Largest divisible pairs subset

Given an array of n distinct elements, find length of the largest subset such that every pair in the subset is such that the larger element of the pair is divisible by smaller element. 

Examples:

Input : arr[] = {10, 5, 3, 15, 20} 
Output : 3 
Explanation: The largest subset is 10, 5, 20.
10 is divisible by 5, and 20 is divisible by 10.

Input : arr[] = {18, 1, 3, 6, 13, 17} 
Output : 4
Explanation: The largest subset is 18, 1, 3, 6,
In the subsequence, 3 is divisible by 1, 
6 by 3 and 18 by 6.

This can be solved using Dynamic Programming. We traverse the sorted array from the end. For every element a[i], we compute dp[i] where dp[i] indicates size of largest divisible subset where a[i] is the smallest element. We can compute dp[i] in array using values from dp[i+1] to dp[n-1]. Finally, we return the maximum value from dp[].

Below is the implementation of the above approach:  

C++

// CPP program to find the largest subset which
// where each pair is divisible.
#include <bits/stdc++.h>
using namespace std;

// function to find the longest Subsequence
int largestSubset(int a[], int n)
{
    // dp[i] is going to store size of largest
    // divisible subset beginning with a[i].
    int dp[n];

    // Since last element is largest, d[n-1] is 1
    dp[n - 1] = 1;

    // Fill values for smaller elements.
    for (int i = n - 2; i >= 0; i--) {

        // Find all multiples of a[i] and consider
        // the multiple that has largest subset
        // beginning with it.
        int mxm = 0;
        for (int j = i + 1; j < n; j++)
            if (a[j] % a[i] == 0 || a[i] % a[j] == 0)
                mxm = max(mxm, dp[j]);

        dp[i] = 1 + mxm;
    }

    // Return maximum value from dp[]
    return *max_element(dp, dp + n);
}

// driver code to check the above function
int main()
{
    int a[] = { 1, 3, 6, 13, 17, 18 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << largestSubset(a, n) << endl;
    return 0;
}

Java

import java.util.Arrays;

// Java program to find the largest
// subset which was each pair
// is divisible.
class GFG {

    // function to find the longest Subsequence
    static int largestSubset(int[] a, int n)
    {
        // dp[i] is going to store size of largest
        // divisible subset beginning with a[i].
        int[] dp = new int[n];

        // Since last element is largest, d[n-1] is 1
        dp[n - 1] = 1;

        // Fill values for smaller elements.
        for (int i = n - 2; i >= 0; i--) {

            // Find all multiples of a[i] and consider
            // the multiple that has largest subset
            // beginning with it.
            int mxm = 0;
            for (int j = i + 1; j < n; j++) {
                if (a[j] % a[i] == 0 || a[i] % a[j] == 0) {
                    mxm = Math.max(mxm, dp[j]);
                }
            }

            dp[i] = 1 + mxm;
        }

        // Return maximum value from dp[]
        return Arrays.stream(dp).max().getAsInt();
    }

    // driver code to check the above function
    public static void main(String[] args)
    {
        int[] a = { 1, 3, 6, 13, 17, 18 };
        int n = a.length;
        System.out.println(largestSubset(a, n));
    }
}

/* This JAVA code is contributed by Rajput-Ji*/

Python3

# Python program to find the largest
# subset where each pair is divisible.

# function to find the longest Subsequence
def largestSubset(a, n):

    # dp[i] is going to store size
    # of largest divisible subset
    # beginning with a[i].
    dp = [0 for i in range(n)]

    # Since last element is largest,
    # d[n-1] is 1
    dp[n - 1] = 1;

    # Fill values for smaller elements
    for i in range(n - 2, -1, -1):

        # Find all multiples of a[i]
        # and consider the multiple
        # that has largest subset
        # beginning with it.
        mxm = 0;
        for j in range(i + 1, n):
            if a[j] % a[i] == 0 or a[i] % a[j] == 0:
                mxm = max(mxm, dp[j])
        dp[i] = 1 + mxm

    # Return maximum value from dp[]
    return max(dp)

# Driver Code
a = [ 1, 3, 6, 13, 17, 18 ]
n = len(a)
print(largestSubset(a, n))

# This code is contributed by
# sahil shelangia

C#

// C# program to find the largest
// subset which where each pair
// is divisible.
using System;
using System.Linq;

public class GFG {

    // function to find the longest Subsequence
    static int largestSubset(int[] a, int n)
    {
        // dp[i] is going to store size of largest
        // divisible subset beginning with a[i].
        int[] dp = new int[n];

        // Since last element is largest, d[n-1] is 1
        dp[n - 1] = 1;

        // Fill values for smaller elements.
        for (int i = n - 2; i >= 0; i--) {

            // Find all multiples of a[i] and consider
            // the multiple that has largest subset
            // beginning with it.
            int mxm = 0;
            for (int j = i + 1; j < n; j++)
                if (a[j] % a[i] == 0 | a[i] % a[j] == 0)
                    mxm = Math.Max(mxm, dp[j]);

            dp[i] = 1 + mxm;
        }

        // Return maximum value from dp[]
        return dp.Max();
    }

    // driver code to check the above function
    static public void Main()
    {
        int[] a = { 1, 3, 6, 13, 17, 18 };
        int n = a.Length;
        Console.WriteLine(largestSubset(a, n));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to find the
// largest subset which
// where each pair is
// divisible.

