Related Articles

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

• Difficulty Level : Medium
• Last Updated : 13 May, 2021

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:
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:

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 ``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);` `    ``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 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 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

 ``
Output:
`3`

Time Complexity: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up