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

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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Output:

3

Time Complexity: O(N)



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Improved By : AnkitRai01, Rajput-Ji