Find the Largest divisor Subset in the Array

Given an array arr[] of N integers, the task is to find the largest subset of arr[] such that in every pair of numbers from that subset, one number is a divisor of the other.

Examples:

Input: arr[] = {1, 2, 3, 4, 5}
Output: 4 2 1
All possible pairs of the subsequence are:
(4, 2) -> 4 % 2 = 0
(4, 1) -> 4 % 1 = 0
and (2, 1) -> 2 % 1 = 0



Input: arr[] = {1, 3, 4, 9}
Output: 1 3 9

Approach: Here the task is to find the largest subset where in every pair of numbers, one is divisible by the other i.e. for the sequence a1, a2, a3 … ak where a1 ≤ a2 ≤ … ≤ ak and ai+1 % ai = 0 for every i. This sequence can be found using dynamic programming (similar to longest increasing subsequence).

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the required subsequence
void findSubSeq(int arr[], int n)
{
  
    // Sort the array
    sort(arr, arr + n);
  
    // Keep a count of the length of the
    // subsequence and the previous element
    int count[n] = { 1 };
    int prev[n] = { -1 };
  
    // Set the initial values
    memset(count, 1, sizeof(count));
    memset(prev, -1, sizeof(prev));
  
    // Maximum length of the subsequence and
    // the last element
    int max = 0;
    int maxprev = -1;
  
    // Run a loop for every element
    for (int i = 0; i < n; i++) {
  
        // Check for all the divisors
        for (int j = i - 1; j >= 0; j--) {
  
            // If the element is a divisor and the length
            // of subsequnce will increase by adding
            // j as previous element of i
            if (arr[i] % arr[j] == 0
                && count[j] + 1 > count[i]) {
  
                // Increase the count
                count[i] = count[j] + 1;
                prev[i] = j;
            }
        }
  
        // Update the max count
        if (max < count[i]) {
            max = count[i];
            maxprev = i;
        }
    }
  
    // Get the last index of the subsequence
    int i = maxprev;
    while (i >= 0) {
  
        // Print the element
        if (arr[i] != -1)
            cout << arr[i] << " ";
  
        // Move the index to the previous element
        i = prev[i];
    }
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(int);
  
    findSubSeq(arr, n);
  
    return 0;
}

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Java

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// Java implementation of the approach 
import java.util.*;
class GFG 
{
      
    // Function to find the required subsequence 
    static void findSubSeq(int arr[], int n) 
    
      
        // Sort the array 
        Arrays.sort(arr); 
      
        // Keep a count of the length of the 
        // subsequence and the previous element 
        int count[] = new int[n]; 
        int prev[] = new int[n]; 
      
        int i, j;
          
        // Set the initial values 
        for(i = 0 ; i < n; i++)
        count[i] = 1;
          
        for(j = 0; j < n; j++)
            prev[j] = -1;
      
        // Maximum length of the subsequence and 
        // the last element 
        int max = 0
        int maxprev = -1
      
        // Run a loop for every element 
        for ( i = 0; i < n; i++)
        
      
            // Check for all the divisors 
            for ( j = i - 1; j >= 0; j--) 
            
      
                // If the element is a divisor and 
                // the length of subsequnce will 
                // increase by adding j as 
                // previous element of i 
                if (arr[i] % arr[j] == 0 && 
                    count[j] + 1 > count[i])
                
      
                    // Increase the count 
                    count[i] = count[j] + 1
                    prev[i] = j; 
                
            
      
            // Update the max count 
            if (max < count[i]) 
            
                max = count[i]; 
                maxprev = i; 
            
        
      
        // Get the last index of the subsequence 
        i = maxprev; 
        while (i >= 0
        
      
            // Print the element 
            if (arr[i] != -1
                System.out.print(arr[i] + " "); 
      
            // Move the index to the previous element 
            i = prev[i]; 
        
    
      
    // Driver code 
    public static void main (String[] args)
    
        int arr[] = { 1, 2, 3, 4, 5 }; 
        int n = arr.length; 
      
        findSubSeq(arr, n); 
    
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 implementation of the approach 
  
# Function to find the required subsequence 
def findSubSeq(arr, n) :
  
    # Sort the array 
    arr.sort(); 
  
    # Keep a count of the length of the 
    # subsequence and the previous element 
    # Set the initial values 
    count = [1] * n; 
    prev = [-1] * n; 
  
    # Maximum length of the subsequence and 
    # the last element 
    max = 0
    maxprev = -1
  
    # Run a loop for every element 
    for i in range(n) :
  
        # Check for all the divisors 
        for j in range(i - 1, -1, -1) :
  
            # If the element is a divisor and the length 
            # of subsequnce will increase by adding 
            # j as previous element of i 
            if (arr[i] % arr[j] == 0 and 
                count[j] + 1 > count[i]) :
  
                # Increase the count 
                count[i] = count[j] + 1
                prev[i] = j; 
  
        # Update the max count 
        if (max < count[i]) :
            max = count[i]; 
            maxprev = i; 
              
    # Get the last index of the subsequence 
    i = maxprev; 
    while (i >= 0) :
  
        # Print the element 
        if (arr[i] != -1) :
            print(arr[i], end = " "); 
  
        # Move the index to the previous element 
        i = prev[i]; 
  
# Driver code 
if __name__ == "__main__"
  
    arr = [ 1, 2, 3, 4, 5 ]; 
    n = len(arr); 
  
    findSubSeq(arr, n); 
      
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach 
using System;
using System.Collections;
  
class GFG 
      
    // Function to find the required subsequence 
    static void findSubSeq(int []arr, int n) 
    
      
        // Sort the array 
        Array.Sort(arr); 
      
        // Keep a count of the length of the 
        // subsequence and the previous element 
        int []count = new int[n]; 
        int []prev = new int[n]; 
      
        int i, j; 
          
        // Set the initial values 
        for(i = 0; i < n; i++) 
        count[i] = 1; 
          
        for(j = 0; j < n; j++) 
            prev[j] = -1; 
      
        // Maximum length of the subsequence  
        // and the last element 
        int max = 0; 
        int maxprev = -1; 
      
        // Run a loop for every element 
        for ( i = 0; i < n; i++) 
        
      
            // Check for all the divisors 
            for ( j = i - 1; j >= 0; j--) 
            
      
                // If the element is a divisor and 
                // the length of subsequnce will 
                // increase by adding j as 
                // previous element of i 
                if (arr[i] % arr[j] == 0 && 
                    count[j] + 1 > count[i]) 
                
      
                    // Increase the count 
                    count[i] = count[j] + 1; 
                    prev[i] = j; 
                
            
      
            // Update the max count 
            if (max < count[i]) 
            
                max = count[i]; 
                maxprev = i; 
            
        
      
        // Get the last index of the subsequence 
        i = maxprev; 
        while (i >= 0) 
        
      
            // Print the element 
            if (arr[i] != -1) 
                Console.Write(arr[i] + " "); 
      
            // Move the index to the previous element 
            i = prev[i]; 
        
    
      
    // Driver code 
    public static void Main () 
    
        int []arr = { 1, 2, 3, 4, 5 }; 
        int n = arr.Length; 
      
        findSubSeq(arr, n); 
    
  
// This code is contributed by AnkitRai01 

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Output:

4 2 1

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