Difference between highest and least frequencies in an array

Given an array, find the difference between highest occurrence and least occurrence of any number in an array

Examples:

Input  : arr[] = [7, 8, 4, 5, 4, 1, 1, 7, 7, 2, 5]
Output : 2
Lowest occurring element (5) occurs once.
Highest occurring element (1 or 7) occurs 3 times

Input  : arr[] = [1, 1, 1, 3, 3, 3]
Output : 0

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to use two loops to count frequency of every element and keep track of maximum and minimum frequencies.

A better solution is to sort the array in O(n log n) and check
consecutive element’s occurrence and compare their count respectively.

C++

// CPP code to find the difference between highest 
// and least frequencies 
#include <bits/stdc++.h> 
using namespace std; 
  
int findDiff(int arr[], int n) 
{ 
    // sort the array 
    sort(arr, arr + n); 
  
    int count = 0, max_count = 0, min_count = n; 
    for (int i = 0; i < (n - 1); i++) { 
  
        // checking consecutive elements 
        if (arr[i] == arr[i + 1]) { 
            count += 1; 
            continue; 
        } 
        else { 
            max_count = max(max_count, count); 
            min_count = min(min_count, count); 
            count = 0; 
        } 
    } 
  
    return (max_count - min_count); 
} 
  
// Driver code 
int main() 
{ 
    int arr[] = { 7, 8, 4, 5, 4, 1, 1, 7, 7, 2, 5 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    cout << findDiff(arr, n) << "\n"; 
    return 0; 
} 

Java

// JAVA Code for Difference between 
// highest and least frequencies 
// in an array 
import java.util.*; 
  
class GFG { 
  
    static int findDiff(int arr[], int n) 
    { 
        // sort the array 
        Arrays.sort(arr); 
  
        int count = 0, max_count = 0, 
            min_count = n; 
  
        for (int i = 0; i < (n - 1); i++) { 
  
            // checking consecutive elements 
            if (arr[i] == arr[i + 1]) { 
                count += 1; 
                continue; 
            } 
            else { 
                max_count = Math.max(max_count, 
                                     count); 
  
                min_count = Math.min(min_count, 
                                     count); 
                count = 0; 
            } 
        } 
  
        return (max_count - min_count); 
    } 
  
    // Driver program to test above function 
    public static void main(String[] args) 
    { 
  
        int arr[] = { 7, 8, 4, 5, 4, 1, 
                      1, 7, 7, 2, 5 }; 
        int n = arr.length; 
  
        System.out.println(findDiff(arr, n)); 
    } 
} 
  
// This code is contributed by Arnav Kr. Mandal. 

Python3

# Python3 code to find the difference  
# between highest nd least frequencies 
  
def findDiff(arr, n): 
      
    # sort the array 
    arr.sort() 
      
    count = 0; max_count = 0; min_count = n 
    for i in range(0, (n-1)): 
  
        # checking consecutive elements 
        if arr[i] == arr[i + 1]: 
            count += 1
            continue
        else: 
            max_count = max(max_count, count) 
            min_count = min(min_count, count) 
            count = 0
    return max_count - min_count 
  
# Driver Code 
arr = [ 7, 8, 4, 5, 4, 1, 1, 7, 7, 2, 5 ] 
n = len(arr) 
print (findDiff(arr, n)) 
  
# This code is contributed by Shreyanshi Arun. 

C#

// C# Code for Difference between 
// highest and least frequencies 
// in an array 
using System; 
  
class GFG { 
  
    static int findDiff(int[] arr, int n) 
    { 
          
        // sort the array 
        Array.Sort(arr); 
  
        int count = 0, max_count = 0, 
            min_count = n; 
  
        for (int i = 0; i < (n - 1); i++) { 
  
            // checking consecutive elements 
            if (arr[i] == arr[i + 1]) { 
                count += 1; 
                continue; 
            } 
            else { 
                max_count = Math.Max(max_count, 
                                    count); 
  
                min_count = Math.Min(min_count, 
                                    count); 
                count = 0; 
            } 
        } 
  
        return (max_count - min_count); 
    } 
  
    // Driver program to test above function 
    public static void Main() 
    { 
  
        int[] arr = { 7, 8, 4, 5, 4, 1, 
                       1, 7, 7, 2, 5 }; 
        int n = arr.Length; 
  
        Console.WriteLine(findDiff(arr, n)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP code to find the  
// difference between highest 
// and least frequencies 
  
// function that  
// returns difference 
function findDiff($arr, $n) 
{ 
      
    // sort the array 
    sort($arr); 
  
    $count = 0; $max_count = 0;  
    $min_count = $n; 
    for ($i = 0; $i < ($n - 1); $i++) 
    { 
  
        // checking consecutive elements 
        if ($arr[$i] == $arr[$i + 1]) 
        { 
            $count += 1; 
            continue; 
        } 
        else
        { 
            $max_count = max($max_count, $count); 
            $min_count = min($min_count, $count); 
            $count = 0; 
        } 
    } 
  
    return ($max_count - $min_count); 
} 
  
// Driver Code 
$arr = array(7, 8, 4, 5, 4, 1, 
             1, 7, 7, 2, 5); 
$n = sizeof($arr); 
  
echo(findDiff($arr, $n) . "\n"); 
  
// This code is contributed by Ajit. 
?> 

Output:

An efficient solution is to use hashing. We count frequencies of all elements and finally traverse the hash table to find maximum and minimum.

