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
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 valueInput: arr[] = {1, 2, 3, 4, 5, 6}
Output: 1
Approach: Store the frequency of every element of the array using the map, and finally count all of that elements whose frequency is equal to their value.
Below is the implementation of the above approach:
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 |
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 |
2
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