Minimum Subsets with Distinct Elements
Last Updated :
27 Feb, 2025
You are given an array of n-element. You have to make subsets from the array such that no subset contain duplicate elements. Find out minimum number of subset possible.
Examples :
Input : arr[] = {1, 2, 3, 4}
Output :1
Explanation : A single subset can contains all values and all values are distinct.
Input : arr[] = {1, 2, 3, 3}
Output : 2
Explanation : We need to create two subsets {1, 2, 3} and {3} [or {1, 3} and {2, 3}] such that both subsets have distinct elements.
[Naive Solution] - Nested Loops - O(n^2) Time and O(1) Space
Let us take a look at few observations.
- If all elements are distinct, we need to make only one subset.
- If all elements are same, we need make n subsets.
- If an element appears twice, and all other are distinct, we need to make two subsets,
Did you see any pattern?
We basically need to find the most frequent element in the array. The result is equal to the frequency of the most frequent element. Since we have to create a subset such that each element in a subset is unique that means that all the repeating elements should be kept in a different set. Hence the maximum no subsets that we require is the frequency of the maximum time occurring element.
Ex -> { 1 , 2 , 1 , 2 , 3 , 3 , 2 , 2 }
here
Frequency of 1 -> 2
Frequency of 2 -> 4
Frequency of 3 -> 2
Since the frequency of 2 is maximum hence we need to have at least 4 subset to keep all the 2 in different subsets and rest of element can be occupied accordingly.
The naive approach involves using two nested loops: the outer loop picks each element, and the inner loop counts the frequency of the picked element. This method is straightforward but inefficient.
C++
// CPP program to find the most frequent element in an array.
#include <bits/stdc++.h>
using namespace std;
int minSubsets(vector<int> &arr)
{
int n = arr.size(), maxcount = 0;
int res;
for (int i = 0; i < n; i++) {
int count = 0;
for (int j = 0; j < n; j++) {
if (arr[i] == arr[j])
count++;
}
if (count > maxcount) {
maxcount = count;
res = arr[i];
}
}
return res;
}
// Driver program
int main()
{
vector<int> arr = { 40, 50, 30, 40, 50, 30, 30 };
cout << minSubsets(arr);
return 0;
}
class GfG {
static int minSubsets(int[] arr) {
int n = arr.length, maxCount = 0, res = arr[0];
for (int i = 0; i < n; i++) {
int count = 0;
for (int j = 0; j < n; j++) {
if (arr[i] == arr[j]) count++;
}
if (count > maxCount) {
maxCount = count;
res = arr[i];
}
}
return res;
}
public static void main(String[] args) {
int[] arr = {40, 50, 30, 40, 50, 30, 30};
System.out.println(minSubsets(arr));
}
}
def min_subsets(arr):
n, max_count, res = len(arr), 0, arr[0]
for i in range(n):
count = sum(1 for j in range(n) if arr[i] == arr[j])
if count > max_count:
max_count = count
res = arr[i]
return res
arr = [40, 50, 30, 40, 50, 30, 30]
print(min_subsets(arr))
function minSubsets(arr) {
let n = arr.length, maxCount = 0, res = arr[0];
for (let i = 0; i < n; i++) {
let count = 0;
for (let j = 0; j < n; j++) {
if (arr[i] === arr[j]) count++;
}
if (count > maxCount) {
maxCount = count;
res = arr[i];
}
}
return res;
}
let arr = [40, 50, 30, 40, 50, 30, 30];
console.log(minSubsets(arr));
[Better Approach] - Using Sorting - O(n Log n) Time and O(1) Space
This method sorts the array first and then finds the maximum frequency by linearly traversing the sorted array. Sorting brings similar elements next to each other, making frequency counting easier.
