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

A simple solution is to use two loops to count the 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. 

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)
{
    // 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.
?>

Javascript

<script>
    // Javascript Code for Difference between
    // highest and least frequencies
    // in an array

    function findDiff(arr, n)
    {

        // sort the array
        arr.sort(function(a, b){return a - b});

        let count = 0, max_count = 0, min_count = n;

        for (let 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);
    }

    let arr = [ 7, 8, 4, 5, 4, 1, 1, 7, 7, 2, 5 ];
    let n = arr.length;

    document.write(findDiff(arr, n));

</script>

Output

2

Time Complexity: O(nlogn) the inbuilt sort function takes N log N time to complete all operations, hence the overall time taken by the algorithm is N log N
Space Complexity: O(1) since no extra array is used so the space taken by the algorithm is constant

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

Javascript

<script>
      // JavaScript code to find the difference between highest
      // and least frequencies
      function findDiff(arr, n)
      {

        // Put all elements in a hash map
        var mp = {};
        for (var i = 0; i < n; i++) {
          if (mp.hasOwnProperty(arr[i])) {
            var val = mp[arr[i]];
            delete mp[arr[i]];
            mp[arr[i]] = val + 1;
          } else {
            mp[arr[i]] = 1;
          }
        }

        // Find counts of maximum and minimum
        // frequent elements
        var max_count = 0,
          min_count = n;
        for (const [key, value] of Object.entries(mp)) {
          max_count = Math.max(max_count, value);
          min_count = Math.min(min_count, value);
        }

        return max_count - min_count;
      }

      // Driver code
      var arr = [7, 8, 4, 5, 4, 1, 1, 7, 7, 2, 5];
      var n = arr.length;
      document.write(findDiff(arr, n));

      // This code is contributed by rdtank.
    </script>

Output

2

Time Complexity: O(n) since one traversal of the array is required to complete all operations hence the overall time required by the algorithm is linear
Auxiliary Space: O(n) since an unordered map is used, in the worst case all elements will be stored inside the unordered map thus the space taken by the algorithm is linear

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