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Minimum number of subsets with distinct elements

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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.

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 subsites 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.

A simple solution is to run two nested loops to count frequency of every element and return the frequency of the most frequent element. Time complexity of this solution is O(n2).

A better solution is to first sort the array and then start count number of repetitions of elements in an iterative manner as all repetition of any number lie beside the number itself. By this method you can find the maximum frequency or repetition by simply traversing the sorted array. This approach will cost O(nlogn) time complexity 

Implementation:

C++




// A sorting based solution to find the
// minimum number of subsets of a set
// such that every subset contains distinct
// elements.
#include <bits/stdc++.h>
using namespace std;
 
// Function to count subsets such that all
// subsets have distinct elements.
int subset(int ar[], int n)
{
    // Take input and initialize res = 0
    int res = 0;
 
    // Sort the array
    sort(ar, ar + n);
 
    // Traverse the input array and
    // find maximum frequency
    for (int i = 0; i < n; i++) {
        int count = 1;
 
        // For each number find its repetition / frequency
        for (; i < n - 1; i++) {
            if (ar[i] == ar[i + 1])
                count++;
            else
                break;
        }
 
        // Update res
        res = max(res, count);
    }
 
    return res;
}
 
// Driver code
int main()
{
    int arr[] = { 5, 6, 9, 3, 4, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << subset(arr, n);
    return 0;
}


Java




// A sorting based solution to find the
// minimum number of subsets of a set
// such that every subset contains distinct
// elements.
import java.util.*;
import java.lang.*;
 
public class GfG{
     
    // Function to count subsets such that all
    // subsets have distinct elements.
    public static int subset(int ar[], int n)
    {
        // Take input and initialize res = 0
        int res = 0;
 
        // Sort the array
        Arrays.sort(ar);
 
        // Traverse the input array and
        // find maximum frequency
        for (int i = 0; i < n; i++) {
            int count = 1;
 
            // For each number find its repetition / frequency
            for (; i < n - 1; i++) {
                if (ar[i] == ar[i + 1])
                    count++;
                else
                    break;
            }
 
            // Update res
            res = Math.max(res, count);
        }
 
        return res;
    }
     
    // Driver function
    public static void main(String argc[])
    {
        int arr[] = { 5, 6, 9, 3, 4, 3, 4 };
        int n = 7;
        System.out.println(subset(arr, n));
    }
     
}
 
/* This code is contributed by Sagar Shukla */


Python3




# A sorting based solution to find the
# minimum number of subsets of a set
# such that every subset contains distinct
# elements.
 
# function to count subsets such that all
# subsets have distinct elements.
def subset(ar, n):
 
    # take input and initialize res = 0
    res = 0
 
    # sort the array
    ar.sort()
 
    # traverse the input array and
    # find maximum frequency
    for i in range(0, n) :
        count = 1
 
        # for each number find its repetition / frequency
        for i in range(n - 1):
            if ar[i] == ar[i + 1]:
                count+=1
            else:
                break
             
        # update res
        res = max(res, count)
     
    return res
 
 
# Driver code
ar = [ 5, 6, 9, 3, 4, 3, 4 ]
n = len(ar)
print(subset(ar, n))
 
# This code is contributed by
# Smitha Dinesh Semwal


C#




// A sorting based solution to find the
// minimum number of subsets of a set
// such that every subset contains distinct
// elements.
using System;
 
public class GfG {
     
    // Function to count subsets such that all
    // subsets have distinct elements.
    public static int subset(int []ar, int n)
    {
        // Take input and initialize res = 0
        int res = 0;
 
        // Sort the array
        Array.Sort(ar);
 
        // Traverse the input array and
        // find maximum frequency
        for (int i = 0; i < n; i++) {
            int count = 1;
 
            // For each number find its
            // repetition / frequency
            for ( ; i < n - 1; i++) {
                if (ar[i] == ar[i + 1])
                    count++;
                else
                    break;
            }
 
            // Update res
            res = Math.Max(res, count);
        }
 
        return res;
    }
     
    // Driver function
    public static void Main()
    {
        int []arr = { 5, 6, 9, 3, 4, 3, 4 };
        int n = 7;
         
        Console.WriteLine(subset(arr, n));
    }
     
}
 
/* This code is contributed by Vt_m */


Javascript




<script>
 
// JavaScript program sorting based solution to find the
// minimum number of subsets of a set
// such that every subset contains distinct
 
