Maximum difference between frequency of two elements such that element having greater frequency is also greater
Last Updated :
12 Feb, 2025
Given an array of n positive integers with many repeating elements. The task is to find the maximum difference between the frequency of any two different elements, such that the element with greater frequency is also greater in value than the second integer.
Examples:
Input : arr[] = { 3, 1, 3, 2, 3, 2 }.
Output : 2
Frequency of 3 = 3.
Frequency of 2 = 2.
Frequency of 1 = 1.
Here difference of frequency of element 3 and 1 is = 3 - 1 = 2.
Also 3 > 1.Method 1 (Use Hashing): The naive approach can be, find the frequency of each element and for each element find the element having lesser value and lesser frequency than the current element.
Below is the implementation of this approach:
C++
// C++ program to find maximum difference
// between frequency of any two element
// such that element with greater frequency
// is also greater in value.
#include<bits/stdc++.h>
using namespace std;
// Return the maximum difference between
// frequencies of any two elements such that
// element with greater frequency is also
// greater in value.
int maxdiff(int arr[], int n)
{
unordered_map<int, int> freq;
// Finding the frequency of each element.
for (int i = 0; i < n; i++)
freq[arr[i]]++;
int ans = 0;
for (int i=0; i<n; i++)
{
for (int j=0; j<n; j++)
{
// finding difference such that element
// having greater frequency is also
// greater in value.
if (freq[arr[i]] > freq[arr[j]] &&
arr[i] > arr[j] )
ans = max(ans, freq[arr[i]]-freq[arr[j]]);
else if (freq[arr[i]] < freq[arr[j]] &&
arr[i] < arr[j] )
ans = max(ans, freq[arr[j]]-freq[arr[i]]);
}
}
return ans;
}
// Driven Program
int main()
{
int arr[] = { 3, 1, 3, 2, 3, 2 };
int n = sizeof(arr)/sizeof(arr[0]);
cout << maxdiff(arr, n) << endl;
return 0;
}
// Java program to find maximum difference
// between frequency of any two element
// such that element with greater frequency
// is also greater in value.
import java.util.*;
class GFG
{
// Return the maximum difference between
// frequencies of any two elements such that
// element with greater frequency is also
// greater in value.
static int maxdiff(int arr[], int n)
{
Map<Integer, Integer> freq = new HashMap<>();
// Finding the frequency of each element.
for (int i = 0; i < n; i++)
freq.put(arr[i],
freq.get(arr[i]) == null ? 1 :
freq.get(arr[i]) + 1);
int ans = 0;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
// finding difference such that element
// having greater frequency is also
// greater in value.
if (freq.get(arr[i]) > freq.get(arr[j]) &&
arr[i] > arr[j])
ans = Math.max(ans, freq.get(arr[i]) -
freq.get(arr[j]));
else if (freq.get(arr[i]) < freq.get(arr[j]) &&
arr[i] < arr[j] )
ans = Math.max(ans, freq.get(arr[j]) -
freq.get(arr[i]));
}
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 3, 1, 3, 2, 3, 2 };
int n = arr.length;
System.out.println(maxdiff(arr, n));
}
}
// This code is contributed by 29AjayKumar
# Python program to find maximum difference
# between frequency of any two element
# such that element with greater frequency
# is also greater in value.
from collections import defaultdict
# Return the maximum difference between
# frequencies of any two elements such that
# element with greater frequency is also
# greater in value.
def maxdiff(arr, n):
freq = defaultdict(lambda: 0)
# Finding the frequency of each element.
for i in range(n):
freq[arr[i]] += 1
ans = 0
for i in range(n):
for j in range(n):
# finding difference such that element
# having greater frequency is also
# greater in value.
if freq[arr[i]] > freq[arr[j]] and arr[i] > arr[j]:
ans = max(ans, freq[arr[i]] - freq[arr[j]])
elif freq[arr[i]] < freq[arr[j]] and arr[i] < arr[j]:
ans = max(ans, freq[arr[j]] - freq[arr[i]])
return ans
arr = [3,1,3,2,3,2]
n = len(arr)
print(maxdiff(arr,n))
# This code is contributed by Shrikant13
// C# program to find maximum difference
// between frequency of any two element
// such that element with greater frequency
// is also greater in value.
using System;
using System.Collections.Generic;
class GFG
{
// Return the maximum difference between
// frequencies of any two elements such that
// element with greater frequency is also
// greater in value.
