Minimum number of subsets with distinct elements

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.

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 */

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
?>

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>

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

#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;
}

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

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.