    // Function to count subsets such that all
    // subsets have distinct elements.
    function subset(ar, n)
    {
     
        // Take input and initialize res = 0
        let res = 0;
   
        // Sort the array
        ar.sort();
   
        // Traverse the input array and
        // find maximum frequency
        for (let i = 0; i < n; i++)
        {
            let count = 1;
   
            // For each number find its repetition / frequency
            for (; i < n - 1; i++)
            {
                if (ar[i] == ar[i + 1])
                    count++;
                else
                    break;
            }
   
            // Update res
            res = Math.max(res, count);
        }
        return res;
    }
      
// Driver Code
        let arr = [ 5, 6, 9, 3, 4, 3, 4 ];
        let n = 7;
        document.write(subset(arr, n));
 
// This code is contributed by chinmoy1997pal.
</script>


PHP




<?php
// A sorting based solution to find the
// minimum number of subsets of a set
// such that every subset contains distinct
// elements.
 
// Function to count subsets such that all
// subsets have distinct elements.
function subset($ar, $n)
{
    // Take input and initialize res = 0
    $res = 0;
 
    // Sort the array
    sort($ar);
 
    // Traverse the input array and
    // find maximum frequency
    for ($i = 0; $i < $n; $i++)
    {
        $count = 1;
 
        // For each number find its
        // repetition / frequency
        for (; $i < $n - 1; $i++)
        {
            if ($ar[$i] == $ar[$i + 1])
                $count++;
            else
                break;
        }
 
        // Update res
        $res = max($res, $count);
    }
 
    return $res;
}
 
// Driver code
$arr = array( 5, 6, 9, 3, 4, 3, 4 );
$n = sizeof($arr);
echo subset($arr, $n);
 
// This code is contributed
// by Sach_Code
?>


Output

2












Time Complexity: O(n2)
Auxiliary Space: O(1)
 
An efficient solution is to use hashing. We count frequencies of all elements in a hash table. Finally we return the key with maximum value in hash table.

Implementation:

C++




// A hashing based solution to find the
// minimum number of subsets of a set
// such that every subset contains distinct
// elements.
#include <bits/stdc++.h>
using namespace std;
 
// Function to count subsets such that all
// subsets have distinct elements.
int subset(int arr[], int n)
{  
    // Traverse the input array and
    // store frequencies of elements
    unordered_map<int, int> mp;   
    for (int i = 0; i < n; i++)
        mp[arr[i]]++;
     
    // Find the maximum value in map.
    int res = 0;
    for (auto x : mp)
       res = max(res, x.second);
 
    return res;
}
 
// Driver code
int main()
{
    int arr[] = { 5, 6, 9, 3, 4, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << subset(arr, n);
    return 0;
}


Java




import java.util.HashMap;
import java.util.Map;
 
// A hashing based solution to find the
// minimum number of subsets of a set
// such that every subset contains distinct
// elements.
class GFG
{
     
// Function to count subsets such that all
// subsets have distinct elements.
static int subset(int arr[], int n)
{
    // Traverse the input array and
    // store frequencies of elements
    HashMap<Integer, Integer> mp = new HashMap<>();
    for (int i = 0; i < n; i++)
        mp.put(arr[i],mp.get(arr[i]) == null?1:mp.get(arr[i])+1);
     
    // Find the maximum value in map.
    int res = 0;
    for (Map.Entry<Integer,Integer> entry : mp.entrySet())
    res = Math.max(res, entry.getValue());
 
    return res;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 5, 6, 9, 3, 4, 3, 4 };
    int n = arr.length;
    System.out.println( subset(arr, n));
 
}
}
 
// This code is contributed by Rajput-Ji


Python3




# A hashing based solution to find the
# minimum number of subsets of a set such 
# that every subset contains distinct
# elements.
 
# Function to count subsets such that
# all subsets have distinct elements.
def subset(arr, n):
     
    # Traverse the input array and
    # store frequencies of elements
    mp = {i:0 for i in range(10)}
    for i in range(n):
        mp[arr[i]] += 1
     
