Given an array arr[] of size N, the task is to split the array into the minimum number of subsets such that every element belongs to exactly one subset and is divisible by the minimum element present in each subset.
Examples:
Input: arr[] = {10, 2, 3, 5, 4, 2}
Output: 3
Explanation:
The three possible groups are:
- {5, 10}, where all the element is divisible by 5(minimum element).
- {2, 2, 4}, where all the element is divisible by 2(minimum element).
- {3}, where all the element is divisible by 3(minimum element).
Input: arr[] = {50, 50, 50, 50, 50}
Output: 1
Approach: The problem can be solved by using Sorting and finding the minimum for each subset. Follow the steps below to solve the problem:
- Sort the array arr[] in ascending order.
- Initialize a variable, say ans, with 0 and an array vis[], for storing the visited array elements.
- Mark all the positions of vis[] array with 0 that represents that the positions that have not been visited.
- Traverse the given array arr[] and perform the following steps:
- If the element arr[i] is not visited, then:
- Consider it as a minimum for the new subset and increment ans by 1.
- Iterate over the range [i + 1, N – 1] using the variable j and if the element arr[j] is not visited and is divisible by arr[i], then set vis[j] = 1.
- Repeat the above steps for each index.
- If the element arr[i] is not visited, then:
- After completing the above steps, print the value of ans as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>#define LL long long#define MM 1000000007using namespace std;// Function to find the minimum number// of subsets into which given array// can be split such that the given// conditions are satisfiedvoid groupDivision(int arr[], int n){ LL z, i, j, ans; // Sort the given array arr[] sort(arr, arr + n); // Initialize answer ans = 0; LL vis[n + 5] = { 0 }; // Iterate for the smaller value // which has not been visited for (i = 0; i < n; i++) { if (!vis[i]) { // Mark all elements that // are divisible by arr[i] for (j = i + 1; j < n; j++) { // If jth index has already // been visited if (vis[j] == 1) continue; if (arr[j] % arr[i] == 0) // Mark the jth index // as visited vis[j] = 1; } // Increment ans by 1 ans++; } } // Print the value of ans cout << ans;}// Driver Codeint main(){ int arr[] = { 10, 2, 3, 5, 4, 2 }; int N = sizeof(arr) / sizeof(arr[0]); groupDivision(arr, N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{static int MM = 1000000007;// Function to find the minimum number// of subsets into which given array// can be split such that the given// conditions are satisfiedstatic void groupDivision(int arr[], int n){ int z, i, j, ans; // Sort the given array arr[] Arrays.sort(arr); // Initialize answer ans = 0; int[] vis = new int[n + 5]; Arrays.fill(vis, 0); // Iterate for the smaller value // which has not been visited for (i = 0; i < n; i++) { if (vis[i] == 0) { // Mark all elements that // are divisible by arr[i] for (j = i + 1; j < n; j++) { // If jth index has already // been visited if (vis[j] == 1) continue; if (arr[j] % arr[i] == 0) // Mark the jth index // as visited vis[j] = 1; } // Increment ans by 1 ans++; } } // Print the value of ans System.out.println(ans);}// Driver Codepublic static void main(String[] args){ int arr[] = { 10, 2, 3, 5, 4, 2 }; int N = arr.length; groupDivision(arr, N);}}// This code is contributed by code_hunt. |
Python3
# Python3 program for the above approachMM = 1000000007# Function to find the minimum number# of subsets into which given array# can be split such that the given# conditions are satisfieddef groupDivision(arr, n): global MM ans = 0 # Sort the given array arr[] arr = sorted(arr) vis = [0]*(n + 5) # Iterate for the smaller value # which has not been visited for i in range(n): if (not vis[i]): # Mark all elements that # are divisible by arr[i] for j in range(i + 1, n): # If jth index has already # been visited if (vis[j] == 1): continue if (arr[j] % arr[i] == 0): # Mark the jth index # as visited vis[j] = 1 # Increment ans by 1 ans += 1 # Print the value of ans print (ans)# Driver Codeif __name__ == '__main__': arr =[10, 2, 3, 5, 4, 2] N = len(arr) groupDivision(arr, N) # This code is contributed by mohit kumar 29. |
C#
// C# program for the above approachusing System;class GFG{static int MM = 1000000007;// Function to find the minimum number// of subsets into which given array// can be split such that the given// conditions are satisfiedstatic void groupDivision(int[] arr, int n){ int z, i, j, ans; // Sort the given array arr[] Array.Sort(arr); // Initialize answer ans = 0; int[] vis = new int[n + 5]; for (i = 0; i < n; i++) { vis[i] = 0; } // Iterate for the smaller value // which has not been visited for (i = 0; i < n; i++) { if (vis[i] == 0) { // Mark all elements that // are divisible by arr[i] for (j = i + 1; j < n; j++) { // If jth index has already // been visited if (vis[j] == 1) continue; if (arr[j] % arr[i] == 0) // Mark the jth index // as visited vis[j] = 1; } // Increment ans by 1 ans++; } } // Print the value of ans Console.Write(ans);}// Driver Codestatic public void Main (){ int[] arr = { 10, 2, 3, 5, 4, 2 }; int N = arr.Length; groupDivision(arr, N);}}// This code is contributed by code_hunt. |
Javascript
<script>// Javascript program for the above approachvar MM = 1000000007;// Creating the bblSort functionfunction bblSort(arr){ for(var i = 0; i < arr.length; i++) { // Last i elements are already in place for(var j = 0; j < (arr.length - i - 1); j++) { // Checking if the item at present iteration // is greater than the next iteration if (arr[j] > arr[j + 1]) { // If the condition is true // then swap them var temp = arr[j] arr[j] = arr[j + 1] arr[j + 1] = temp } } } // Return the sorted array return (arr);}// Function to find the minimum number// of subsets into which given array// can be split such that the given// conditions are satisfiedfunction groupDivision(arr, n){ var z, i, j, ans; // Sort the given array arr arr = bblSort(arr); // Initialize answer ans = 0; var vis = Array(n + 5).fill(0); // Iterate for the smaller value // which has not been visited for(i = 0; i < n; i++) { if (vis[i] == 0) { // Mark all elements that // are divisible by arr[i] for(j = i + 1; j < n; j++) { // If jth index has already // been visited if (vis[j] == 1) continue; if (arr[j] % arr[i] == 0) // Mark the jth index // as visited vis[j] = 1; } // Increment ans by 1 ans++; } } // Print the value of ans document.write(ans);}// Driver Codevar arr = [ 10, 2, 3, 5, 4, 2 ];var N = arr.length;groupDivision(arr, N);// This code is contributed by gauravrajput1</script> |
Output:
3
Time Complexity: O(N2)
Auxiliary Space: O(N)
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