Given an array of integers **arr[]** of size **N**, the task is to count all the elements of the array which have a frequency equals to its value.

**Examples:**

Input:arr[] = {3, 2, 2, 3, 4, 3}Output:2

Frequency of element 2 is 2

Frequency of element 3 is 3

Frequency of element 4 is 1

2 and 3 are elements which have same frequency as it’s value

Input:arr[] = {1, 2, 3, 4, 5, 6}Output:1

**Approach:** Store the frequency of every element of the array using the map, and finally count all of that elements whose frequency is equal to their value.

Below is the implementation of the above approach:

## C++

`// C++ program to count the elements` `// having frequency equals to its value` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the count` `int` `find_maxm(` `int` `arr[], ` `int` `n)` `{` ` ` `// Hash map for counting frquency` ` ` `map<` `int` `, ` `int` `> mpp;` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `// Counting freq of each element` ` ` `mpp[arr[i]] += 1;` ` ` `}` ` ` `int` `ans = 0;` ` ` `for` `(` `auto` `x : mpp) {` ` ` `int` `value = x.first;` ` ` `int` `freq = x.second;` ` ` `// Check if value equls to frequency` ` ` `// and increment the count` ` ` `if` `(value == freq) {` ` ` `ans++;` ` ` `}` ` ` `}` ` ` `return` `ans;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 3, 2, 2, 3, 4, 3 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `// Function call` ` ` `cout << find_maxm(arr, n);` ` ` `return` `0;` `}` |

## Java

`// Java program to count the elements` `// having frequency equals to its value` `import` `java.util.*;` `class` `GFG{` ` ` `// Function to find the count` `static` `int` `find_maxm(` `int` `arr[], ` `int` `n)` `{` ` ` `// Hash map for counting frquency` ` ` `HashMap<Integer,Integer> mp = ` `new` `HashMap<Integer,Integer>();` ` ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) {` ` ` ` ` `// Counting freq of each element` ` ` `if` `(mp.containsKey(arr[i])){` ` ` `mp.put(arr[i], mp.get(arr[i])+` `1` `);` ` ` `}` `else` `{` ` ` `mp.put(arr[i], ` `1` `);` ` ` `}` ` ` `}` ` ` ` ` `int` `ans = ` `0` `;` ` ` `for` `(Map.Entry<Integer,Integer> x : mp.entrySet()){` ` ` `int` `value = x.getKey();` ` ` `int` `freq = x.getValue();` ` ` ` ` `// Check if value equls to frequency` ` ` `// and increment the count` ` ` `if` `(value == freq) {` ` ` `ans++;` ` ` `}` ` ` `}` ` ` ` ` `return` `ans;` `}` ` ` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `arr[] = { ` `3` `, ` `2` `, ` `2` `, ` `3` `, ` `4` `, ` `3` `};` ` ` `int` `n = arr.length;` ` ` ` ` `// Function call` ` ` `System.out.print(find_maxm(arr, n));` `}` `}` `// This code is contributed by Princi Singh` |

## Python3

`# Python3 program to count the elements` `# having frequency equals to its value` `# Function to find the count` `def` `find_maxm(arr, n):` ` ` ` ` `# Hash map for counting frquency` ` ` `mpp ` `=` `{}` ` ` ` ` `for` `i ` `in` `range` `(` `0` `, n):` ` ` ` ` `# Counting freq of each element` ` ` `if` `arr[i] ` `in` `mpp:` ` ` `mpp[arr[i]] ` `=` `mpp[arr[i]] ` `+` `1` ` ` `else` `:` ` ` `mpp[arr[i]] ` `=` `1` ` ` ` ` `ans ` `=` `0` ` ` ` ` `for` `key ` `in` `mpp:` ` ` `value ` `=` `key` ` ` `freq ` `=` `mpp[key]` ` ` ` ` `# Check if value equls to frequency` ` ` `# and increment the count` ` ` `if` `value ` `=` `=` `freq:` ` ` `ans ` `=` `ans ` `+` `1` ` ` ` ` `return` `ans` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` ` ` `arr ` `=` `[ ` `3` `, ` `2` `, ` `2` `, ` `3` `, ` `4` `, ` `3` `]` ` ` `n ` `=` `len` `(arr)` ` ` ` ` `# Function call` ` ` `print` `(find_maxm(arr, n))` ` ` `# This code is contributed by akhilsaini` |

## C#

`// C# program to count the elements` `// having frequency equals to its value` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{` ` ` `// Function to find the count` `static` `int` `find_maxm(` `int` `[]arr, ` `int` `n)` `{` ` ` `// Hash map for counting frquency` ` ` `Dictionary<` `int` `,` `int` `> mp = ` `new` `Dictionary<` `int` `,` `int` `>();` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` ` ` `// Counting freq of each element` ` ` `if` `(mp.ContainsKey(arr[i])){` ` ` `mp[arr[i]] = mp[arr[i]] + 1;` ` ` `}` `else` `{` ` ` `mp.Add(arr[i], 1);` ` ` `}` ` ` `}` ` ` ` ` `int` `ans = 0;` ` ` `foreach` `(KeyValuePair<` `int` `,` `int` `> x ` `in` `mp){` ` ` `int` `value = x.Key;` ` ` `int` `freq = x.Value;` ` ` ` ` `// Check if value equls to frequency` ` ` `// and increment the count` ` ` `if` `(value == freq) {` ` ` `ans++;` ` ` `}` ` ` `}` ` ` ` ` `return` `ans;` `}` ` ` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `[]arr = { 3, 2, 2, 3, 4, 3 };` ` ` `int` `n = arr.Length;` ` ` ` ` `// Function call` ` ` `Console.Write(find_maxm(arr, n));` `}` `}` `// This code is contributed by PrinciRaj1992` |

**Output:**

2

**Method #2:Using collections.Counter()**

We can solve this problem quickly using python **Counter()** method. Approach is very simple.

- First create a dictionary using Counter method having elements as keys and their frequencies as values
- count all of that elements whose frequency is equal to their value(key)

Below is the implementation of above approach:

## Python3

`# Python3 program to count the elements` `# having frequency equals to its value` `# importing counter from collections` `from` `collections ` `import` `Counter` `# Function to find the count` `def` `findElements(arr, n):` ` ` ` ` `# Now create dictionary using counter method` ` ` `# which will have elements as key and their` ` ` `# frequencies as values` ` ` `Element_Counter ` `=` `Counter(arr)` ` ` `ans ` `=` `0` ` ` `for` `key ` `in` `Element_Counter:` ` ` `value ` `=` `key` ` ` `freq ` `=` `Element_Counter[key]` ` ` `# Check if value equls to frequency` ` ` `# and increment the count` ` ` `if` `value ` `=` `=` `freq:` ` ` `ans ` `=` `ans ` `+` `1` ` ` `return` `ans` `# Driver code` `arr ` `=` `[` `3` `, ` `2` `, ` `2` `, ` `3` `, ` `4` `, ` `3` `]` `n ` `=` `len` `(arr)` `# Function call` `print` `(findElements(arr, n))` `# This code is contributed by vikkycirus` |

#### Output:

2

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