    # Find the maximum value in map.
    res = 0
    for key, value in mp.items():
        res = max(res, value)
 
    return res
 
# Driver code
if __name__ == '__main__':
    arr = [5, 6, 9, 3, 4, 3, 4]
    n = len(arr)
    print(subset(arr, n))
 
# This code is contributed by
# Surendra_Gangwar


C#




// A hashing based solution to find the
// minimum number of subsets of a set
// such that every subset contains distinct
// elements.
using System;
using System.Collections.Generic;
 
class GFG
{
     
// Function to count subsets such that all
// subsets have distinct elements.
static int subset(int []arr, int n)
{
    // Traverse the input array and
    // store frequencies of elements
    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 the maximum value in map.
    int res = 0;
    foreach(KeyValuePair<int, int> entry in mp)
        res = Math.Max(res, entry.Value);
 
    return res;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 5, 6, 9, 3, 4, 3, 4 };
    int n = arr.Length;
    Console.WriteLine(subset(arr, n));
}
}
 
// This code is contributed by Rajput-Ji


Javascript




<script>
      // A hashing based solution to find the
      // minimum number of subsets of a set
      // such that every subset contains distinct
      // elements.
      // Function to count subsets such that all
      // subsets have distinct elements.
      function subset(arr, n) {
        // Traverse the input array and
        // store frequencies of elements
        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 the maximum value in map.
        var res = 0;
        for (const [key, value] of Object.entries(mp)) {
          res = Math.max(res, value);
        }
        return res;
      }
 
      // Driver code
      var arr = [5, 6, 9, 3, 4, 3, 4];
      var n = arr.length;
      document.write(subset(arr, n));
       
      // This code is contributed by rdtank.
    </script>


Output

2












Time Complexity: O(n)
Auxiliary Space: O(n)

Example in c:

Approach:

Initialize a hash set to store distinct elements and a variable, count, to zero.

Iterate over all elements of the array.

For each element, check if it is already present in the hash set using the following conditions:

If the element is not present, add it to the hash set.
If the element is present, increment the count and reset the hash set.
Finally, return the count.

C++




// C++ code for the above approach
#include <iostream>
 
using namespace std;
 
struct node {
    int data;
    struct node* next;
};
 
struct node* create_node(int data) {
    struct node* new_node = (struct node*) malloc(sizeof(struct node));
    new_node->data = data;
    new_node->next = NULL;
    return new_node;
}
 
struct set {
    struct node* head;
};
 
struct set* create_set() {
    struct set* new_set = (struct set*) malloc(sizeof(struct set));
    new_set->head = NULL;
    return new_set;
}
 
void add_to_set(struct set* set, int data) {
    if (set->head == NULL) {
        set->head = create_node(data);
        return;
    }
 
    struct node* current = set->head;
    while (current != NULL) {
        if (current->data == data) {
            return;
        }
        current = current->next;
    }
 
    struct node* new_node = create_node(data);
    new_node->next = set->head;
    set->head = new_node;
}
 
void delete_set(struct set* set) {
    struct node* current = set->head;
    while (current != NULL) {
        struct node* temp = current;
        current = current->next;
        free(temp);
    }
    free(set);
}
 
int count_min_subsets_with_distinct_elements(int arr[], int n) {
    struct set* set = create_set();
    int count = 0;
 
    for (int i = 0; i < n; i++) {
        if (set->head == NULL || !(set, arr[i])) {
            add_to_set(set, arr[i]);
        } else {
            count++;
            delete_set(set);
            set = create_set();
            add_to_set(set, arr[i]);
        }
    }
 
    delete_set(set);
 
    if (count == 0) {
        return 1;
    } else {
        return count + 1;
    }
}
 
int main() {
    int arr[] = {1, 2, 3, 1, 2, 3, 4, 5};
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Minimum number of subsets with distinct elements: " << count_min_subsets_with_distinct_elements(arr, n) << endl;
    return 0;
}
 
// Contributed by adityasharmadev01


C




#include <stdio.h>
#include <stdlib.h>
 
struct node {
    int data;
    struct node *next;
};
 
struct node* create_node(int data) {
    struct node* new_node = (struct node*) malloc(sizeof(struct node));
    new_node->data = data;
    new_node->next = NULL;
    return new_node;
}
 
struct set {
    struct node *head;
};
 
struct set* create_set() {
    struct set* new_set = (struct set*) malloc(sizeof(struct set));
    new_set->head = NULL;
    return new_set;
}
 
void add_to_set(struct set *set, int data) {
    if (set->head == NULL) {
        set->head = create_node(data);
        return;
    }
 
    struct node *current = set->head;
    while (current != NULL) {
        if (current->data == data) {
            return;
        }
        current = current->next;
    }
 
    struct node *new_node = create_node(data);
    new_node->next = set->head;
    set->head = new_node;
}
 
