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 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;} |
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 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#
// 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 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.?> |
Output:
4
Time Complexity: O(n*n)
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