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 

Approach:  Follow the steps below to solve the problem: 

1. Store the frequencies of every array element with the value less than equal to the given array size in freq[] array.
2. The reason for the above step is that the frequency of an element can be at most equal to the size of the given array. Hence, there is no need to store the frequency of an element, whose value is greater than the 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++ program of the 
// above approach

#include <bits/stdc++.h>
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
    int freq[n+1] ={0};

    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 Code
int main()
{
    int arr[] = { 3, 1, 1, 3, 2, 2, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);

    solve(arr, N);
    return 0;
}

// Java program of the 
// above approach
class GFG{

// Function to find the integer
// which has a frequency in the
// array equal to its value.
static void solve(int arr[], 
                  int n)
{
  // Store frequency of array
  // elements
  int []freq = new int[n + 1];

  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
  System.out.print(count + "\n");
}

// Driver Code
public static void main(String[] args)
{
  int arr[] = {3, 1, 1, 3, 2, 2, 3};
  int N = arr.length;
  solve(arr, N);
}
}

// This code is contributed by shikhasingrajput

# Python3 program of the
# above approach

# Function to find the integer
# which has a frequency in the
# array equal to its value.
def solve(arr, n):
  
    # Store frequency of array
    # elements
    freq = [0] * (n + 1);

    for i in range(n):
      
        # Store the frequency
        # only if arr[i]<=n
        if (arr[i] <= n):
            freq[arr[i]] += 1;

    # Initially the count is zero
    count = 0;

    for i in range(1, n + 1):
      
        # If the freq[i] is equal
        # to i, then increment
        # the count by 1
        if (i == freq[i]):
            count += 1;

    # Print final count
    print(count , "");


# Driver Code
if __name__ == '__main__':
  
    arr = [3, 1, 1, 3, 
           2, 2, 3];
    N = len(arr);
    solve(arr, N);

# This code is contributed by shikhasingrajput

// C# program of the 
// above approach
using System;
class GFG{

// Function to find the integer
// which has a frequency in the
// array equal to its value.
static void solve(int []arr, 
                  int n)
{
  // Store frequency of array
  // elements
  int []freq = new int[n + 1];

  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 readonly count
  Console.Write(count + "\n");
}

// Driver Code
public static void Main(String[] args)
{
  int []arr = {3, 1, 1, 3, 2, 2, 3};
  int N = arr.Length;
  solve(arr, N);
}
}

// This code is contributed by shikhasingrajput 

<script>

// Javascript program for
// the above approach

// Function to find the integer
// which has a frequency in the
// array equal to its value.
function solve(arr, n)
{

  // Store frequency of array
  // elements
  let freq = Array.from({length: n+1}, (_, i) => 0);
 
  for (let 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
  let count = 0;
 
  for (let 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
  document.write(count + "<br/>");
}

// Driver code
    
  let arr = [ 3, 1, 1, 3, 2, 2, 3 ];
  let N = arr.length;
  solve(arr, N);
 
 // This code is contributed by souravghosh0416.
</script>

2

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

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