void delete_set(struct set *set) {
    struct node *current = set->head;
    while (current != NULL) {
        struct node *temp = current;
        current = current->next;
        free(temp);
    }
    free(set);
}
 
int count_min_subsets_with_distinct_elements(int arr[], int n) {
    struct set *set = create_set();
    int count = 0;
 
    for (int i = 0; i < n; i++) {
        if (set->head == NULL || !(set, arr[i])) {
            add_to_set(set, arr[i]);
        } else {
            count++;
            delete_set(set);
            set = create_set();
            add_to_set(set, arr[i]);
        }
    }
 
    delete_set(set);
 
    if (count == 0) {
        return 1;
    } else {
        return count + 1;
    }
}
 
int main() {
    int arr[] = {1, 2, 3, 1, 2, 3, 4, 5};
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("Minimum number of subsets with distinct elements: %d\n", count_min_subsets_with_distinct_elements(arr, n));
    return 0;
}


Java




import java.util.HashSet;
import java.util.Set;
 
public class Main {
    // Function to find the minimum number of subsets with
    // distinct elements
    public static int
    minSubsetsWithDistinctElements(int[] arr)
    {
        Set<Integer> uniqueElements = new HashSet<>();
        int count = 0; // Counter to keep track of subsets
                       // with distinct elements
 
        for (int i = 0; i < arr.length; i++) {
            if (!uniqueElements.contains(arr[i])) {
                // If the element is not in the set, add it
                // to the set
                uniqueElements.add(arr[i]);
            }
            else {
                // If the element is already in the set,
                // it's a duplicate, so we:
                // 1. Increase the count by the size of the
                // current set (distinct elements)
                count += uniqueElements.size();
                // 2. Clear the set to start counting the
                // next subset of distinct elements
                uniqueElements.clear();
                // 3. Add the current element to the cleared
                // set to continue counting
                uniqueElements.add(arr[i]);
            }
        }
 
        // After the loop, add any remaining distinct
        // elements in the last subset
        return count + uniqueElements.size();
    }
 
    public static void main(String[] args)
    {
        int[] arr = { 1, 2, 3, 1, 2, 3, 4, 5 };
        int result = minSubsetsWithDistinctElements(arr);
        System.out.println(
            "Minimum number of subsets with distinct elements: "
            + result);
    }
}


Python




# Function to find the minimum number of subsets with distinct elements
def min_subsets_with_distinct_elements(arr):
    unique_elements = set()
    count = 0  # Counter to keep track of subsets with distinct elements
 
    for element in arr:
        if element not in unique_elements:
            # If the element is not in the set, add it to the set
            unique_elements.add(element)
        else:
            # If the element is already in the set, it's a duplicate, so we:
            # 1. Increase the count by the size of the current set (distinct elements)
            count += len(unique_elements)
            # 2. Clear the set to start counting the next subset of distinct elements
            unique_elements.clear()
            # 3. Add the current element to the cleared set to continue counting
            unique_elements.add(element)
 
    # After the loop, add any remaining distinct elements in the last subset
    return count + len(unique_elements)
 
 
# Main function
if __name__ == "__main__":
    arr = [1, 2, 3, 1, 2, 3, 4, 5]
    result = min_subsets_with_distinct_elements(arr)
    print("Minimum number of subsets with distinct elements:", result)


C#




using System;
 
public class Program {
    public class Node {
        public int data;
        public Node next;
    }
 
    public static Node create_node(int data) {
        Node new_node = new Node();
        new_node.data = data;
        new_node.next = null;
        return new_node;
    }
 
    public class Set {
        public Node head;
    }
 
    public static Set create_set() {
        Set new_set = new Set();
        new_set.head = null;
        return new_set;
    }
 
    public static void add_to_set(Set set, int data) {
        if (set.head == null) {
            set.head = create_node(data);
            return;
        }
        Node current = set.head;
        while (current != null) {
            if (current.data == data) {
                return;
            }
            current = current.next;
        }
        Node new_node = create_node(data);
        new_node.next = set.head;
        set.head = new_node;
    }
 
    public static void delete_set(Set set) {
        Node current = set.head;
        while (current != null) {
            Node temp = current;
            current = current.next;
            temp = null;
        }
        set.head = null;
    }
 
    public static int count_min_subsets_with_distinct_elements(int[] arr, int n) {
        Set set = create_set();
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (set.head == null || !set_has(set, arr[i])) {
                add_to_set(set, arr[i]);
            } else {
                count++;
                delete_set(set);
                set = create_set();
                add_to_set(set, arr[i]);
            }
        }
        delete_set(set);
        if (count == 0) {
            return 1;
        } else {
            return count + 1;
        }
    }
 
