Largest divisible pairs subset - GeeksforGeeks

21 Apr, 2021
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++
Java
Python3
C#
PHP
Javascript
// 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 Subsequenceint 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 functionint main(){    int a[] = { 1, 3, 6, 13, 17, 18 };    int n = sizeof(a) / sizeof(a[0]);    cout << largestSubset(a, n) << endl;    return 0;}
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*/
# Python program to find the largest# subset where each pair is divisible. # function to find the longest Subsequencedef 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 Codea = [ 1, 3, 6, 13, 17, 18 ]n = len(a)print(largestSubset(a, n)) # This code is contributed by# sahil shelangia
// 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 program to find the// largest subset which// where each pair is// divisible. // function to find the// longest Subsequencefunction 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.?>
<script> // Javascript program to find the largest// subset which was each pair// is divisible. // Function to find the longest Subsequencefunction 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 codelet 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(n*n) 
vt_m
sahilshelangia
Rajput-Ji
amethyststan
princiraj1992
vinayaksjcebhat
sanjoy_62
divisibility
subset
Arrays
Dynamic Programming
Arrays
Dynamic Programming
subset
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