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 codeint 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 codear = [ 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 codeint 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 codepublic 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 codeif __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 codepublic 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;} |
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.
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] * n # 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 codeif __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) |
Javascript
// Function to calculate the minimum number of subsets with distinct elementsfunction 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 casesconst 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.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