    public static bool set_has(Set set, int data) {
        Node current = set.head;
        while (current != null) {
            if (current.data == data) {
                return true;
            }
            current = current.next;
        }
        return false;
    }
 
    public static void Main() {
        int[] arr = {1, 2, 3, 1, 2, 3, 4, 5};
        int n = arr.Length;
        Console.WriteLine("Minimum number of subsets with distinct elements: " + count_min_subsets_with_distinct_elements(arr, n));
    }
}
// This code is contributed by shiv1o43g


Javascript




function GFG(arr) {
    let uniqueElements = new Set();
    let count = 0;
    for (let i = 0; i < arr.length; i++) {
        if (!uniqueElements.has(arr[i])) {
            uniqueElements.add(arr[i]);
        } else {
            count += uniqueElements.size;
            uniqueElements.clear();
            uniqueElements.add(arr[i]);
        }
    }
    return count + uniqueElements.size;
}
const arr = [1, 2, 3, 1, 2, 3, 4, 5];
const result = GFG(arr);
console.log("Minimum number of subsets with distinct elements:", result);


Output

Minimum number of subsets with distinct elements: 8












The time complexity of this approach is O(n^2) since we use a loop within a loop to check for distinct elements. 

The space complexity is O(n) since we use a hash set to store distinct elements.

Approach:

At first, Initialize a DP array of size n, where n is the length of the input array. Now, we know dp[i] represents the minimum number of subsets required up to the i-th element of the given array.
Initialize a set to keep track of distinct elements seen so far. This set will be used to check for duplicate elements.
Initialize dp[0] as 1, since the first element itself forms a subset. Iterate through the array starting from the second element (i = 1) to the last element (i = n-1):Check if the current element is already present in the set.
If it is not present, add it to the set and set dp[i] = dp[i-1].
If it is present, increment dp[i] by 1 to account for a new subset required.
Clear the set and add the current element to start a new subset.
The final answer will be dp[n-1], representing the minimum number of subsets required to cover all the distinct elements.

C++




#include <iostream>
#include <unordered_set>
using namespace std;
 
int GFG(int arr[], int n) {
    int dp[n];
    // Using an unordered_set to track distinct elements
    unordered_set<int> distinctElements;  
    dp[0] = 1;
    distinctElements.insert(arr[0]);
    for (int i = 1; i < n; i++) {
        if (distinctElements.find(arr[i]) == distinctElements.end()) {
            // If the element is not in the set
            // add it and continue with previous subset count
            distinctElements.insert(arr[i]);
            dp[i] = dp[i - 1];
        } else {
            // If the element is already in the set
            // start a new subset and clear the set
            dp[i] = dp[i - 1] + 1;
            distinctElements.clear();
            distinctElements.insert(arr[i]);
        }
    }
    return dp[n - 1];
}
int main() {
    int arr[] = {1, 2, 3, 4};
    int n = sizeof(arr) / sizeof(arr[0]);
    int minSubsets = GFG(arr, n);
    cout << "Minimum number of subsets: " << minSubsets << endl;
    int arr2[] = {1, 2, 3, 3};
    int n2 = sizeof(arr2) / sizeof(arr2[0]);
    minSubsets = GFG(arr2, n2);
    cout << "Minimum number of subsets: " << minSubsets << endl;
    return 0;
}


Java




/*package whatever //do not write package name here */
 
import java.io.*;
import java.util.HashSet;
import java.util.Set;
 
class GFG {
    public static int minSubsetsDistinctElements(int[] arr) {
        int n = arr.length;
        int[] dp = new int[n];
 
        Set<Integer> distinctElements = new HashSet<>();
        dp[0] = 1;
        distinctElements.add(arr[0]);
 
        for (int i = 1; i < n; i++) {
            if (!distinctElements.contains(arr[i])) {
                distinctElements.add(arr[i]);
                dp[i] = dp[i - 1];
            } else {
                dp[i] = dp[i - 1] + 1;
                distinctElements.clear();
                distinctElements.add(arr[i]);
            }
        }
 
        return dp[n - 1];
    }
 
    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 4};
        int minSubsets = minSubsetsDistinctElements(arr);
        System.out.println("Minimum number of subsets: " + minSubsets);
 
        arr = new int[]{1, 2, 3, 3};
        minSubsets = minSubsetsDistinctElements(arr);
        System.out.println("Minimum number of subsets: " + minSubsets);
    }
}
 
