给定一个大小为N的整数arr[]数组，任务是计算数组中频率等于其值的所有元素。
例子：
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
方法：使用映射存储数组中每个元素的频率，最后计算所有频率等于它们的值的元素。
下面是上述方法的实现：
C++
// C++ program to count the elements
// having frequency equals to its value
#include
using namespace std;
// Function to find the count
int find_maxm(int arr[], int n)
{
// Hash map for counting frequency
map 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 equals 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 frequency
HashMap mp = new HashMap();
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 x : mp.entrySet()){
int value = x.getKey();
int freq = x.getValue();
// Check if value equals 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 frequency
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 equals 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 akhilsainiC#
// 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 frequency
Dictionary mp = new Dictionary();
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 x in mp){
int value = x.Key;
int freq = x.Value;
// Check if value equals 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 Javascript
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 equals 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输出：
2方法#2：使用 collections.Counter()
我们可以使用Python Counter()方法快速解决这个问题。方法很简单。
- 首先使用 Counter 方法创建一个字典，将元素作为键，将它们的频率作为值
- 计算所有频率等于它们的值（键）的元素
下面是上述方法的实现：
蟒蛇3
# 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 equals 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
输出：
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