计算频率等于其值的元素数量
原文：https://www.geeksforgeeks.org/count-the-elements-having-frequency-equals-to-its-value/
给定一个大小为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。
示例：
输入：
arr[] = {3, 2, 2, 3, 4, 3}输出：2
元素 2 的频率为 2，
元素 3 的频率为 3，
元素 4 的频率为 1，
2 和 3 是具有与其值相同频率的元素。
输入：
arr[] = {1, 2, 3, 4, 5, 6}输出：1
方法：使用映射存储数组的每个元素的频率，最后计数频率等于其值的所有那些元素。
下面是上述方法的实现：
C++
// C++ program to count the elements
// having frequency equals to its value
#include <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
输出：
2
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