//This code is contributed by Sovi


Python3




def minSubsetsDistinctElements(arr):
    n = len(arr)
    dp = [0] * # Create an array to store the minimum subset count for each element
 
    distinctElements = set()  # Create a set to store distinct elements
    dp[0] = 1  # Initialize the first element with 1
    distinctElements.add(arr[0])
 
    for i in range(1, n):  # Iterate through the array starting from the second element
        if arr[i] not in distinctElements: 
            distinctElements.add(arr[i]) 
            dp[i] = dp[i - 1# The minimum subset count
                               #remains the same as the previous element
        else:
            dp[i] = dp[i - 1] + 1
            # Clear the set to start tracking distinct elements again
            distinctElements.clear()
             # Add the current element as a distinct element
            distinctElements.add(arr[i])
 
    return dp[n - 1]
 
# Driver code
if __name__ == "__main__":
    arr = [1, 2, 3, 4]
    minSubsets = minSubsetsDistinctElements(arr)
    print("Minimum number of subsets:", minSubsets)
 
    arr = [1, 2, 3, 3]
    minSubsets = minSubsetsDistinctElements(arr)
    print("Minimum number of subsets:", minSubsets)


C#




using System;
using System.Collections.Generic;
 
class Program {
    static int GFG(int[] arr, int n)
    {
        int[] dp = new int[n];
        // Using a HashSet to track distinct elements
        HashSet<int> distinctElements = new HashSet<int>();
        dp[0] = 1;
        distinctElements.Add(arr[0]);
 
        for (int i = 1; i < n; i++) {
            if (!distinctElements.Contains(arr[i])) {
                // If the element is not in the set
                // add it and continue with the previous
                // subset count
                distinctElements.Add(arr[i]);
                dp[i] = dp[i - 1];
            }
            else {
                // If the element is already in the set
                // start a new subset and clear the set
                dp[i] = dp[i - 1] + 1;
                distinctElements.Clear();
                distinctElements.Add(arr[i]);
            }
        }
        return dp[n - 1];
    }
 
    static void Main()
    {
        int[] arr = { 1, 2, 3, 4 };
        int n = arr.Length;
        int minSubsets = GFG(arr, n);
        Console.WriteLine("Minimum number of subsets: "
                          + minSubsets);
 
        int[] arr2 = { 1, 2, 3, 3 };
        int n2 = arr2.Length;
        minSubsets = GFG(arr2, n2);
        Console.WriteLine("Minimum number of subsets: "
                          + minSubsets);
    }
}


Javascript




// Function to calculate the minimum number of subsets with distinct elements
function minSubsetsDistinctElements(arr) {
    const n = arr.length;
    const dp = new Array(n); // Create an array to store the dynamic programming values
 
    const distinctElements = new Set(); // Create a set to keep track of distinct elements
    dp[0] = 1; // Initialize the first value in dp as 1 because there's one subset with the first element
    distinctElements.add(arr[0]); // Add the first element to the set of distinct elements
 
    // Iterate through the array starting from the second element
    for (let i = 1; i < n; i++) {
        if (!distinctElements.has(arr[i])) {
            // If the current element is not in the set of distinct elements
            distinctElements.add(arr[i]); // Add it to the set
            dp[i] = dp[i - 1]; // The number of subsets remains the same as the previous element
        } else {
            // If the current element is already in the set (not distinct)
            dp[i] = dp[i - 1] + 1; // Increment the number of subsets
            distinctElements.clear(); // Clear the set to start counting distinct elements for the next subset
            distinctElements.add(arr[i]); // Add the current element to the set
        }
    }
 
    return dp[n - 1]; // Return the minimum number of subsets
}
 
// Test cases
const arr1 = [1, 2, 3, 4];
const minSubsets1 = minSubsetsDistinctElements(arr1);
console.log("Minimum number of subsets:", minSubsets1);
 
const arr2 = [1, 2, 3, 3];
const minSubsets2 = minSubsetsDistinctElements(arr2);
console.log("Minimum number of subsets:", minSubsets2);


Output

Minimum number of subsets: 1
Minimum number of subsets: 2












 Time complexity : O(n), where n is the size of the input array.

 Auxiliary Space: O(n) as we have used the DP array to store intermediate results.



Last Updated : 13 Dec, 2023
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