C++
// CPP program to find the most frequent element
#include <bits/stdc++.h>
using namespace std;
int minSubsets(vector<int>& arr)
{
// Sort the array
sort(arr.begin(), arr.end());
// Find the max frequency using linear traversal
int max_count = 1, res = arr[0], curr_count = 1;
for (int i = 1; i < arr.size(); i++) {
if (arr[i] == arr[i - 1])
curr_count++;
else
curr_count = 1;
if (curr_count > max_count) {
max_count = curr_count;
res = arr[i - 1];
}
}
return res;
}
// Driver program
int main()
{
vector<int> arr = { 40,50,30,40,50,30,30};
cout << minSubsets(arr);
return 0;
}
// Java program to find the most frequent element
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
public class minSubsets {
public static int minSubsets(int[] arr) {
// Sort the array
Arrays.sort(arr);
// Find the max frequency using linear traversal
int max_count = 1, res = arr[0], curr_count = 1;
for (int i = 1; i < arr.length; i++) {
if (arr[i] == arr[i - 1])
curr_count++;
else
curr_count = 1;
if (curr_count > max_count) {
max_count = curr_count;
res = arr[i - 1];
}
}
return res;
}
// Driver program
public static void main(String[] args) {
int[] arr = { 40, 50, 30, 40, 50, 30, 30 };
System.out.println(minSubsets(arr));
}
}
def min_subsets(arr):
# Sort the array
arr.sort()
# Find the max frequency using linear traversal
max_count = 1
res = arr[0]
curr_count = 1
for i in range(1, len(arr)):
if arr[i] == arr[i - 1]:
curr_count += 1
else:
curr_count = 1
if curr_count > max_count:
max_count = curr_count
res = arr[i - 1]
return res
# Driver program
arr = [40, 50, 30, 40, 50, 30, 30]
print(min_subsets(arr))
// C# program to find the most frequent element
using System;
using System.Linq;
public class GfG {
public static int minSubsets(int[] arr) {
// Sort the array
Array.Sort(arr);
// Find the max frequency using linear traversal
int max_count = 1, res = arr[0], curr_count = 1;
for (int i = 1; i < arr.Length; i++) {
if (arr[i] == arr[i - 1])
curr_count++;
else
curr_count = 1;
if (curr_count > max_count) {
max_count = curr_count;
res = arr[i - 1];
}
}
return res;
}
// Driver program
public static void Main() {
int[] arr = { 40, 50, 30, 40, 50, 30, 30 };
Console.WriteLine(minSubsets(arr));
}
}
// JavaScript program to find the most frequent element
function minSubsets(arr) {
// Sort the array
arr.sort((a, b) => a - b);
// Find the max frequency using linear traversal
let max_count = 1, res = arr[0], curr_count = 1;
for (let i = 1; i < arr.length; i++) {
if (arr[i] === arr[i - 1])
curr_count++;
else
curr_count = 1;
if (curr_count > max_count) {
max_count = curr_count;
res = arr[i - 1];
}
}
return res;
}
// Driver program
const arr = [40, 50, 30, 40, 50, 30, 30];
console.log(minSubsets(arr));
[Expected Approach] - Using Hashing - O(n) Time and O(n) Space
Using a hash table, this approach stores each element's frequency and then finds the element with the maximum frequency.
C++
// CPP program to find the most frequent element
// in an array.
#include <bits/stdc++.h>
using namespace std;
int minSubsets(int arr[], int n)
{
// Insert all elements in hash.
unordered_map<int, int> freq;
for (int i = 0; i < n; i++)
freq[arr[i]]++;
// find the max frequency
int max_count = 0, res = -1;
for (auto i : freq) {
if (max_count < i.second) {
res = i.first;
max_count = i.second;
}
}
return res;
}
int main()
{
int arr[] = {40,50,30,40,50,30,30 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << minSubsets(arr, n);
return 0;
}
// Java program to find the most frequent element
// in an array
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
class GFG {
static int minSubsets(int arr[], int n)
{
// Insert all elements in hash
Map<Integer, Integer> hp
= new HashMap<Integer, Integer>();
for (int i = 0; i < n; i++) {
int key = arr[i];
if (hp.containsKey(key)) {
int freq = hp.get(key);
freq++;
hp.put(key, freq);
}
else {
hp.put(key, 1);
}
}
// find max frequency.
int max_count = 0, res = -1;
for (Entry<Integer, Integer> val : hp.entrySet()) {
if (max_count < val.getValue()) {
res = val.getKey();
max_count = val.getValue();
}
}
return res;
}
public static void main(String[] args)
{
int arr[] = { 40, 50, 30, 40, 50, 30, 30 };
int n = arr.length;
System.out.println(minSubsets(arr, n));
}
}
# Python3 program to find the most
# frequent element in an array.
import math as mt
def minSubsets(arr, n):
# Insert all elements in Hash.
Hash = dict()
for i in range(n):
if arr[i] in Hash.keys():
Hash[arr[i]] += 1
else:
Hash[arr[i]] = 1
# find the max frequency
max_count = 0
res = -1
for i in Hash:
if (max_count < Hash[i]):
res = i
max_count = Hash[i]
return res
arr = [ 40,50,30,40,50,30,30]
n = len(arr)
print(minSubsets(arr, n))
// C# program to find the most
// frequent element in an array
using System;
using System.Collections.Generic;
class GFG
{
static int minSubsets(int []arr,
int n)
{
// Insert all elements in hash
Dictionary<int, int> hp =
new Dictionary<int, int>();
for (int i = 0; i < n; i++)
{
int key = arr[i];
if(hp.ContainsKey(key))
{
int freq = hp[key];
freq++;
hp[key] = freq;
}
else
hp.Add(key, 1);
}
// find max frequency.
int min_count = 0, res = -1;
foreach (KeyValuePair<int,
int> pair in hp)
{
if (min_count < pair.Value)
{
res = pair.Key;
min_count = pair.Value;
}
}
return res;
}
static void Main ()
{
int []arr = new int[]{40,50,30,40,50,30,30};
int n = arr.Length;
Console.Write(minSubsets(arr, n));
}
}
// Javascript program to find
// the most frequent element
// in an array.
function minSubsets(arr, n)
{
// Insert all elements in hash.
var hash = new Map();
for (var i = 0; i < n; i++) {
if (hash.has(arr[i]))
hash.set(arr[i], hash.get(arr[i]) + 1)
else hash.set(arr[i], 1)
}
// find the max frequency
var max_count = 0, res = -1;
hash.forEach((value, key) => {
if (max_count < value) {
res = key;
max_count = value;
}
});
return res;
}
var arr = [ 40, 50, 30, 40, 50, 30, 30 ];
var n = arr.length;
console.log(minSubsets(arr, n));
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