// function to find the
// longest Subsequence
function largestSubset($a, $n)
{
    // dp[i] is going to
    // store size of largest
    // divisible subset
    // beginning with a[i].
    $dp = array();

    // Since last element is
    // largest, d[n-1] is 1
    $dp[$n - 1] = 1;

    // Fill values for
    // smaller elements.
    for ($i = $n - 2; $i >= 0; $i--)
    {

        // Find all multiples of
        // a[i] and consider
        // the multiple that
        // has largest subset
        // beginning with it.
        $mxm = 0;
        for ($j = $i + 1; $j < $n; $j++)
            if ($a[$j] % $a[$i] == 0 or $a[$i] % $a[$j] == 0)
                $mxm = max($mxm, $dp[$j]);

        $dp[$i] = 1 + $mxm;
    }

    // Return maximum value
    // from dp[]
    return max($dp);
}

    // Driver Code
    $a = array(1, 3, 6, 13, 17, 18);
    $n = count($a);
    echo largestSubset($a, $n);

// This code is contributed by anuj_67.
?>

Javascript

<script>

// Javascript program to find the largest
// subset which was each pair
// is divisible.

// Function to find the longest Subsequence
function largestSubset(a, n)
{

    // dp[i] is going to store size of largest
    // divisible subset beginning with a[i].
    let dp = [];

    // Since last element is largest, d[n-1] is 1
    dp[n - 1] = 1;

    // Fill values for smaller elements.
    for(let i = n - 2; i >= 0; i--)
    {

        // Find all multiples of a[i] and consider
        // the multiple that has largest subset
        // beginning with it.
        let mxm = 0;
        for(let j = i + 1; j < n; j++)
        {
            if (a[j] % a[i] == 0 ||
                a[i] % a[j] == 0)
            {
                mxm = Math.max(mxm, dp[j]);
            }
        }
        dp[i] = 1 + mxm;
    }

    // Return maximum value from dp[]
    return Math.max(...dp);
}

// Driver code
let a = [ 1, 3, 6, 13, 17, 18 ];
let n = a.length;

document.write(largestSubset(a, n));

// This code is contributed by sanjoy_62

</script>

Output

4

Time Complexity: O(n²)
Space Complexity: O(n)

Largest divisible pairs subset in c:

Approach

1.Create an array dp of size n to store the size of the largest divisible subset ending with each element of the array arr.

2. Initialize all elements of the dp array to 1, as every element is a divisible subset of itself.

3. For every element arr[i] in the array, check all previous elements arr[j] (where j < i) to see if arr[i] is divisible by arr[j] or arr[j] is divisible by arr[i].

4. If arr[i] is divisible by arr[j] or arr[j] is divisible by arr[i], then we can extend the divisible subset ending with arr[j] by including the arr[i] element as well. We will choose the arr[j] element which gives the largest divisible subset ending with arr[j].

5. Update the dp array with the size of the largest divisible subset ending with arr[i].

6. Keep track of the maximum element in the dp array, which gives the largest divisible subset among all elements.

7. Return the maximum element in the dp array.

8. In the main() function, create an array arr of integers and call the largest_divisible_pairs_subset() function with the array and its size.

9. Print the result returned by the largest_divisible_pairs_subset() function, which gives the size of the largest divisible subset.

C

#include <stdio.h>
#include <stdlib.h>

int largest_divisible_pairs_subset(int arr[], int n) {
    int dp[n];
    int i, j, max_dp = 1;

    // Initialize dp array
    for (i = 0; i < n; i++) {
        dp[i] = 1;
    }

    // Compute dp array
    for (i = 1; i < n; i++) {
        for (j = 0; j < i; j++) {
            if (arr[i] % arr[j] == 0 || arr[j] % arr[i] == 0) {
                dp[i] = dp[j] + 1 > dp[i] ? dp[j] + 1 : dp[i];
            }
        }
        max_dp = dp[i] > max_dp ? dp[i] : max_dp;
    }

    return max_dp;
}

int main() {
    int arr[] = {3, 5, 10, 20, 21, 33};
    int n = sizeof(arr)/sizeof(arr[0]);
    printf("%d", largest_divisible_pairs_subset(arr, n));
    return 0;
}

C++

#include <bits/stdc++.h>
using namespace std;

int largest_divisible_pairs_subset(int arr[], int n) {
    vector<int> dp(n, 1);
    int max_dp = 1;

    // Compute dp array
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (arr[i] % arr[j] == 0 || arr[j] % arr[i] == 0) {
                dp[i] = max(dp[j] + 1, dp[i]);
            }
        }
        max_dp = max(dp[i], max_dp);
    }

    return max_dp;
}

int main() {
    int arr[] = {3, 5, 10, 20, 21, 33};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << largest_divisible_pairs_subset(arr, n);
    return 0;
}

Javascript

function largest_divisible_pairs_subset(arr, n) {
    const dp = new Array(n).fill(1);
    let max_dp = 1;

    // Compute dp array
    for (let i = 1; i < n; i++) {
        for (let j = 0; j < i; j++) {
            if (arr[i] % arr[j] === 0 || arr[j] % arr[i] === 0) {
                dp[i] = Math.max(dp[j] + 1, dp[i]);
            }
        }
        max_dp = Math.max(dp[i], max_dp);
    }

    return max_dp;
}

const arr = [3, 5, 10, 20, 21, 33];
const n = arr.length;
console.log(largest_divisible_pairs_subset(arr, n));

Output

3

Time Complexity:
The time complexity of the above algorithm is O(n^2), as we need to check all previous elements for each element.

Auxiliary Space:
The space complexity of the above algorithm is O(n), as we are only using an array of size n to store the largest divisible subset ending with each element.