# Count the elements having frequency equals to its value | Set 2

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
Explanation :
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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach :
Follow the steps below to solve the problem:

1. Store the frequencies of every array element with value less than equal to array size in a Hashmap.
2. The reason for above step is that the frequency of an element can be atmost equal to the size of the given array. Hence, there is no need to store frequency of an element, whose value is greater than given array size.
3. Count the integers having frequencies equal to its value.
4. Print the final count.

Below is the implementation of the above approach:

## C++

 `#include ` `using` `namespace` `std; ` ` `  `// Function to find the integer ` `// which has a frequency in the ` `// array equal to its value. ` `void` `solve(``int` `arr[], ``int` `n) ` `{ ` `    ``// Store frequency of array ` `    ``// elements ` `    ``map<``int``, ``int``> freq; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Store the frequency ` `        ``// only if arr[i]<=n ` `        ``if` `(arr[i] <= n) ` `            ``freq[arr[i]]++; ` `    ``} ` ` `  `    ``// Initially the count is zero ` `    ``int` `count = 0; ` ` `  `    ``for` `(``int` `i = 1; i <= n; i++) { ` `        ``// If the freq[i] is equal ` `        ``// to i, then increment ` `        ``// the count by 1 ` `        ``if` `(i == freq[i]) { ` `            ``count++; ` `        ``} ` `    ``} ` ` `  `    ``// Print the final count ` `    ``cout << count << ``"\n"``; ` `} ` ` `  `// Driver Program ` `int` `main() ` `{ ` `    ``int` `arr[] = { 3, 1, 1, 3, 2, 2, 3 }; ` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``solve(arr, N); ` `    ``return` `0; ` `} `

Output:

```2
```

Time Complexity: O(N)
Auxillary Space Complexity: O(N)

Refer to the previous post for O(N*log(N)) approach.

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