Minimum deletions required such that any number X will occur exactly X times

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Given an array arr[] of N integers, the task is to find the minimum deletions required such that frequency of arr[i] is exactly arr[i] in the array for all possible values of i.
Examples:

Input: arr[] = {1, 2, 2, 3, 3}
Output: 2
Frequency(1) = 1
Frequency(2) = 2
Frequency(3) = 2, frequency can’t be increased
So, delete every occurrence of 3.
Input: arr[] = {2, 3, 2, 3, 4, 4, 4, 4, 5}
Output: 3

Approach: There are two cases:

If frequency of X is greater than or equal o X then we delete extra frequencies of X to get exactly X elements of value X.
If frequency of X is less than X then we delete all of the occurrences of X as it is impossible to get extra element of value X.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the minimum
// deletions required
int MinDeletion(int a[], int n)
{
 
    // To store the frequency of
    // the array elements
    unordered_map<int, int> map;
 
    // Store frequency of each element
    for (int i = 0; i < n; i++)
        map[a[i]]++;
 
    // To store the minimum deletions required
    int ans = 0;
 
    for (auto i : map) {
 
        // Value
        int x = i.first;
 
        // It's frequency
        int frequency = i.second;
 
        // If number less than or equal
        // to it's frequency
        if (x <= frequency) {
 
            // Delete extra occurrences
            ans += (frequency - x);
        }
 
        // Delete every occurrence of x
        else
            ans += frequency;
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int a[] = { 2, 3, 2, 3, 4, 4, 4, 4, 5 };
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << MinDeletion(a, n);
 
    return 0;
}

Java

// Java Implementation of above approach 
import java.util.*;
 
class GFG 
{
     
// Function to return the minimum
// deletions required
static int MinDeletion(int a[], int n)
{
 
    // To store the frequency of
    // the array elements
    Map<Integer,Integer> mp = new HashMap<>();
 
    // Store frequency of each element
    for (int i = 0 ; i < n; i++)
    {
        if(mp.containsKey(a[i]))
        {
            mp.put(a[i], mp.get(a[i])+1);
        }
        else
        {
            mp.put(a[i], 1);
        }
    }
    // To store the minimum deletions required
    int ans = 0;
 
    for (Map.Entry<Integer,Integer> i : mp.entrySet()) 
    {
 
        // Value
        int x = i.getKey();
 
        // It's frequency
        int frequency = i.getValue();
 
        // If number less than or equal
        // to it's frequency
        if (x <= frequency) 
        {
 
            // Delete extra occurrences
            ans += (frequency - x);
        }
 
        // Delete every occurrence of x
        else
            ans += frequency;
    }
 
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int a[] = { 2, 3, 2, 3, 4, 4, 4, 4, 5 };
    int n = a.length;
 
    System.out.println(MinDeletion(a, n));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach 
 
# Function to return the minimum 
# deletions required 
def MinDeletion(a, n) :
 
    # To store the frequency of 
    # the array elements 
    map = dict.fromkeys(a, 0); 
 
    # Store frequency of each element 
    for i in range(n) : 
        map[a[i]] += 1; 
 
    # To store the minimum deletions required 
    ans = 0; 
 
    for key,value in map.items() :
 
        # Value 
        x = key; 
 
        # It's frequency 
        frequency = value; 
 
        # If number less than or equal 
        # to it's frequency 
        if (x <= frequency) :
 
            # Delete extra occurrences 
            ans += (frequency - x); 
 
        # Delete every occurrence of x 
        else :
            ans += frequency; 
 
    return ans; 
 
 
# Driver code 
if __name__ == "__main__" : 
 
    a = [ 2, 3, 2, 3, 4, 4, 4, 4, 5 ];
    n = len(a); 
 
    print(MinDeletion(a, n)); 
 
# This code is contributed by AnkitRai01

C#

// C# Implementation of above approach
using System;
using System.Collections.Generic; 
 
class GFG 
{
     
// Function to return the minimum
// deletions required
static int MinDeletion(int []a, int n)
{
 
    // To store the frequency of
    // the array elements
    Dictionary<int,
               int> mp = new Dictionary<int,
                                        int>();
 
    // Store frequency of each element
    for (int i = 0 ; i < n; i++)
    {
        if(mp.ContainsKey(a[i]))
        {
            var val = mp[a[i]];
            mp.Remove(a[i]);
            mp.Add(a[i], val + 1); 
        }
        else
        {
            mp.Add(a[i], 1);
        }
    }
     
    // To store the minimum deletions required
    int ans = 0;
 
    foreach(KeyValuePair<int, int> i in mp)
    {
 
        // Value
        int x = i.Key;
 
        // It's frequency
        int frequency = i.Value;
 
        // If number less than or equal
        // to it's frequency
        if (x <= frequency) 
        {
 
            // Delete extra occurrences
            ans += (frequency - x);
        }
 
        // Delete every occurrence of x
        else
            ans += frequency;
    }
 
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    int []a = { 2, 3, 2, 3, 4, 4, 4, 4, 5 };
    int n = a.Length;
 
    Console.WriteLine(MinDeletion(a, n));
}
}
 
// This code is contributed by PrinciRaj1992 

Javascript

<script>
// javaScript implementation of the approach
 
// Function to return the minimum
// deletions required
function MinDeletion( a, n){
    // To store the frequency of
    // the array elements
    let map = new Map();
     
    // Store frequency of each element
    for (let i = 0; i < n; i++){
        if(map[a[i]])
            map[a[i]]++;
        else
            map[a[i]] = 1
     }
 
    // To store the minimum deletions required
    let ans = 0;
    for(var m in map){
         
        // Value
        let x = m;
 
        // It's frequency
        let frequency = map[m];
 
        // If number less than or equal
        // to it's frequency
        if (x <= frequency) {
 
            // Delete extra occurrences
            ans += (frequency - x);
        }
 
        // Delete every occurrence of x
        else
            ans += frequency;
    };
 
    return ans;
}
 
// Driver code
let a = [ 2, 3, 2, 3, 4, 4, 4, 4, 5 ];
let n = a.length;
document.write( MinDeletion(a, n));
 
// This code is contributed by rohitsingh07052.
</script>

Output:

Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(n), where n is the size of the given array.

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