Below is the implementation.

C++

// CPP code to find the difference between highest 
// and least frequencies 
#include <bits/stdc++.h> 
using namespace std; 
  
int findDiff(int arr[], int n) 
{ 
    // Put all elements in a hash map 
    unordered_map<int, int> hm; 
    for (int i = 0; i < n; i++) 
        hm[arr[i]]++; 
  
    // Find counts of maximum and minimum 
    // frequent elements 
    int max_count = 0, min_count = n; 
    for (auto x : hm) { 
        max_count = max(max_count, x.second); 
        min_count = min(min_count, x.second); 
    } 
  
    return (max_count - min_count); 
} 
  
// Driver 
int main() 
{ 
    int arr[] = { 7, 8, 4, 5, 4, 1, 1, 7, 7, 2, 5 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    cout << findDiff(arr, n) << "\n"; 
    return 0; 
} 

Java

// Java code to find the difference between highest 
// and least frequencies 
import java.util.*; 
  
class GFG 
{ 
  
static int findDiff(int arr[], int n) 
{ 
    // Put all elements in a hash map 
    Map<Integer,Integer> mp = new HashMap<>(); 
    for (int i = 0 ; i < n; i++) 
    { 
        if(mp.containsKey(arr[i])) 
        { 
            mp.put(arr[i], mp.get(arr[i])+1); 
        } 
        else
        { 
            mp.put(arr[i], 1); 
        } 
    } 
  
    // Find counts of maximum and minimum 
    // frequent elements 
    int max_count = 0, min_count = n; 
    for (Map.Entry<Integer,Integer> x : mp.entrySet()) 
    { 
        max_count = Math.max(max_count, x.getValue()); 
        min_count = Math.min(min_count, x.getValue()); 
    } 
  
    return (max_count - min_count); 
} 
  
// Driver code 
public static void main(String[] args)  
{ 
    int arr[] = { 7, 8, 4, 5, 4, 1, 1, 7, 7, 2, 5 }; 
    int n = arr.length; 
    System.out.println(findDiff(arr, n)); 
} 
} 
  
/* This code is contributed by PrinciRaj1992 */

Python3

# Python code to find the difference between highest  
# and least frequencies  
  
from collections import defaultdict 
def findDiff(arr,n): 
  
    # Put all elements in a hash map 
    mp = defaultdict(lambda:0) 
    for i in range(n): 
        mp[arr[i]]+=1
  
    # Find counts of maximum and minimum  
    # frequent elements  
    max_count=0;min_count=n 
    for key,values in mp.items(): 
        max_count= max(max_count,values) 
        min_count = min(min_count,values) 
  
    return max_count-min_count 
  
  
# Driver code 
arr = [ 7, 8, 4, 5, 4, 1, 1, 7, 7, 2, 5] 
n = len(arr) 
print(findDiff(arr,n)) 
  
# This code is contributed by Shrikant13 

C#

// C# code to find the difference between highest 
// and least frequencies 
using System; 
using System.Collections.Generic; 
  
class GFG 
{ 
  
static int findDiff(int []arr, int n) 
{ 
    // Put all elements in a hash map 
    Dictionary<int,int> mp = new Dictionary<int,int>(); 
    for (int i = 0 ; i < n; i++) 
    { 
        if(mp.ContainsKey(arr[i])) 
        { 
            var val = mp[arr[i]]; 
            mp.Remove(arr[i]); 
            mp.Add(arr[i], val + 1);  
        } 
        else
        { 
            mp.Add(arr[i], 1); 
        } 
    } 
  
    // Find counts of maximum and minimum 
    // frequent elements 
    int max_count = 0, min_count = n; 
    foreach(KeyValuePair<int, int> entry in mp) 
    { 
        max_count = Math.Max(max_count, entry.Value); 
        min_count = Math.Min(min_count, entry.Value); 
    } 
  
    return (max_count - min_count); 
} 
  
// Driver code 
public static void Main(String[] args)  
{ 
    int []arr = { 7, 8, 4, 5, 4, 1, 1, 7, 7, 2, 5 }; 
    int n = arr.Length; 
    Console.WriteLine(findDiff(arr, n)); 
} 
} 
  
// This code is contributed by Princi Singh 

Output:

Time Complexity : O(n)

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

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