static int maxdiff(int[] arr, int n)
{
Dictionary<int,
int> freq = new Dictionary<int,
int>();
// Finding the frequency of each element.
for (int i = 0; i < n; i++)
{
if (freq.ContainsKey(arr[i]))
freq[arr[i]]++;
else
freq.Add(arr[i], 1);
}
int ans = 0;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
// finding difference such that element
// having greater frequency is also
// greater in value.
if (freq[arr[i]] > freq[arr[j]] && arr[i] > arr[j])
ans = Math.Max(ans, freq[arr[i]] - freq[arr[i]]);
else if (freq[arr[i]] < freq[arr[j]] && arr[i] < arr[j])
ans = Math.Max(ans, freq[arr[j]] - freq[arr[i]]);
}
}
return ans;
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 3, 1, 3, 2, 3, 2 };
int n = arr.Length;
Console.WriteLine(maxdiff(arr, n));
}
}
// This code is contributed by
// sanjeev2552
// JavaScript program to find maximum difference
// between frequency of any two elements
// such that the element with greater frequency
// is also greater in value.
function maxdiff(arr, n) {
let freq = new Map();
// Finding the frequency of each element.
for (let i = 0; i < n; i++)
freq.set(arr[i], freq.get(arr[i]) == null ? 1 : freq.get(arr[i]) + 1);
let ans = 0;
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
// Finding difference such that element
// having greater frequency is also greater in value.
if (freq.get(arr[i]) > freq.get(arr[j]) && arr[i] > arr[j])
ans = Math.max(ans, freq.get(arr[i]) - freq.get(arr[j]));
else if (freq.get(arr[i]) < freq.get(arr[j]) && arr[i] < arr[j])
ans = Math.max(ans, freq.get(arr[j]) - freq.get(arr[i]));
}
}
return ans;
}
// Driver code
let arr = [3, 1, 3, 2, 3, 2];
let n = arr.length;
console.log(maxdiff(arr, n));
Time Complexity: O(n2).
Auxiliary Space: O(n)
Method 2 (Use Hashing and Sorting): The idea is to find all the distinct elements and store them in an array, say dist[ ]. Sort the distinct element array dist[] in increasing order. Now for any distinct element at index i, for all index j such that i > j > 0, find the element between index 0 to i-1 having a minimum frequency. We can find the frequency of an element in the same way as method 1, i.e., storing frequencies in a hash table.
So do this for all i and find the maximum difference. To find the minimum frequency for all i maintain a prefix minimum.
Below is the representation of this approach:
C++
// Efficient C++ program to find maximum
// difference between frequency of any two
// elements such that element with greater
// frequency is also greater in value.
#include<bits/stdc++.h>
using namespace std;
// Return the maximum difference between
// frequencies of any two elements such that
// element with greater frequency is also
// greater in value.
int maxdiff(int arr[], int n)
{
unordered_map<int, int> freq;
int dist[n];
// Finding the frequency of each element.
int j = 0;
for (int i = 0; i < n; i++)
{
if (freq.find(arr[i]) == freq.end())
dist[j++] = arr[i];
freq[arr[i]]++;
}
// Sorting the distinct element
sort(dist, dist + j);
int min_freq = n+1;
// Iterate through all sorted distinct elements.
// For each distinct element, maintaining the
// element with minimum frequency than that
// element and also finding the maximum
// frequency difference
int ans = 0;
for (int i=0; i<j; i++)
{
int cur_freq = freq[dist[i]];
ans = max(ans, cur_freq - min_freq);
min_freq = min(min_freq, cur_freq);
}
return ans;
}
// Driven Program
int main()
{
int arr[] = { 3, 1, 3, 2, 3, 2 };
int n = sizeof(arr)/sizeof(arr[0]);
cout << maxdiff(arr, n) << endl;
return 0;
}
// Efficient Java program to find maximum
// difference between frequency of any two
// elements such that element with greater
// frequency is also greater in value.
import java.util.*;
class GFG
{
// Return the maximum difference between
// frequencies of any two elements such that
// element with greater frequency is also
// greater in value.
static int maxdiff(int arr[], int n)
{
HashMap<Integer,Integer> freq= new HashMap<>();
int []dist = new int[n];
// Finding the frequency of each element.
int j = 0;
for (int i = 0; i < n; i++)
{
dist[j++] = arr[i];
if (!freq.containsKey(arr[i]))
freq.put(arr[i], 1);
else
freq.put(arr[i], freq.get(arr[i]) + 1);
}
// Sorting the distinct element
Arrays.sort(dist);
int min_freq = n + 1;
// Iterate through all sorted distinct elements.
