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 arrayint 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 the 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 Codeint 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 approachimport java.util.*;class GFG{// Function to find the mode of the// frequency of the arraystatic 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 the 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 Codepublic 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 approachfrom collections import defaultdictimport sys# Function to find the mode of the# frequency of the arraydef 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 the 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 Codeif __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 approachusing System;using System.Collections.Generic;class GFG{// Function to find the mode of the// frequency of the arraystatic 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 the 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 Codepublic 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 |
Javascript
<script>// Javascript program of the// above approach// Function to find the mode of the// frequency of the arrayfunction countFreq(arr,n){ // Stores the frequencies // of array elements let mp1 = new Map(); // Traverse through array // elements and // count frequencies for(let i = 0; i < n; ++i) { if (mp1.has(arr[i])) { mp1.set(arr[i], mp1.get(arr[i]) + 1); } else { mp1.set(arr[i], 1); } } // Stores the frequencies's of // frequencies of array elements let mp2 = new Map(); for(let [key, value] of mp1.entries()) { if (mp2.has(value)) { mp2.set(value, mp2.get(value) + 1); } else { mp2.set(value, 1); } } // Stores the minimum value let M = Number.MIN_VALUE; // Find the Mode in 2nd map for(let [key, value] of mp2.entries()) { M = Math.max(M, value); } // Search for this Mode for(let [key, value] of mp2.entries()) { // When mode is find then return // to main function. if (M == value) { return key; } } // If mode is not found return 0;}// Driver Codelet arr = [ 6, 10, 3, 10, 8, 3, 6, 4];let n = arr.length;document.write(countFreq(arr, n)); // This code is contributed by avanitrachhadiya2155</script> |
Output:
2
Time Complexity: O(N)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
