    ``return` `cntMinRem;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``2``, ``4``, ``1``, ``4``, ``2` `};` `    ``System.out.println(min_elements(``        ``arr, arr.length));``}``}` `// This code is contributed by grand_master`

Python3

 `# Python3 program to implement``# the above approach` `# Function to find the minimum count of``# elements required to be removed such``# that frequency of arr[i] equal to arr[i]``def` `min_elements(arr, N) :``    ` `    ``# Stores frequency of each``    ``# element of the array``    ``mp ``=` `{};` `    ``# Traverse the array``    ``for` `i ``in` `range``(N) :` `        ``# Update frequency``        ``# of arr[i]``        ``if` `arr[i] ``in` `mp :``            ``mp[arr[i]] ``+``=` `1``;``        ``else` `:``            ``mp[arr[i]] ``=` `1``;` `    ``# Stores minimum count of removals``    ``cntMinRem ``=` `0``;` `    ``# Traverse the map``    ``for` `it ``in` `mp :` `        ``# Stores key value``        ``# of the map``        ``i ``=` `it;` `        ``# If frequency of i is``        ``# less than i``        ``if` `(mp[i] < i) :` `            ``# Update cntMinRem``            ``cntMinRem ``+``=` `mp[i];` `        ``# If frequency of i is``        ``# greater than i``        ``elif` `(mp[i] > i) :` `            ``# Update cntMinRem``            ``cntMinRem ``+``=` `(mp[i] ``-` `i);``            ` `    ``return` `cntMinRem;` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr ``=` `[ ``2``, ``4``, ``1``, ``4``, ``2` `];``    ``N ``=` `len``(arr);``    ``print``(min_elements(arr, N));``    ` `    ``# This code is contributed by AnkThon`

C#

 `// C# program to implement``// the above approach``using` `System;``using` `System.Collections.Generic; ``class` `GFG ``{``    ` `    ``// Function to find the minimum count of``    ``// elements required to be removed such``    ``// that frequency of arr[i] equal to arr[i]``    ``static` `int` `min_elements(``int``[] arr, ``int` `N)``    ``{``      ` `        ``// Stores frequency of each``        ``// element of the array``        ``Dictionary<``int``, ``int``> mp = ``new` `Dictionary<``int``, ``int``>();  ``     ` `        ``// Traverse the array``        ``for` `(``int` `i = 0; i < N; i++)``        ``{``     ` `            ``// Update frequency``            ``// of arr[i]``            ``if``(mp.ContainsKey(arr[i]))``            ``{``                ``mp[arr[i]]++;``            ``}``            ``else``            ``{``                ``mp[arr[i]] = 1;``            ``}``        ``}``     ` `        ``// Stores minimum count of removals``        ``int` `cntMinRem = 0;``     ` `        ``// Traverse the map``        ``foreach` `(KeyValuePair<``int``, ``int``> it ``in` `mp) ``        ``{``     ` `            ``// Stores key value``            ``// of the map``            ``int` `i = it.Key;``     ` `            ``// If frequency of i is``            ``// less than i``            ``if` `(mp[i] < i) ``            ``{``     ` `                ``// Update cntMinRem``                ``cntMinRem += mp[i];``            ``}``     ` `            ``// If frequency of i is``            ``// greater than i``            ``else` `if` `(mp[i] > i) ``            ``{``     ` `                ``// Update cntMinRem``                ``cntMinRem += (mp[i] - i);``            ``}``        ``}     ``        ``return` `cntMinRem;``    ``}` `  ``// Driver code``  ``static` `void` `Main()``  ``{``    ``int``[] arr = { 2, 4, 1, 4, 2 }; ``    ``int` `N = arr.Length;``    ``Console.Write(min_elements(arr, N));``  ``}``}` `// This code is contributed by divyesh072019`

Javascript

 ``

Output:
`2`

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

Approach 2: No Extra Space:

The above approach uses a unordered_map and has auxiliary space O(N).

• To optimize the given code in O(1) space, we can make use of the fact that the input array elements are positive integers. We can make use of the input array itself to store the frequency of each element.
• We can iterate over the array and for each element arr[i], we can increment the frequency of element (arr[i] % N) by N. Here, N is the size of the input array. This will not change the value of the element and will help us keep track of the frequency of the element in the array.
• After we have updated the frequency of each element, we can iterate over the array again and count the number of elements for which the frequency is less than or greater than the element value. We can return the sum of these counts as the minimum count of elements required to be removed.