// For each distinct element, maintaining the
// element with minimum frequency than that
// element and also finding the maximum
// frequency difference
int ans = 0;
for (int i = 0; i < j; i++)
{
int cur_freq = freq.get(dist[i]);
ans = Math.max(ans, cur_freq - min_freq);
min_freq = Math.min(min_freq, cur_freq);
}
return ans;
}
// Driven Program
public static void main(String[] args)
{
int arr[] = { 3, 1, 3, 2, 3, 2 };
int n = arr.length;
System.out.print(maxdiff(arr, n) +"\n");
}
}
// This code is contributed by Rajput-Ji
# Efficient Python3 program to find maximum
# difference between frequency of any two
# elements such that element with greater
# frequency is also greater in value.
# Return the maximum difference between
# frequencies of any two elements such that
# element with greater frequency is also
# greater in value.
def maxdiff(arr, n):
freq = {}
dist = [0] * n
# Finding the frequency of each element.
j = 0
for i in range(n):
if (arr[i] not in freq):
dist[j] = arr[i]
j += 1
freq[arr[i]] = 0
if (arr[i] in freq):
freq[arr[i]] += 1
dist = dist[:j]
# Sorting the distinct element
dist.sort()
min_freq = n + 1
# Iterate through all sorted distinct elements.
# For each distinct element, maintaining the
# element with minimum frequency than that
# element and also finding the maximum
# frequency difference
ans = 0
for i in range(j):
cur_freq = freq[dist[i]]
ans = max(ans, cur_freq - min_freq)
min_freq = min(min_freq, cur_freq)
return ans
# Driven Program
arr = [3, 1, 3, 2, 3, 2]
n = len(arr)
print(maxdiff(arr, n))
# This code is contributed by SHUBHAMSINGH10
// Efficient C# program to find maximum
// difference between frequency of any two
// elements such that element with greater
// frequency is also greater in value.
using System.Collections.Generic;
using System;
class GFG{
// Return the maximum difference between
// frequencies of any two elements such
// that element with greater frequency
// is also greater in value.
static int maxdiff(int []arr, int n)
{
Dictionary<int,
int> freq = new Dictionary<int,
int>();
List<int> dist = new List<int>();
// Finding the frequency of each element.
int j = 0;
for(int i = 0; i < n; i++)
{
if (freq.ContainsKey(arr[i]) == false)
{
dist.Add(arr[i]);
j++;
}
if (freq.ContainsKey(arr[i]))
freq[arr[i]]++;
else
freq[arr[i]]=1;
}
// Sorting the distinct element
dist.Sort();
int min_freq = n + 1;
// Iterate through all sorted distinct elements.
// For each distinct element, maintaining the
// element with minimum frequency than that
// element and also finding the maximum
// frequency difference
int ans = 0;
for(int i = 0; i < j; i++)
{
int cur_freq = freq[dist[i]];
ans = Math.Max(ans, cur_freq - min_freq);
min_freq = Math.Min(min_freq, cur_freq);
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 3, 1, 3, 2, 3, 2 };
int n = arr.Length;
Console.WriteLine(maxdiff(arr, n));
}
}
// This code is contributed by Stream_Cipher
// Efficient JavaScript program to find maximum
// difference between frequency of any two
// elements such that the element with greater
// frequency is also greater in value.
function maxdiff(arr, n) {
var freq = new Map();
var dist = Array(n);
// Finding the frequency of each element.
var j = 0;
for (var i = 0; i < n; i++) {
if (!freq.has(arr[i]))
dist[j++] = arr[i];
if (freq.has(arr[i]))
freq.set(arr[i], freq.get(arr[i]) + 1);
else
freq.set(arr[i], 1);
}
// Sorting the distinct elements
dist.sort((a, b) => a - b);
var min_freq = n + 1;
var ans = 0;
// Iterate through all sorted distinct elements.
for (var i = 0; i < j; i++) {
var cur_freq = freq.get(dist[i]);
ans = Math.max(ans, cur_freq - min_freq);
min_freq = Math.min(min_freq, cur_freq);
}
return ans;
}
// Driver Code
var arr = [3, 1, 3, 2, 3, 2];
var n = arr.length;
console.log(maxdiff(arr, n));
Time Complexity: O(n log n).
Auxiliary Space: O(n)
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