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++
#include <bits/stdc++.h>
using namespace std;
int findDiff( int arr[], int n)
{
sort(arr, arr + n);
int count = 0, max_count = 0, min_count = n;
for ( int i = 0; i < (n - 1); i++) {
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);
}
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
import java.util.*;
class GFG {
static int findDiff( int arr[], int n)
{
Arrays.sort(arr);
int count = 0 , max_count = 0 ,
min_count = n;
for ( int i = 0 ; i < (n - 1 ); i++) {
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);
}
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));
}
}
|
Python3
def findDiff(arr, n):
arr.sort()
count = 0 ; max_count = 0 ; min_count = n
for i in range ( 0 , (n - 1 )):
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
arr = [ 7 , 8 , 4 , 5 , 4 , 1 , 1 , 7 , 7 , 2 , 5 ]
n = len (arr)
print (findDiff(arr, n))
|
C#
using System;
class GFG {
static int findDiff( int [] arr, int n)
{
Array.Sort(arr);
int count = 0, max_count = 0,
min_count = n;
for ( int i = 0; i < (n - 1); i++) {
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);
}
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));
}
}
|
PHP
<?php
function findDiff( $arr , $n )
{
sort( $arr );
$count = 0; $max_count = 0;
$min_count = $n ;
for ( $i = 0; $i < ( $n - 1); $i ++)
{
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 );
}
$arr = array (7, 8, 4, 5, 4, 1,
1, 7, 7, 2, 5);
$n = sizeof( $arr );
echo (findDiff( $arr , $n ) . "\n" );
?>
|
Javascript
<script>
function findDiff(arr, n)
{
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++) {
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>
|
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++
#include <bits/stdc++.h>
using namespace std;
int findDiff( int arr[], int n)
{
unordered_map< int , int > hm;
for ( int i = 0; i < n; i++)
hm[arr[i]]++;
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);
}
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
import java.util.*;
class GFG
{
static int findDiff( int arr[], int n)
{
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 );
}
}
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);
}
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));
}
}
|
Python3
from collections import defaultdict
def findDiff(arr,n):
mp = defaultdict( lambda : 0 )
for i in range (n):
mp[arr[i]] + = 1
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
arr = [ 7 , 8 , 4 , 5 , 4 , 1 , 1 , 7 , 7 , 2 , 5 ]
n = len (arr)
print (findDiff(arr,n))
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static int findDiff( int []arr, int n)
{
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);
}
}
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);
}
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));
}
}
|
Javascript
<script>
function findDiff(arr, n)
{
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;
}
}
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;
}
var arr = [7, 8, 4, 5, 4, 1, 1, 7, 7, 2, 5];
var n = arr.length;
document.write(findDiff(arr, n));
</script>
|
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
This article is contributed by Himanshu Ranjan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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