Here is the optimized code:

C++

 `#include ``using` `namespace` `std;` `int` `min_elements(``int` `arr[], ``int` `N) {``    ``// update the frequency of each element in the array``    ``for` `(``int` `i = 0; i < N; i++) {``        ``arr[arr[i] % N] += N;``    ``}` `    ``int` `cntMinRem = 0;``    ``// count the number of elements for which the frequency is less than or greater than the element value``    ``for` `(``int` `i = 0; i < N; i++) {``        ``if` `(arr[i] / N < i) {``            ``cntMinRem += arr[i] / N;``        ``}``        ``if` `(arr[i] / N > i + 1) {``            ``cntMinRem += (arr[i] / N - i - 1);``        ``}``    ``}` `    ``return` `cntMinRem;``}` `int` `main() {``    ``int` `arr[] = { 2, 4, 1, 4, 2 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << min_elements(arr, N);``    ``return` `0;``}`

Java

 `/*package whatever //do not write package name here */``import` `java.util.*;` `public` `class` `Main {``    ``public` `static` `int` `minElements(``int``[] arr, ``int` `N) {``        ``// update the frequency of each element in the array``        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``arr[arr[i] % N] += N;``        ``}` `        ``int` `cntMinRem = ``0``;``        ``// count the number of elements for which the frequency ``       ``// is less than or greater than the element value``        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``if` `(arr[i] / N < i) {``                ``cntMinRem += arr[i] / N;``            ``}``            ``if` `(arr[i] / N > i + ``1``) {``                ``cntMinRem += (arr[i] / N - i - ``1``);``            ``}``        ``}` `        ``return` `cntMinRem;``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``int``[] arr = { ``2``, ``4``, ``1``, ``4``, ``2` `};``        ``int` `N = arr.length;``        ``System.out.println(minElements(arr, N));``    ``}``}`

Python3

 `def` `min_elements(arr, N):``    ``# update the frequency of each element in the array``    ``for` `i ``in` `range``(N):``        ``arr[arr[i] ``%` `N] ``+``=` `N` `    ``cntMinRem ``=` `0``    ``# count the number of elements for which the frequency is less than or greater than the element value``    ``for` `i ``in` `range``(N):``        ``if` `arr[i] ``/``/` `N < i:``            ``cntMinRem ``+``=` `arr[i] ``/``/` `N``        ``if` `arr[i] ``/``/` `N > i ``+` `1``:``            ``cntMinRem ``+``=` `arr[i] ``/``/` `N ``-` `i ``-` `1` `    ``return` `cntMinRem` `arr ``=` `[``2``, ``4``, ``1``, ``4``, ``2``]``N ``=` `len``(arr)``print``(min_elements(arr, N))`

C#

 `using` `System;` `public` `class` `Program {``    ``public` `static` `int` `MinElements(``int``[] arr, ``int` `N)``    ``{``        ``// update the frequency of each element in the array``        ``for` `(``int` `i = 0; i < N; i++) {``            ``arr[arr[i] % N] += N;``        ``}``        ``int` `cntMinRem = 0;``        ``// count the number of elements for which the``        ``// frequency is less than or greater than the``        ``// element value``        ``for` `(``int` `i = 0; i < N; i++) {``            ``if` `(arr[i] / N < i) {``                ``cntMinRem += arr[i] / N;``            ``}``            ``if` `(arr[i] / N > i + 1) {``                ``cntMinRem += (arr[i] / N - i - 1);``            ``}``        ``}` `        ``return` `cntMinRem;``    ``}` `    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr = { 2, 4, 1, 4, 2 };``        ``int` `N = arr.Length;``        ``Console.WriteLine(MinElements(arr, N));``    ``}``}`

Javascript

 `function` `min_elements(arr, N) {``    ``// update the frequency of each element in the array``    ``for` `(let i = 0; i < N; i++) {``        ``arr[arr[i] % N] += N;``    ``}` `    ``let cntMinRem = 0;``    ``// count the number of elements for which the ``    ``// frequency is less than or greater than the element value``    ``for` `(let i = 0; i < N; i++) {``        ``if` `(arr[i] / N < i) {``            ``cntMinRem += Math.floor(arr[i] / N);``        ``}``        ``if` `(arr[i] / N > i + 1) {``            ``cntMinRem += Math.floor(arr[i] / N) - i - 1;``        ``}``    ``}` `    ``return` `cntMinRem;``}` `let arr = [2, 4, 1, 4, 2];``let N = arr.length;``console.log(min_elements(arr, N));`

OUTPUT:

`2`

Time Complexity: O(N)
Auxiliary Space: O(1)

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