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Size of the largest divisible subset in an Array
  • Last Updated : 22 Jan, 2020

Given an array arr[] of size N. The task is to find the size of the set of numbers from the given array such that each number divides another or is divisible by another.

Examples:

Input : arr[] = {3, 4, 6, 8, 10, 18, 21, 24}
Output : 3
One of the possible set with maximum size is {3, 6, 18}

Input : arr[] = {2, 3, 4, 8, 16}
Output : 5

Approach:



  1. Let’s take all the numbers in increasing order.
  2. Note that set X = xi, …, ?xk} is acceptable iff xi divides xi+1 for (1 ≤ i ≤ k – 1).
  3. Therefore, dp[x] is equal to the length of the longest suitable increasing subsequence starting in a number x.
  4. DP Relation: dp[x] = max(dp[x], 1 + dp[y]) if x divides y.

Below is the implementation of the above approach:

CPP

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// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
  
#define N 1000005
  
// Function to find the size of the 
//largest divisible subarray
int maximum_set(int a[], int n)
{
    int dp[N] = { 0 };
  
    // Mark all elements of the array
    for (int i = 0; i < n; i++)
        dp[a[i]] = 1;
  
    int ans = 1;
  
    // Traverse reverse
    for (int i = N - 1; i >= 1; i--) {
  
        if (dp[i] != 0) {
            // For all multiples of i
            for (int j = 2 * i; j < N; j += i) {
                dp[i] = max(dp[i], 1 + dp[j]);
                ans = max(ans, dp[i]);
            }
        }
    }
  
    // Return the required answer
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 3, 4, 6, 8, 10, 18, 21, 24 };
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function call
    cout << maximum_set(arr, n);
  
    return 0;
}

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Java

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// Java implementation of the above approach 
class GFG
{
      
    final static int N = 1000005 ;
      
    // Function to find the size of the 
    //largest divisible subarray 
    static int maximum_set(int a[], int n) 
    
        int dp[] = new int[N] ; 
      
        // Mark all elements of the array 
        for (int i = 0; i < n; i++) 
            dp[a[i]] = 1
      
        int ans = 1
      
        // Traverse reverse 
        for (int i = N - 1; i >= 1; i--) 
        
      
            if (dp[i] != 0
            
                // For all multiples of i 
                for (int j = 2 * i; j < N; j += i) 
                
                    dp[i] = Math.max(dp[i], 1 + dp[j]); 
                    ans = Math.max(ans, dp[i]); 
                
            
        
      
        // Return the required answer 
        return ans; 
    
      
    // Driver code 
    public static void main (String[] args)
    
        int arr[] = { 3, 4, 6, 8, 10, 18, 21, 24 }; 
      
        int n = arr.length; 
      
        // Function call 
        System.out.println(maximum_set(arr, n)); 
    
}
  
// This code is contributed by AnkitRai01

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Python

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# Python3 implementation of the above approach
  
N = 1000005
  
# Function to find the size of the
# largest divisible subarray
def maximum_set(a, n):
    dp = [0 for i in range(N)]
  
    # Mark all elements of the array
    for i in a:
        dp[i] = 1
  
    ans = 1
  
    # Traverse reverse
    for i in range(N - 1, 0, -1):
  
        if (dp[i] != 0):
              
            # For all multiples of i
            for j in range(2 * i, N, i):
                dp[i] = max(dp[i], 1 + dp[j])
                ans = max(ans, dp[i])
  
    # Return the required answer
    return ans
  
# Driver code
  
arr = [3, 4, 6, 8, 10, 18, 21, 24]
  
n = len(arr)
  
# Function call
print(maximum_set(arr, n))
  
# This code is contributed by mohit kumar 29

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C#

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// C# implementation of the above approach 
using System;
  
class GFG 
    static int N = 1000005 ; 
      
    // Function to find the size of the 
    //largest divisible subarray 
    static int maximum_set(int []a, int n) 
    
        int []dp = new int[N] ; 
      
        // Mark all elements of the array 
        for (int i = 0; i < n; i++) 
            dp[a[i]] = 1; 
      
        int ans = 1; 
      
        // Traverse reverse 
        for (int i = N - 1; i >= 1; i--) 
        
      
            if (dp[i] != 0) 
            
                // For all multiples of i 
                for (int j = 2 * i; j < N; j += i) 
                
                    dp[i] = Math.Max(dp[i], 1 + dp[j]); 
                    ans = Math.Max(ans, dp[i]); 
                
            
        
      
        // Return the required answer 
        return ans; 
    
      
    // Driver code 
    public static void Main() 
    
        int []arr = { 3, 4, 6, 8, 10, 18, 21, 24 }; 
        int n = arr.Length; 
      
        // Function call 
        Console.WriteLine(maximum_set(arr, n)); 
    
  
// This code is contributed by AnkitRai01 

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Output:

3

Time Complexity: O(n*sqrt(n))

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