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: 2 
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 <bits/stdc++.h>
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<Integer,
        Integer> mp = new HashMap<Integer,
                                  Integer>();

    // 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

<script>

// Javascript 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]
function min_elements(arr, N)
{
    // Stores frequency of each
    // element of the array
    var mp = new Map();

    // Traverse the array
    for (var i = 0; i < N; i++) {

        // Update frequency
        // of arr[i]
        if(mp.has(arr[i]))
        {
            mp.set(arr[i], mp.get(arr[i])+1);
        }
        else
        {
            mp.set(arr[i], 1);
        }
    }

    // Stores minimum count of removals
    var cntMinRem = 0;

    // Traverse the map
    mp.forEach((value, key) => {

        // Stores key value
        // of the map
        var i = key;

        // If frequency of i is
        // less than i
        if (mp.get(i) < i) {

            // Update cntMinRem
            cntMinRem += mp.get(i);
        }

        // If frequency of i is
        // greater than i
        else if (mp.get(i) > i) {

            // Update cntMinRem
            cntMinRem += (mp.get(i) - i);
        }
    });

    return cntMinRem;
}

// Driver Code
var arr = [2, 4, 1, 4, 2];
var N = arr.length;
document.write( min_elements(arr, N));

</script>

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 <iostream>
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)