Maximum element in an array which is equal to its frequency
Given an array of integers arr[] of size N, the task is to find the maximum element in the array whose frequecny equals to it’s value
Examples:
Input: arr[] = {3, 2, 2, 3, 4, 3}
Output: 3
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 and 3 is the maximum.Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: 1
Approach: Store the frequency of every element of the array using the map, and finally find out the maximum of those element whose frequency is equal to their value.
Below is the implementation of the above approach:
CPP
// C++ program to find the maximum element // whose frequency equals to it’s value #include <bits/stdc++.h> using namespace std; // Function to find the maximum element // whose frequency equals to it’s value 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 it is the maximum element or not if (value == freq) { ans = max(ans, value); } } return ans; } // Driver code int main() { int arr[] = { 3, 2, 2, 3, 4, 3 }; // Size of array int n = sizeof(arr) / sizeof(arr[0]); // Function call cout << find_maxm(arr, n); return 0; } |
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Java
// Java program to find the maximum element // whose frequency equals to it’s value import java.util.*; class GFG{ // Function to find the maximum element // whose frequency equals to it’s value 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 it is the maximum element or not if (value == freq) { ans = Math.max(ans, value); } } return ans; } // Driver code public static void main(String[] args) { int arr[] = { 3, 2, 2, 3, 4, 3 }; // Size of array int n = arr.length; // Function call System.out.print(find_maxm(arr, n)); } } // This code is contributed by 29AjayKumar |
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Python3
# Python3 program to find the maximum element # whose frequency equals to it’s value # Function to find the maximum element # whose frequency equals to it’s value 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.update( {arr[i] : mpp[arr[i]] + 1} ) else: mpp[arr[i]] = 1 ans = 0 for value,freq in mpp.items(): # Check if value equls to frequency # and it is the maximum element or not if (value == freq): ans = max(ans, value) return ans # Driver code arr = [ 3, 2, 2, 3, 4, 3 ] # Size of array n = len(arr) # Function call print(find_maxm(arr, n)) # This code is contributed by Sanjit_Prasad |
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C#
// C# program to find the maximum element // whose frequency equals to it’s value using System; using System.Collections.Generic; class GFG{ // Function to find the maximum element // whose frequency equals to it’s value 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 it is the maximum element or not if (value == freq) { ans = Math.Max(ans, value); } } return ans; } // Driver code public static void Main(String[] args) { int []arr = { 3, 2, 2, 3, 4, 3 }; // Size of array int n = arr.Length; // Function call Console.Write(find_maxm(arr, n)); } } // This code is contributed by Rajput-Ji |
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Output:
3
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