Mode of frequencies of given array elements

Given an array arr[], the task is to find the mode of frequencies of elements of the given array.

Examples:

Input: arr[] = {6, 10, 3, 10, 8, 3, 6, 4}, N = 8
Output: 2
Explanation:
Here (3, 10 and 6) have frequency 2, while (4 and 8) have frequency 1.
Three numbers have frequency 2, while 2 numbers have frequency 1.
Thus, the mode of the frequencies is 2.

Input: arr[] = {5, 9, 2, 9, 7, 2, 5, 3, 1}, N = 9
Output:1

Approach: The idea is to find the frequency of the frequencies of all array elements. Finally, compute the mode of the frequencies.

Below is the implementation of the above approach:

C++

// C++ program of the 
// above approach 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find the mode of the 
// frequency of the array 
int countFreq(int arr[], int n) 
{ 
  
    // Stores the frequencies 
    // of array elements 
    unordered_map<int, int> mp1; 
  
    // Traverse through array 
    // elements and 
    // count frequencies 
    for (int i = 0; i < n; ++i) { 
        mp1[arr[i]]++; 
    } 
  
    // Stores the frequencies's of 
    // frequencies of array elements 
    unordered_map<int, int> mp2; 
  
    for (auto it : mp1) { 
        mp2[it.second]++; 
    } 
  
    // Stores teh minimum value 
    int M = INT_MIN; 
  
    // Find the Mode in 2nd map 
    for (auto it : mp2) { 
        M = max(M, it.second); 
    } 
  
    // search for this Mode 
    for (auto it : mp2) { 
        // When mode is find then return 
        // to main function. 
        if (M == it.second) { 
            return it.first; 
        } 
    } 
  
    // If mode is not found 
    return 0; 
} 
  
// Driver Code 
int main() 
{ 
    int arr[] = { 6, 10, 3, 10, 8, 3, 6, 4 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    cout << countFreq(arr, n); 
    return 0; 
}

Java

// Java program of the 
// above approach 
import java.util.*; 
  
class GFG{ 
  
// Function to find the mode of the 
// frequency of the array 
static int countFreq(int arr[], int n) 
{ 
      
    // Stores the frequencies 
    // of array elements 
    HashMap<Integer, 
            Integer> mp1 = new HashMap<Integer, 
                                       Integer>(); 
  
    // Traverse through array 
    // elements and 
    // count frequencies 
    for(int i = 0; i < n; ++i) 
    { 
        if (mp1.containsKey(arr[i])) 
        { 
            mp1.put(arr[i], mp1.get(arr[i]) + 1); 
        } 
        else
        { 
            mp1.put(arr[i], 1); 
        } 
    } 
  
    // Stores the frequencies's of 
    // frequencies of array elements 
    HashMap<Integer, 
            Integer> mp2 = new HashMap<Integer, 
                                       Integer>(); 
  
    for(Map.Entry<Integer, 
                  Integer> it : mp1.entrySet()) 
    { 
        if (mp2.containsKey(it.getValue())) 
        { 
            mp2.put(it.getValue(), 
            mp2.get(it.getValue()) + 1); 
        } 
        else
        { 
            mp2.put(it.getValue(), 1); 
        } 
    } 
  
    // Stores teh minimum value 
    int M = Integer.MIN_VALUE; 
  
    // Find the Mode in 2nd map 
    for(Map.Entry<Integer, 
                  Integer> it : mp2.entrySet()) 
    { 
        M = Math.max(M, it.getValue()); 
    } 
  
    // Search for this Mode 
    for(Map.Entry<Integer, 
                  Integer> it : mp2.entrySet()) 
    { 
          
        // When mode is find then return 
        // to main function. 
        if (M == it.getValue())  
        { 
            return it.getKey(); 
        } 
    } 
  
    // If mode is not found 
    return 0; 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    int arr[] = { 6, 10, 3, 10, 8, 3, 6, 4 }; 
    int n = arr.length; 
      
    System.out.print(countFreq(arr, n)); 
} 
} 
  
// This code is contributed by Amit Katiyar

Python3

# Python3 program of the 
# above approach 
from collections import defaultdict 
import sys 
  
# Function to find the mode of the 
# frequency of the array 
def countFreq(arr, n): 
  
  # Stores the frequencies 
  # of array elements 
  mp1 = defaultdict (int) 
  
  # Traverse through array 
  # elements and 
  # count frequencies 
  for i in range (n): 
    mp1[arr[i]] += 1
  
    # Stores the frequencies's of 
    # frequencies of array elements 
    mp2 = defaultdict (int) 
  
    for it in mp1: 
      mp2[mp1[it]] += 1
  
      # Stores teh minimum value 
      M = -sys.maxsize - 1
  
      # Find the Mode in 2nd map 
      for it in mp2: 
        M = max(M, mp2[it]) 
  
        # search for this Mode 
        for it in mp2: 
            
          # When mode is find then return 
          # to main function. 
            
          if (M == mp2[it]): 
            return it 
  
          # If mode is not found 
          return 0
  
# Driver Code 
if __name__ == "__main__": 
  
  arr = [6, 10, 3, 10, 
         8, 3, 6, 4] 
  n = len(arr) 
  print (countFreq(arr, n)) 
      
# This code is contributed by Chitranayal

C#

// C# program of the 
// above approach 
using System; 
using System.Collections.Generic; 
  
class GFG{ 
  
// Function to find the mode of the 
// frequency of the array 
static int countFreq(int []arr, int n) 
{ 
      
    // Stores the frequencies 
    // of array elements 
    Dictionary<int, 
               int> mp1 = new Dictionary<int, 
                                         int>(); 
  
    // Traverse through array 
    // elements and 
    // count frequencies 
    for(int i = 0; i < n; ++i) 
    { 
        if (mp1.ContainsKey(arr[i])) 
        { 
            mp1[arr[i]] = mp1[arr[i]] + 1; 
        } 
        else
        { 
            mp1.Add(arr[i], 1); 
        } 
    } 
  
    // Stores the frequencies's of 
    // frequencies of array elements 
    Dictionary<int, 
               int> mp2 = new Dictionary<int, 
                                         int>(); 
  
    foreach(KeyValuePair<int, int> it in mp1) 
    { 
        if (mp2.ContainsKey(it.Value)) 
        { 
            mp2[it.Value] = 
            mp2[it.Value] + 1; 
        } 
        else
        { 
            mp2.Add(it.Value, 1); 
        } 
    } 
  
    // Stores teh minimum value 
    int M = int.MinValue; 
  
    // Find the Mode in 2nd map 
    foreach(KeyValuePair<int, int> it in mp2) 
    { 
        M = Math.Max(M, it.Value); 
    } 
  
    // Search for this Mode 
    foreach(KeyValuePair<int, int> it in mp2) 
    { 
          
        // When mode is find then return 
        // to main function. 
        if (M == it.Value)  
        { 
            return it.Key; 
        } 
    } 
  
    // If mode is not found 
    return 0; 
} 
  
// Driver Code 
public static void Main(String[] args) 
{ 
    int []arr = { 6, 10, 3, 10, 8, 3, 6, 4 }; 
    int n = arr.Length; 
      
    Console.Write(countFreq(arr, n)); 
} 
} 
  
// This code is contributed by Amit Katiyar 

Output:

Time Complexity: O(N)
Auxiliary Space: O(N)

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My Personal Notes

C++

Java

Python3

C#

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