计算频率等于其值的元素|套装2

给定大小为N的整数arr []数组，任务是对频率等于其值的频率的所有元素进行计数。
例子：

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 

方法：请按照以下步骤解决问题：

1. 将值小于等于给定数组大小的每个数组元素的频率存储在freq []数组中。
2. 进行上述步骤的原因是，一个元素的频率最多可以等于给定数组的大小。因此，不需要存储其值大于给定数组大小的元素的频率。
3. 计算频率等于其值的整数。
4. 打印最终计数。

下面是上述方法的实现：

C++

// C++ program of the
// above approach
 
#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
    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

// 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

# 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#

// 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

输出

2

时间复杂度： O(N)
辅助空间： O(N)

有关O(N * log(N))方法，请参阅上一篇文章。